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Example 3.5.1. Someone gives us a pencil and two unmarked sticks of lengths \( {52}\mathrm{\\;{cm}} \) and \( {20}\mathrm{\\;{cm}} \) respectively (see Figure 3.5.1). We are told to make measuring sticks by using the pencil to make markings on the sticks. Question: what is the smallest length that we can accurately measure? Clearly we can measure \( {20}\mathrm{\\;{cm}} \) lengths with the shorter rod, but is it possible to make smaller measurements? | Here's one way to look at the situation. Imagine for a moment that we lay the \( {20}\mathrm{\\;{cm}} \) measuring stick next to the \( {52}\mathrm{\\;{cm}} \) stick such that the ends line up. At that point we could make a \( {20}\mathrm{\\;{cm}} \) mark on the \( {52}\mathrm{\\;{cm}} \) stick (see Figure 3.5.2). At this point we move the \( {20}\mathrm{\\;{cm}} \) stick further down the the \( {52}\mathrm{\\;{cm}} \) stick such that one end is on the pencil mark, and and make another mark. Now there are two \( {20}\mathrm{\\;{cm}} \) sections marked on the \( {52}\mathrm{\\;{cm}} \) stick, as shown in Figure 3.5.3. Since we know the sum of the marked sections is \( {40}\mathrm{\\;{cm}} \), and the length of the large stick is \( {52}\mathrm{\\;{cm}} \), the remainder of the distance must be \( {12}\mathrm{\\;{cm}} \), as shown in Figure 3.5.4. So we've actually made progress. At the beginning we were only able to measure lengths larger than \( {20}\mathrm{\\;{cm}} \) : but now we can measure \( {12}\mathrm{\\;{cm}} \) with the latest mark we’ve made. But let’s not stop there. We can use the \( {12}\mathrm{\\;{cm}} \) section to divide up the \( {20}\mathrm{\\;{cm}} \) stick. This will subdivide the \( {20}\mathrm{\\;{cm}} \) stick into a \( {12}\mathrm{\\;{cm}} \) section and a \( 8\mathrm{\\;{cm}} \) section, as shown in Figure 3.5.5. Now we’re rolling! Let’s subdivide the \( {12}\mathrm{\\;{cm}} \) section using the \( 8\mathrm{\\;{cm}} \) section. This will produce an \( 8\mathrm{\\;{cm}} \) section and a \( 4\mathrm{\\;{cm}} \) section (see Figure 3.5.6). Now if we try to use the \( 4\mathrm{\\;{cm}} \) section to subdivide any of the other sections, we will no longer have a remainder. This is because \( 4\mathrm{\\;{cm}} \) evenly divides all the other lengths we have created, as shown in Figure 3.5.7. | Yes |
Let's use algebraic language to express the two-sticks algorithm applied to 52 and 20. Let's start by setting this up as a division problem with a remainder (recall Proposition 3.2.3), since this is effectively what is being done in the stick example above. | \[ 52 = 20 \cdot q_1 + r_1 \] where \( q_1 \) and \( r_1 \) are integers (we put the subscript ’ 1 ’ on the variables \( q_1 \) and \( r_1 \) because we’re going to repeat the process). By division with remainder we find \( q_1 = 2 \) and \( r_1 = 12 \). Now we repeat the process, but this time dividing the remainder 12 into the smaller stick length 20 : \[ 20 = 12 \cdot q_2 + r_2 \] which yields \( q_2 = 1, r_2 = 8 \). Here we go again, this time dividing the second remainder 8 into the first remainder 12: \[ 12 = 8 \cdot q_3 + r_3 \] This yields \( q_3 = 1, r_3 = 4 \). One more time, this time dividing the new remainder 4 into the previous remainder 8: \[ 8 = 4 \cdot q_4 + r_4 \] This yields \( q_4 = 2, r_4 = 0 \). Now notice that 8 is divisible by 4. In the equation before that, we have \( 12 = 4 \cdot 2 + 4 \). Since the right hand side is a sum of multiples of 4, the left hand side must also be a multiple of 4. In the next equation up \( 20 = 12 \cdot x + 8 \) again, the right hand side is a sum of multiples of 4, so the left hand side must also be a multiple of 4. Continuing this logic upward shows that all intervals created along the way are divisible by 4. Hence the algorithm has generated a divisor of the original lengths 52 and 20. In summary, the last nonzero remainder gave us the gcd. | Yes |
Proposition 3.5.5. The Euclidean algorithm applied to two integers will give the gcd of those two integers. | Proof. This proof is broken up into two parts, (A) and (B). Part (A) shows that the algorithm always produces a divisor of the two given integers. Part (B) shows that the produced divisor is indeed the gcd.\n\n(A) Given integers \( a \) and \( b \) and \( a > b \) if we were to plug them into the Euclidean Algorithm we get:\n\n\[ a = b \cdot {q}_{1} + {r}_{1} \]\n\n\[ b = {r}_{1} \cdot {q}_{2} + {r}_{2} \]\n\n\[ {r}_{1} = {r}_{2} \cdot {q}_{3} + {r}_{3} \]\n\n\[ \vdots \]\n\nuntil there is an equation with no remainder left.\n\n\[ {r}_{k - 2} = {r}_{k - 1} \cdot {q}_{k - 1} + {r}_{k} \]\n\n\[ {r}_{k - 1} = {r}_{k} \cdot {q}_{k} + 0 \]\n\nIt is clear that \( {r}_{k} \) divides \( {r}_{k - 1} \) . Consider the next equation up.\n\n\[ {r}_{k - 2} = {r}_{k - 1} \cdot {q}_{k - 1} + {r}_{k} = {r}_{k} \cdot {q}_{k - 1} \cdot {q}_{k} + {r}_{k} \]\n\nThis shows that \( {r}_{k} \) divides the right hand side, so \( {r}_{k} \) must divide \( {r}_{k - 2} \) . In the next equation up, the right can be set up as multiples of \( {r}_{k} \) which means the next \( r \) term is divisible by \( {r}_{k} \) Continue all the way to the top and it must be that \( {r}_{k} \) divides both \( a \) and \( b \)\n\n(B) Now suppose there is another number \( c \) that divides \( a \) and \( b \) such that \( {a}_{1} \cdot c = a \) and \( {b}_{1} \cdot c = b \) . We can rewrite the initial equation of the algorithm as follows.\n\n\[ {a}_{1} \cdot c = \left( {{b}_{1} \cdot c}\right) \cdot {q}_{1} + {r}_{1} \Rightarrow {a}_{1} \cdot c - \left( {{b}_{1} \cdot c}\right) \cdot {q}_{1} = {r}_{1} \]\n\nThis shows that \( c \) must divide \( {r}_{1} \) . Consider the next equation.\n\n\[ {b}_{1} \cdot c = \left( {r}_{1}\right) \cdot {q}_{2} + {r}_{2} \Rightarrow {b}_{1} \cdot c - \left( {r}_{1}\right) \cdot {q}_{2} = {r}_{2} \]\n\nSince \( c \) divides both \( {r}_{1} \) and \( {b}_{1} \) then \( c \) must divide \( {r}_{2} \) also. Repeat all the way to the bottom and \( c \) will have to divide \( {r}_{k} \).\n\nSince \( c \) divides \( {r}_{k}, c \) is no larger than \( {r}_{k} \) . So all divisors of \( a \) and \( b \) must be no larger than \( {r}_{k} \) . From part (A) we know that \( {r}_{k} \) divides both \( a \) and \( b \) . Therefore \( {r}_{k} \) must be the gcd of \( a \) and \( b \) . | Yes |
We'll give another example, giving just the computations and no other words. We find integer solutions to \( {1053x} + {863y} = {245} \) as follows: | \[ {1053} = {863} + {190} \Rightarrow {190} = \left( {1, - 1}\right) \] \[ {863} = 4 \cdot {190} + {103} \Rightarrow {103} = \left( {0,1}\right) - 4 \cdot \left( {1, - 1}\right) = \left( {-4,5}\right) \] \[ {190} = {103} + {87} \Rightarrow {87} = \left( {1, - 1}\right) - \left( {-4,5}\right) = \left( {5, - 6}\right) \] \[ {103} = {87} + {16} \Rightarrow {16} = \left( {-4,5}\right) - \left( {5, - 6}\right) = \left( {-9,{11}}\right) \] \[ {87} = 5 \cdot {16} + 7 \Rightarrow 7 = \left( {5, - 6}\right) - 5 \cdot \left( {-9,{11}}\right) = \left( {{50}, - {61}}\right) \] \[ {16} = 2 \cdot 7 + 2 \Rightarrow 2 = \left( {-9,{11}}\right) - 2 \cdot \left( {{50}, - {61}}\right) = \left( {-{109},{133}}\right) \] \[ 7 = 3 \cdot 2 + 1 \Rightarrow 1 = \left( {{50}, - {61}}\right) - 3 \cdot \left( {-{109},{133}}\right) = \left( {{377} - {460}}\right) . \] This means that: \( {377} \cdot {1053} - {460} \cdot {863} = 1 \) (You may check this on a calculator.) Now we may multiply both sides by 245 , which gives: \[ \left( {{245} \cdot {377}}\right) \cdot {1053} - \left( {{245} \cdot {460}}\right) \cdot {863} = {245}. \] Thus \( x = \left( {{245} \cdot {377}}\right) = {92365} \) and \( y = - \left( {{245} \cdot {460}}\right) = - {112700} \), so that \[ {1053} \cdot {92365} - {863} \cdot {112700} = {245} \] is an integer solution. | Yes |
Proposition 3.5.16. Given the Diophantine equation \( {an} + {bm} = c \), where \( a, b, c \) are integers. Then the equation has integer solutions for \( n \) and \( m \) if and only if \( c \) is a multiple of the gcd of \( a \) and \( b \) . | Proof. Since this is an \ | No |
Example 3.5.24. The Cayley table for \( {\mathbb{Z}}_{3} \smallsetminus \{ 0\} \) is:\n\n<table><thead><tr><th>\( \odot \)</th><th>1</th><th>2</th></tr></thead><tr><td>1</td><td>1</td><td>2</td></tr><tr><td>2</td><td>2</td><td>1</td></tr></table> | Notice that each column has 1, meaning that each element has an inverse. It is also closed, associative and has an identity. Thus \( {\mathbb{Z}}_{3} \smallsetminus \{ 0\} \) is a group under \( \odot \) . | Yes |
The Cayley table for \( {\mathbb{Z}}_{4} \smallsetminus \{ 0\} \) is\n\n<table><thead><tr><th>\( \odot \)</th><th>1</th><th>2</th><th>3</th></tr></thead><tr><td>1</td><td>1</td><td>2</td><td>3</td></tr><tr><td>2</td><td>2</td><td>0</td><td>2</td></tr><tr><td>3</td><td>3</td><td>2</td><td>1</td></tr></table>\n\nNotice that the 2 column does not have a 1 in it, meaning that 2 does not have an inverse in \( {\mathbb{Z}}_{4} \). Thus, \( {\mathbb{Z}}_{4} \smallsetminus \{ 0\} \) is not a group under \( \odot \) . | The fact that 2 has no inverse is due to 2 being a divisor of 4. This makes all integer multiples of 2 to cycle between the values 0 and 2 (mod 4). | Yes |
Proposition 3.5.28. If \( p \) is a prime number, then all elements in \( {\mathbb{Z}}_{p} \smallsetminus \{ 0\} \) have an inverse under multiplication \( {\;\operatorname{mod}\;p} \) . | Proof. Let \( a, p \) be known integers where \( a < p \) and \( p \) is prime. There exists an inverse to \( a \) under multiplication \( \left( {\;\operatorname{mod}\;p}\right) \) when there is a solution \( k \) to the equation \( {ak} = 1\left( {\;\operatorname{mod}\;p}\right) \) where \( k \) is an integer. By Proposition 3.5.20, this equation can be solved if and only if the gcd of \( a \) and \( p \) is equal to 1 . Since \( p \) is prime and \( a < p \) then the gcd of \( a \) and \( p \) must be 1 . | Yes |
Let’s find the digit \( {d}_{6} \) for the number \( n = {1928307465} \) (we may note in this case \( {d}_{6} = 8 \) ). | First, we can remove the digits above \( {d}_{6} \) digit taking \( n \) modulo \( {10}^{7} \) :\n\n\[ \n{\;\operatorname{mod}\;\left( {n,{10}^{7}}\right) } = {8307465}.\n\]\n\nOn the other hand, we can obtain all digits below \( {d}_{6} \) by taking \( n \) modulo \( {10}^{6} \) :\n\n\[ \n{\;\operatorname{mod}\;\left( {n,{10}^{6}}\right) } = {307465}\n\]\n\nNow subtracting the two we get:\n\n\[ \n{\;\operatorname{mod}\;\left( {n,{10}^{7}}\right) } - {\;\operatorname{mod}\;\left( {n,{10}^{6}}\right) } = {8000000}\n\]\n\nFrom this point, we easily obtain \( {d}_{6} \) by dividing by \( {10}^{6} \) . So in summary, we\n\nhave:\n\[ \n{d}_{6} = \frac{{\;\operatorname{mod}\;\left( {n,{10}^{7}}\right) } - {\;\operatorname{mod}\;\left( {n,{10}^{6}}\right) }}{{10}^{6}}\n\]\n\nThis formula can be generalized to find the digit \( {d}_{k} \) for any positive integer \( n \) :\n\n\[ \n{d}_{k} = \frac{{\;\operatorname{mod}\;\left( {n,{10}^{k + 1}}\right) } - {\;\operatorname{mod}\;\left( {n,{10}^{k}}\right) }}{{10}^{k}}\n\] | Yes |
Find \( {d}_{-3} \) of the decimal number \( x = {0.17428} \) | Since we’re looking for \( {d}_{-3} \), Let’s multiply \( x \) by \( {10}^{3} \). \[ {0.17428} \cdot {10}^{3} = {174.28} \] Then take the floor: \[ \lfloor {174.28}\rfloor = {174} \] Finally, take the modulus base 10 (which is the 1's place of the number, as we've seen before): \[ {\;\operatorname{mod}\;\left( {{174},{10}}\right) } = 4 \] This gives us the correct value of \( {d}_{-3} \). | Yes |
Proposition 4.1.13. Let \( n > 2 \) be an integer such that \( \gcd \left( {n,{10}}\right) = 1 \) . Then there exist a positive integer \( m \) such that \( {\;\operatorname{mod}\;\left( {{10}^{m}, n}\right) } = 1 \) . | Proof. Consider the infinite sequence: \( {\;\operatorname{mod}\;\left( {{10}, n}\right) },{\;\operatorname{mod}\;\left( {{10}^{2}, n}\right) },{\;\operatorname{mod}\;\left( {{10}^{3}, n}\right) },\ldots \) . All of these numbers are between 1 and \( n - 1 \) . Since the sequence is infinite and only can take at most \( n - 1 \) values, it follows there must be at least two values that are equal, so \( {\;\operatorname{mod}\;\left( {{10}^{k}, n}\right) } = {\;\operatorname{mod}\;\left( {{10}^{j}, n}\right) } \), where \( k > j \) . But,\n\n\[ \n{\;\operatorname{mod}\;\left( {{10}^{k}, n}\right) } = {\;\operatorname{mod}\;\left( {{10}^{j} \cdot {10}^{k - j}, n}\right) }\;\text{[basic arithmetic]} \n\]\n\n\[ \n= {\;\operatorname{mod}\;\left( {{10}^{j}, n}\right) } \odot {\;\operatorname{mod}\;\left( {{10}^{k - j}, n}\right) }\;\text{ [Proposition 3.4.4]. } \n\]\n\nSince \( {\;\operatorname{mod}\;\left( {{10}^{k}, n}\right) } = {\;\operatorname{mod}\;\left( {{10}^{j}, n}\right) } \), it follows by substitution that\n\n\[ \n{\;\operatorname{mod}\;\left( {{10}^{j}, n}\right) } = {\;\operatorname{mod}\;\left( {{10}^{j}, n}\right) } \odot {\;\operatorname{mod}\;\left( {{10}^{k - j}, n}\right) } \n\]\n\nWhich implies \( {\;\operatorname{mod}\;\left( {{10}^{k - j}, n}\right) } = 1 \) . So if we set \( m = k - j \), we have \( {\;\operatorname{mod}\;\left( {{10}^{m}, n}\right) } = 1 \), and the proof is finished. | Yes |
Proposition 4.1.15. Let \( n > 1 \) be a positive integer with \( \gcd \left( {{10}, n}\right) = 1 \) , and let \( m \) be the multiplicative order of \( n\left( {\;\operatorname{mod}\;{10}}\right) \) . Then the decimal expansion of \( \frac{1}{n} \) repeats every \( m \) digits. | Proof.\n\nGiven that \( m \) is the multiplicative order of \( n{\;\operatorname{mod}\;{10}} \), from Definition 4.1.14, we get \( {\;\operatorname{mod}\;\left( {\left( {{10}^{m} - 1}\right), n}\right) } = 0 \) . In other words, \( {10}^{m} - 1 \) is divisible by \( n \), so that \( \frac{{10}^{m} - 1}{n} \) is an integer. Letting \( k = \frac{{10}^{m} - 1}{n} \), it follows that:\n\n\[ \frac{1}{n} = \frac{k}{{10}^{m} - 1} \]\n\nsubstitution\n\n\[ = \frac{k}{{10}^{m}}\left( \frac{1}{1 - {10}^{-m}}\right) \]\n\nfactor\n\n\[ = \frac{k}{{10}^{m}}\left( {1 + {10}^{-m} + \cdots }\right) \]\ngeometric series\n\n\( = k \cdot {10}^{-m} + k \cdot {10}^{-{2m}} + \cdots \; \) distributive law and algebra\n\nNext from the definition of \( k \), we may conclude that \( k < {10}^{m} \) (verify this). So the nonzero decimal digits of \( k \cdot {10}^{-m} \) are all in the \( {10}^{ - }1, k \cdot {10}^{-m} \) decimal places, the nonzero decimal digits of \( k \cdot {10}^{-{2m}} \) are all in the \( {10}^{-m - 1}, k \) . \( {10}^{-{2m}} \) decimal places, the nonzero decimal digits of \( k \cdot {10}^{-{3m}} \) are all in the \( {10}^{-{2m} - 1}, k \cdot {10}^{-{3m}} \) decimal places, and so on. In other words, no two of the terms \( k \cdot {10}^{-m}, k \cdot {10}^{-{2m}},\cdots \) have nonzero digits in the same decimal place. This means that \( k \) is the repeating sequence in the repeating decimal (possibly padded by some zeros, in case \( k \) has less than \( m \) nonzero digits), and that the fraction repeats every \( m \) digits. | Yes |
Is 6472 divisible by 11 ? | In the following argument we use the fact that \( {10} \equiv - 1\left( {\;\operatorname{mod}\;{11}}\right) \), which means that we can replace 10\n\nwith -1 whenever we are taking mod's base 10 .\n\n\[ \n{\;\operatorname{mod}\;\left( {{6472},{11}}\right) } = {\;\operatorname{mod}\;\left( {6 \cdot {10}^{3} + 4 \cdot {10}^{2} + 7 \cdot {10} + 2 \cdot 1,{11}}\right) } \]\n\n\[ \n= {\;\operatorname{mod}\;\left( {6 \cdot {\left( -1\right) }^{3} + 4 \cdot {\left( -1\right) }^{2} + 7 \cdot \left( {-1}\right) + 2,{11}}\right) } \]\n\n\[ \n= {\;\operatorname{mod}\;\left( {-6 + 4 - 7 + 2,{11}}\right) } = {\;\operatorname{mod}\;\left( {-7,{11}}\right) } = 4 \]\n\nSince \( {\;\operatorname{mod}\;\left( {{6472},{11}}\right) } \neq 0,{6472} \) is not divisible by 11 . | Yes |
Proposition 4.1.22. A number is divisible by 11 if and only if the alternating sums of the digits is divisible by 11. (Note: alternating sums is where the signs of the number alternate when summing.) | Proof. Given an integer with digits \( {d}_{0}\ldots {d}_{n} \) where the number is writeen as \( {d}_{n}{d}_{n - 1}\ldots {d}_{1}{d}_{0} \) we can write\n\n\[ n = {d}_{m} \cdot {10}^{m} + {d}_{m - 1} \cdot {10}^{m - 1} + \cdots + {d}_{0} \cdot {10}^{0} \]\n\n it follows that:\n\n\[ {\;\operatorname{mod}\;\left( {n,{11}}\right) } \]\n\n\[ = {\;\operatorname{mod}\;\left( {{d}_{m} \cdot {10}^{m} + {d}_{m - 1} \cdot {10}^{m - 1} + \cdots + {d}_{0} \cdot {10}^{0},{11}}\right) }\;\text{[substitution]} \]\n\n\[ = {\;\operatorname{mod}\;\left( {{d}_{m} \cdot {\left( -1\right) }^{m} + {d}_{m - 1} \cdot {\left( -1\right) }^{m - 1} + \cdots + {d}_{0} \cdot {\left( -1\right) }^{0},{11}}\right) }\;\left\lbrack {{\;\operatorname{mod}\;\left( {{10},{11}}\right) } = - 1}\right\rbrack \]\n\n\[ = {\;\operatorname{mod}\;\left( {{\left( -1\right) }^{m}\left( {{d}_{m} - {d}_{m - 1} + \cdots + {d}_{0} \cdot 1}\right) ,{11}}\right) }\;\text{[factor out}\left. {\left( -1\right) }^{m}\right\rbrack \]\n\nTherefore, \( {\;\operatorname{mod}\;\left( {n,{11}}\right) } = 0 \) if and only if the alternating sums of the digits of the number \( {d}_{n}\ldots {d}_{0} \) is divisible by 11. | Yes |
Find 137 in base 6. | \[ {a}_{0} = {137};{d}_{0} = {\;\operatorname{mod}\;\left( {{137},6}\right) } = 5 \] \[ {a}_{1} = \frac{{137} - 5}{6} = {22};{d}_{1} = {\;\operatorname{mod}\;\left( {{22},6}\right) } = 4 \] \[ {a}_{2} = \frac{{22} - 4}{6} = 3;{d}_{2} = {\;\operatorname{mod}\;\left( {3,6}\right) } = 3 \] \[ {a}_{3} = \frac{3 - 3}{6} = 0 \] Since \( {a}_{3} = 0 \) we can stop. To write the solution take the moduli in reverse order. Therefore, 137 in base 6 is 345 . | Yes |
Find 121 in base 3. | Once again using the recursive method\n\n\[ \n{a}_{0} = {121};\;{d}_{0} = {\;\operatorname{mod}\;\left( {{121},3}\right) } = 1 \]\n\n\[ \n{a}_{1} = \frac{{121} - 1}{3} = {40};\;{d}_{1} = {\;\operatorname{mod}\;\left( {{40},3}\right) } = 1 \]\n\n\[ \n{a}_{2} = \frac{{40} - 1}{3} = {13};\;{d}_{2} = {\;\operatorname{mod}\;\left( {{13},3}\right) } = 1 \]\n\n\[ \n{a}_{3} = \frac{{13} - 1}{3} = 4;\;{d}_{3} = {\;\operatorname{mod}\;\left( {4,3}\right) } = 1 \]\n\n\[ \n{a}_{4} = \frac{4 - 1}{3} = 1;\;{d}_{4}{\;\operatorname{mod}\;\left( {1,3}\right) } = 1 \]\n\n\[ \n{a}_{5} = \frac{1 - 1}{3} = 0 \]\n\n\nSince \( {a}_{5} = 0 \) we do not have to continue. To write the solution take the moduli in reverse order. Therefore, 121 in base 3 is 11111 . | Yes |
Find the 5th digit of 65432 in base 3. (This is the coefficient of \( {3}^{4} \) in the base 3 representation). | \[ {d}_{4} = \frac{{\;\operatorname{mod}\;\left( {{65432},{3}^{5}}\right) } - {\;\operatorname{mod}\;\left( {{65432},{3}^{4}}\right) }}{{3}^{4}} \] | Yes |
Example 4.2.5. Find 31 in base 2. | \[ {a}_{0} = {31};{d}_{0} = {\;\operatorname{mod}\;\left( {{31},2}\right) } = 1;\;{a}_{1} = \frac{{31} - 1}{2} = {15};{b}_{1} = {\;\operatorname{mod}\;\left( {{15},2}\right) } = 1 \] \[ {a}_{2} = \frac{{15} - 1}{2} = {15};{b}_{2} = {\;\operatorname{mod}\;\left( {{14},2}\right) } = 0;\;{a}_{3} = \frac{{14} - 0}{2} = {15};{b}_{3} = {\;\operatorname{mod}\;\left( {7,2}\right) } = 1 \] \[ {a}_{4} = \frac{7 - 1}{2} = {15};{b}_{4} = {\;\operatorname{mod}\;\left( {6,2}\right) } = 0;\;{a}_{5} = \frac{6 - 0}{2} = {15};{b}_{5} = {\;\operatorname{mod}\;\left( {3,2}\right) } = 1 \] \[ {a}_{6} = \frac{3 - 1}{2} = {15};{b}_{6} = {\;\operatorname{mod}\;\left( {1,2}\right) } = 1;\;{a}_{7} = \frac{1 - 1}{2} = 0;{b}_{7} = {\;\operatorname{mod}\;\left( {0,2}\right) } = 0 \] Therefore \( N = {31} \) written in base 2 is 0110101 . If stored as a 2-byte integer, \( N \) would be represented as \( {0b000000000110101} \) (the ’0b’ prefix indicates that the number is a binary number). | Yes |
Example 4.2.8. Convert 121 in base 3 to a number in base 10. | \[ {\left( {121}\right) }_{3} = 1 \cdot {3}^{2} + 2 \cdot {3}^{1} + 1 \cdot {3}^{0} = 1 \cdot 9 + 2 \cdot 3 + 1 \cdot 1 = {\left( {16}\right) }_{10} \] | Yes |
Example 4.2.9. Convert 4752 in base 8 to a number in base 10 | \[ {\left( {4752}\right) }_{8} = 4 \cdot {8}^{3} + 7 \cdot {8}^{2} + 5 \cdot {8}^{1} + 2 \cdot {8}^{0} = 4 \cdot {512} + 7 \cdot {64} + 5 \cdot 8 + 2 \cdot 1 = {\left( {2538}\right) }_{10} \] | Yes |
Example 4.2.12. Convert 1011 in base 2 to a number in base 10. | \[ {\left( {1011}\right) }_{2} = 1 \cdot {2}^{3} + 0 \cdot {2}^{2} + 1 \cdot {2}^{1} + 1 \cdot {2}^{0} = 1 \cdot 8 + 0 \cdot 4 + 1 \cdot 2 + 1 \cdot 1 = {\left( {11}\right) }_{10} \] | Yes |
(a) \( \mathop{\bigcup }\limits_{{i = 1}}^{n}\{ i\} \) | (a) \( \mathop{\bigcup }\limits_{{i = 1}}^{n}\{ i\} = \{ 1\} \bigcup \{ 2\} \bigcup \{ 3\} \bigcup \ldots \bigcup \{ n\} \n\n\( = \{ 1,\ldots, n\} \; \) [list of elements]\n\n\( = \) all integers from 1 to \( n \) . [property] | Yes |
Example 5.1.19. Let \( \mathbb{N} \) be the universal set, and suppose that\n\n\[ A = \{ x \in \mathbb{N} : x\text{ is divisible by }2\} \]\n\n\[ B = \{ x \in \mathbb{N} : x\text{ is divisible by }3\} \]\n\n\[ C = \{ x \in \mathbb{N} : x\text{ is divisible by }6\} \]\n\n\[ D = \{ \text{the odd natural numbers}\} \]\n\nThen specify the following sets:\n\n## (a) \( A \cap B \) | (a)\n\n\[ A \cap B = \{ x \in \mathbb{N} : x\text{ is divisible by }2\text{ and }x\text{ is divisible by }3\} \]\n\n\[ = \{ x \in \mathbb{N} : x\text{ is divisible by }6\} \]\n\n\[ = C \] | Yes |
Proposition 5.2.1. Given any sets \( A, B \), It is always true that\n\n\[ A \cap B \subset A \\text{ and } A \subset A \cup B. \]\n | Proof. The style of proof we'll use here is often described as element by element, because the proofs make use of the definitions of \( A \cap B \) and \( A \cup B \) in terms of their elements.\n\nFirst, suppose that \( x \) is an element of \( A \cap B \) . we then have:\n\n\[ x \in A \cap B \] [supposition]\n\n\[ \\Rightarrow x \in A \\text{and} x \in B \] [def. of \( \\cap \) ]\n\n\[ \\Rightarrow x \in A. \] [logic]\n\nSince every element of \( A \cap B \) is an element of \( A \), it follows by the definition of \( \\subset \) that \( A \cap B \\subset A \). | No |
Proposition 5.2.3. Let \( A, B \), and \( C \) be subsets of a universal set \( U \) . Then\n\n1. \( A \cup {A}^{\prime } = U \) and \( A \cap {A}^{\prime } = \varnothing \) | Proof. We'll prove parts (1), (2), (5), and (7), and leave the rest to you!\n\n(1) From our definitions we have:\n\n\[ A \cup {A}^{\prime } = \left\{ {x : x \in A\text{ or }x \in {A}^{\prime }}\right\} \]\n[def. of \( \cup \) ]\n\n\[ = \{ x : x \in A\text{ or }x \notin A\} \]\n[def. of complement]\n\nBut every \( x \in U \) must satisfy either \( x \in A \) or \( x \notin A \) . It follows that \( A \cup {A}^{\prime } \) includes all elements of \( U \) ; so \( A \cup {A}^{\prime } = U \) .\n\nWe also have\n\n\[ A \cap {A}^{\prime } = \left\{ {x : x \in A\text{ and }x \in {A}^{\prime }}\right\} \]\n[def. of \( \cap \) ]\n\n\[ = \{ x : x \in A\text{ and }x \notin A\} \]\n[def. of complement]\n\nBut there is no element \( x \) that is both in \( A \) and not in \( A \), it follows that there are no elements in \( A \cap {A}^{\prime } \) ; so \( A \cap {A}^{\prime } = \varnothing \) . | No |
Example 5.2.8. Prove that\n\n\[ \left( {A \smallsetminus B}\right) \cap \left( {B \smallsetminus A}\right) = \varnothing . \] | Proof. To see that this is true, observe that\n\n\[ \left( {A \smallsetminus B}\right) \cap \left( {B \smallsetminus A}\right) = \left( {A \cap {B}^{\prime }}\right) \cap \left( {B \cap {A}^{\prime }}\right) \] [definition of \( \smallsetminus \) ]\n\n\[ = A \cap {A}^{\prime } \cap B \cap {B}^{\prime }\; \] [by Proposition 5.2.3 parts 5 and 6]\n\n\[ = \varnothing \cap \varnothing \] [by Proposition 5.2.3 part 1]\n\n\[ = \varnothing \text{.} \] | Yes |
Proposition 6.1.8. Given any sets \( A \) and \( B \), then:\n\n\[ \left| {A \times B}\right| = \left| A\right| \cdot \left| B\right| \] | Proof. We can prove this formula by some creative arranging. Suppose the sets \( A \) and \( B \) have \( m \) and \( n \) elements, respectively. We may list these elements as follows:\n\n\[ A = \left\{ {{a}_{1},{a}_{2},{a}_{3},\ldots ,{a}_{m}}\right\} \text{ and }B = \left\{ {{b}_{1},{b}_{2},{b}_{3},\ldots ,{b}_{n}}\right\} .\n\nIt follows that the elements of \( A \times B \) are:\n\n\[ \left( {{a}_{1},{b}_{1}}\right) ,\;\left( {{a}_{1},{b}_{2}}\right) ,\;\left( {{a}_{1},{b}_{3}}\right) ,\;\cdots \;\left( {{a}_{1},{b}_{n}}\right) ,\n\n\[ \left( {{a}_{2},{b}_{1}}\right) ,\;\left( {{a}_{2},{b}_{2}}\right) ,\;\left( {{a}_{2},{b}_{3}}\right) ,\;\cdots \;\left( {{a}_{2},{b}_{n}}\right) ,\n\n\[ \left( {{a}_{3},{b}_{1}}\right) ,\;\left( {{a}_{3},{b}_{2}}\right) ,\;\left( {{a}_{3},{b}_{3}}\right) ,\;\cdots \;\left( {{a}_{3},{b}_{n}}\right) ,\n\n\[ \ldots \;\ldots \;\ldots \;\ldots\n\n\[ \cdots \;\cdots \;\cdots \;\cdots\n\n\[ \left( {{a}_{m},{b}_{1}}\right) ,\;\left( {{a}_{m},{b}_{2}}\right) ,\;\left( {{a}_{m},{b}_{3}}\right) ,\;\cdots \;\left( {{a}_{m},{b}_{n}}\right) .\n\nIn the above table that represents the elements of \( A \times B \) :\n\n- each row has exactly n elements, and\n\n- there are \( \mathrm{m} \) rows,\n\nIt follows that the number of entries in the table is \( m \cdot n \) . | Yes |
For the function \( f\left( x\right) = {x}^{3} \), the input \( x \) can be any real number. Plugging a value for \( x \) into the formula yields an output value, which is also a real number. For example, using \( x = 2 \) as the input yields the output value \( f\left( 2\right) = {2}^{3} = 8 \) . | In Example 6.2.1, any real number can be used as the input \( x \), so the domain is \( \mathbb{R} \), the set of all real numbers. Similarly, any output is a real number, so the codomain can also be taken as \( \mathbb{R} \) . | Yes |
Example 6.2.3. \( g\left( x\right) = 1/x \) is not a function from \( \mathbb{R} \) to \( \mathbb{R} \) . This is because 0 is an element of \( \mathbb{R} \), but the formula does not define a value for \( g\left( 0\right) \) . Thus, 0 cannot be in the domain of \( g \) . | To correct this problem, one could say that \( g \) is a function from the set \( \{ x \in \mathbb{R} \mid x \neq 0\} \) of nonzero real numbers, to \( \mathbb{R} \) . \( \blacklozenge \) | Yes |
Example 6.2.13. Suppose that the function \( f \) is defined by \( f\left( x\right) = {x}^{2} \), on the domain \( \{ 0,1,2,4\} \) . Then | 1. to represent \( f \) as a set of ordered pairs, each element of the domain must appear exactly once as a first coordinate, with the corresponding output given in the second coordinate. Since there are four elements in the domain, there will be four ordered pairs: \( \{ \left( {0,0}\right) ,\left( {1,1}\right) ,\left( {2,4}\right) ,\left( {4,{16}}\right) \} \) ;\n\n2. to give a table for \( f \), we include one row for every element of the domain. The table will be:\n\n<table><thead><tr><th>\( n \)</th><th>\( f\left( n\right) \)</th></tr></thead><tr><td>0</td><td>0</td></tr><tr><td>1</td><td>1</td></tr><tr><td>2</td><td>4</td></tr><tr><td>4</td><td>16</td></tr></table>\n\n3. if we are asked what is \( f\left( 3\right) \), the answer is that \( f\left( 3\right) \) is undefined, because 3 is not in the domain of \( f \) . Even though we know that \( {3}^{2} = 9 \), the formula we gave for \( f \) only applies to elements that are in the domain of \( f \) ! It is not true that \( f\left( 3\right) = 9 \) ;\n\n4. the range of \( f \) is the set of possible outputs: in this case, \( \{ 0,1,4,{16}\} \) ;\n\n5. if we are asked what is \( f\left( 2\right) \), the answer is \( f\left( 2\right) = 4 \) ;\n\n6. is \( f \) a function from \( \{ n \in \mathbb{N} \mid n \leq 4\} \) to \( \{ 0,1,4,{16}\} \) ? The answer is no, because the first set is \( \{ 0,1,2,3,4\} \), which includes the value 3, but 3 is not in the domain of \( f \) .\n\n7. is \( f \) a function from \( \{ 0,1,2,4\} \) to \( \{ n \in \mathbb{N} \mid n \leq {16}\} \) ? The answer is yes; even though the second set has many values that are not in the range, it is a possible codomain for \( f \) . A codomain can be any set that contains all of the elements of the range. | Yes |
Suppose Inspector Gadget knows two facts:\n\n1. Alice is the thief's wife, and\n\n2. Alice is Bob's wife.\n\nThen the inspector can arrest Bob for theft, because a person cannot (legally) be the wife of more than one husband. | According to U.S. law as of 2017. | No |
Is Temp a one-to-one function? | Not at all: it's very likely that at any given time, at least two points on the equator have exactly the same temperature (to arbitrary precision). Another way to say this is that at any given time, there exists a temperature \( b \) for which we can find two points on earth \( x \) and \( y \) such that \( \operatorname{Temp}\left( \mathrm{x}\right) = \operatorname{Temp}\left( \mathrm{y}\right) = \mathrm{b} \). | Yes |
(a) \( f : \mathbb{R} \rightarrow \mathbb{R} \), defined by \( f\left( x\right) = x + 1 \) . | Let's go back to the definition of one-to-one. Suppose we know that \( f\left( x\right) = f\left( y\right) \), where \( x, y \) are real numbers. Can we conclude that \( x = y \) ? If so, then that means that \( f \) is one-to-one.\n\nSo let’s follow through on this. \( f\left( x\right) = f\left( y\right) \) means that \( x + 1 = y + 1 \) . Subtracting 1 from both sides of the equation, we find that indeed, \( x = y \) . Hence, \( f \) is one-to-one, according to the definition. | Yes |
Let \( f : \mathbb{N} \rightarrow \mathbb{N} \) be defined by \( f\left( n\right) = {\left( n - 2\right) }^{2} + 1 \) . Is \( f \) one-to-one? | First let’s try to prove that \( f \) is one-to-one. Start with arbitrary elements \( m, n \in \mathbb{N} \), and suppose that \( f\left( m\right) = f\left( n\right) \) . By the definition of \( f \), this means that \( {\left( m - 2\right) }^{2} + 1 = {\left( n - 2\right) }^{2} + 1 \), or \( {\left( m - 2\right) }^{2} = {\left( n - 2\right) }^{2} \) . Two numbers have the same square, if and only if they are equal in absolute value, so it follows that \( m - 2 = \pm \left( {n - 2}\right) \) . There are now two cases:\n\n- If \( m - 2 = + \left( {n - 2}\right) \) then adding 2 to each side, we get \( m = n \) .\n\n- If \( m - 2 = - \left( {n - 2}\right) = - n + 2 \), then adding 2 to each side, we get\n\n\( m = - n + 4 \) .\n\nSince \( m, n \in \mathbb{N} \), it’s not hard to see that if \( n \geq 4 \), then \( - n + 4 \) is not a natural number. But if \( n \) is \( 1,2,3 \) then \( - n + 4 \in \mathbb{N} \) . For example \( n = 1 \) gives \( m = 3 \), which suggests that \( f\left( 1\right) = f\left( 3\right) \) . We may indeed check that \( f\left( 1\right) = f\left( 3\right) \) .\n\nNow the great thing about cases where \( f \) is not one-to-one is that the writeup of the solution is very simple. All you have to do is give one example of two different values that return the same function value. In the current example we have:\n\nSolution: \( f \) is not one-to-one because \( f\left( 1\right) = 2 \) and \( f\left( 3\right) = 2 \) . | Yes |
Example 6.3.16. We know from calculus that the function \( {e}^{x} : \mathbb{R} \rightarrow \mathbb{R} \) is a strictly increasing function since its derivative is always positive. In mathematical terms, we can say\n\n\[ x > y\\text{ implies }{e}^{x} > {e}^{y}. \]\n\nWe can use this fact and Definition 6.3.15 to prove that \( {e}^{x} \) is a one-to-one function as follows: | Take any two real numbers \( {x}_{1} \) and \( {x}_{2} \) where \( {x}_{1} \\neq {x}_{2} \) . If \( {x}_{1} > {x}_{2} \), then by the above equation it follows that \( {e}^{{x}_{1}} > {e}^{{x}_{2}} \) . On the other hand, if \( {x}_{1} < {x}_{2} \) , then by the above equation it follows that \( {e}^{{x}_{1}} < {e}^{{x}_{2}} \) . In either case, we have \( {e}^{{x}_{1}} \\neq {e}^{{x}_{2}} \) . By Definition 6.3.15, it follows that \( {e}^{x} \) must be one-to-one. | Yes |
Define \( g : \mathbb{R} \rightarrow \mathbb{R} \) by \( g\left( x\right) = {5x} - 2 \) . Determine whether \( g \) is onto. | Given \( y \in \mathbb{R} \), let \( x = \left( {y + 2}\right) /5 \) . Since the reals are closed under addition and non-zero division, it follows that \( x \in \mathbb{R} \) . Then\n\n\[ g\left( x\right) = {5x} - 2 = 5\left( \frac{y + 2}{5}\right) - 2 = \left( {y + 2}\right) - 2 = y.\]\n\nTherefore \( g \) is onto. | Yes |
Define \( h : \left\lbrack {0,2}\right\rbrack \rightarrow \left\lbrack {-7, - 1}\right\rbrack \) by \( h\left( x\right) = - {3x} - 1 \) . Determine whether \( h \) is onto. | Proof. Given \( y \in \mathbb{R} \), let \( x = \frac{y + 1}{-3} \) . By basic algebra, \( - 7 \leq y \leq - 1 \Rightarrow 0 \leq \) \( \frac{y + 1}{-3} \leq 2 \), so \( x \) is in the domain of \( h \) . Also,\n\n\[ h\left( x\right) = - 3\left( \frac{y + 1}{-3}\right) - 1 = y. \]\n\nTherefore \( h \) is onto. | Yes |
Example 6.4.16. Define \( f : \mathbb{C} \rightarrow \mathbb{C} \) by \( f\left( z\right) = {z}^{2} \) . Determine whether \( f \) is onto. | Proof. Given \( z = r\operatorname{cis}\theta \in \mathbb{C} \), let \( w = \sqrt{r}\operatorname{cis}\left( {\theta /2}\right) \) . By the definition of polar form, \( w \in \mathbb{C} \) and we have\n\n\[ f\left( w\right) = {\left( \sqrt{r}\operatorname{cis}\left( \theta /2\right) \right) }^{2} = {\left( \sqrt{r}\right) }^{2}\operatorname{cis}\left( {{2\theta }/2}\right) = r\operatorname{cis}\theta = z, \]\n\nwhere we have used De Moivre’s Theorem. It follows that \( f \) is onto. \( ▱\blacklozenge \) | Yes |
Consider a hypothetical country \( \mathrm{Z} \), in which\n\n- every person is married to at least one other person (no singles),\n\n- everyone is married to at most one other person (no polygamists or polyandrists), and\n\n- every marriage is between a man and a woman (no same-sex marriages).\n\nLet \( \mathsf{{Men}} = \{ \) male inhabitants of \( \;\mathsf{Z}\} \), and \( \mathsf{{Women}} = \{ \) female inhabitants of \( \;\mathsf{Z}\} \) . Then the function wife: Men \( \rightarrow \) Women is a bijection, since: | - Two different men cannot have the same wife, so we know that wife is one-to-one.\n\n- Every woman is the wife of some man (because everyone is married), so wife is also onto.\n\nSimilarly, the function husband: Women \( \rightarrow \) Men is also a bijection. | Yes |
Example 6.5.6. Define \( f : \left\lbrack {1,3}\right\rbrack \rightarrow \left\lbrack {-2,8}\right\rbrack \) by \( f\left( x\right) = {5x} - 7 \) . Then \( f \) is a bijection. | Proof. It suffices to show that \( f \) is both one-to-one and onto:\n\n- (one-to-one) Given \( {x}_{1},{x}_{2} \in \mathbb{R} \), such that \( f\left( {x}_{1}\right) = f\left( {x}_{2}\right) \), we have\n\n\[ 5{x}_{1} - 7 = 5{x}_{2} - 7 \]\n\nAdding 7 to both sides and dividing by 5 , we have\n\n\[ \frac{\left( {5{x}_{1} - 7}\right) + 7}{5} = \frac{\left( {5{x}_{2} - 7}\right) + 7}{5} \]\n\nWhich implies \( {x}_{1} = {x}_{2} \) . So \( f \) is one-to-one.\n\n- (onto) Given \( y \in \mathbb{R} \), let \( x = \left( {y + 7}\right) /5 \) . Then\n\n\[ f\left( x\right) = {5x} - 7 = 5\left( \frac{y + 7}{5}\right) - 7 = \left( {y + 7}\right) - 7 = y. \]\n\nWe need to verify that \( x \) is in the domain of \( f \) for every \( y \) is in the codomain:\n\n\[ - 2 \leq y \leq 8 \Rightarrow 5 \leq y + 7 \leq {15}\;\text{ [basic algebra] } \]\n\n\[ \Rightarrow 1 \leq \frac{y + 7}{5} \leq 3\;\text{ [basic algebra] } \]\n\n\[ \Rightarrow x \in \left\lbrack {1,3}\right\rbrack \;\text{ [substitution] } \]\n\nSo \( f \) is onto.\n\nSince \( f \) is both one-to-one and onto, we conclude that \( f \) is a bijection.\n\n\( \blacklozenge \) | Yes |
Example 6.6.4. Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) and \( g : \mathbb{R} \rightarrow \mathbb{R} \) by \( f\left( x\right) = {3x} \) and \( g\left( x\right) = {x}^{2} \) . Then \( g \circ f \) and \( f \circ g \) are functions from \( \mathbb{R} \) to \( \mathbb{R} \) . For all \( x \in \mathbb{R} \) , we have | \[ g \circ f\left( x\right) = g\left( {f\left( x\right) }\right) = g\left( {3x}\right) = {\left( 3x\right) }^{2} = 9{x}^{2} \] and \[ f \circ g\left( x\right) = f\left( {g\left( x\right) }\right) = f\left( {x}^{2}\right) = 3\left( {x}^{2}\right) = 3{x}^{2}. \] Notice that (in this example) \( f \circ g \neq g \circ f \), so composition is not commutative. \( \blacklozenge \) | Yes |
Example 6.6.7. Figure 6.6.1 provides an arrow diagram to illustrate the composition \( g \circ f \) . | - Starting from any point of \( A \), follow the arrow (for the function \( f \) that starts there to arrive at some point of \( B \) .\n- Then follow the arrow (for the function \( g \) ) that starts there to arrive at a point of \( C \) .\nFor example, the \( f \)-arrow from \( a \) leads to \( m \) and the \( g \)-arrow from \( m \) leads to \( u \). So \( g \circ f\left( a\right) = u \). Notice how we write the result as \( g \circ f \) with \( g \) on the left and \( f \) on the right even though \( f \) appears on the left in Figure 6.6.1. This is an unfortunate consequence of the fact that when we calculate \( g\left( {f\left( x\right) }\right) \) we work right to left, computing \( f\left( x\right) \) first and applying \( g \) to the result. | Yes |
Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \), where \( A \subset C \). Show that if\n\n\[ g \circ f\left( a\right) = a\text{, for every}a \in A\text{,} \]\n\nthen \( f \) is one-to-one. | Proof. Given that \( g \circ f\left( a\right) = a \), for every \( a \in A \), by the definition of composition, this means that, for any \( {a}_{1},{a}_{2} \in A \) we have\n\n\[ g\left( {f\left( {a}_{1}\right) }\right) = {a}_{1}\text{ and }g\left( {f\left( {a}_{2}\right) }\right) = {a}_{2}. \]\n\nNow suppose \( f\left( {a}_{1}\right) = f\left( {a}_{2}\right) \). Then by the definition of a function,\n\n\[ g\left( {f\left( {a}_{1}\right) }\right) = g\left( {f\left( {a}_{2}\right) }\right) \]\n\nBy our original hypothesis we then get \( {a}_{1} = {a}_{2} \), and thus \( f \) is one-to-one. | Yes |
Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) . Show that if \( f \) and \( g \) are onto, then \( g \circ f \) is onto. | Proof. Let \( c \) be an arbitrary element of \( C \) . Since \( g \) is onto, there exists a \( b \) in \( B \) such that \( g\left( b\right) = c \) . Since \( f \) is onto, there exists a \( a \) in \( A \) such that \( f\left( a\right) = b \) . It follows that \( g \circ f\left( a\right) = g\left( {f\left( a\right) }\right) = g\left( b\right) = c \) . Since \( c \) is an arbitrary element of \( C \), this implies that \( g \circ f \) is onto. | Yes |
Example 6.6.17. Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) . Show that if \( g \circ f \) is one-to-one, and the range of \( f \) is \( B \), then \( g \) is one-to-one. | Proof. Suppose \( {b}_{1} \) and \( {b}_{2} \) are distinct elements of \( B \) . Since the range of \( f \) is \( B \), it follows that there exist \( {a}_{1} \neq {a}_{2} \) such that \( f\left( {a}_{1}\right) = {b}_{1} \) and \( f\left( {a}_{2}\right) = {b}_{2} \) . Since \( g \circ f \) is one-to-one, it follows that \( g \circ f\left( {a}_{1}\right) \neq g \circ f\left( {a}_{2}\right) \) . But by definition of \( \circ, g \circ f\left( {a}_{1}\right) = g\left( {f\left( {a}_{1}\right) }\right) = g\left( {b}_{1}\right) \) ; and similarly \( g \circ f\left( {a}_{2}\right) = g\left( {b}_{2}\right) \) . By substitution, it follows that \( g\left( {b}_{1}\right) \neq g\left( {b}_{2}\right) \) . Thus distinct elements of \( B \) always map to distinct elements of \( C \) under the function \( g \) : which is the same as saying that \( g \) is one-to-one. | Yes |
Example 6.6.20. Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) and \( g \) is not onto. Then \( g \circ f \) is not onto. | Proof. We will prove by contradiction. Suppose on the contrary that \( g \circ f \) is onto. Then given any \( c \in C \), there exists a \( a \in A \) such that \( g \circ f\left( a\right) = c \) . By the definition of composition, it follows that \( g\left( b\right) = c \) where \( {bf}\left( a\right) \) . Since \( c \) was an arbitrary element of \( C \), this implies that \( g \) is onto, which contradicts our original condition that \( g \) is not onto. This is a contradiction. Therefore our supposition must be false, and \( g \circ f \) is not onto. | Yes |
In Example 6.5.6, we showed that \( f\left( x\right) = {5x} - 7 \) is a bijection. A quick look at the proof reveals that the formula\n\n\[ x = \frac{y + 7}{5} \]\n\nplays a key role. This formula is obtained by replacing \( f\left( x\right) \) in \( f\left( x\right) = {5x} - 7 \) with \( y \), and solving for \( x \) . | In order to see \( x = \frac{y + 7}{5} \) as an \ | No |
Example 6.7.2. Let \( f : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) be defined by: \( f\left( x\right) = {x}^{2} \) . We may define \( g : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) by: \( g\left( y\right) = \sqrt{y} \) . Note that in this case the domains and ranges are restricted to positive real numbers. Given this restriction, by the definition of square root we have | \[ y = {x}^{2} \Leftrightarrow x = \sqrt{y} \] In view of the definitions of \( f \) and \( g \), we may see that this is the same formula as in the previous example: \( y = f\left( x\right) \Leftrightarrow x = g\left( y\right) \). | Yes |
Proposition 6.7.4. Suppose that \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) are functions such that\n\n\[ \forall x \in X,\forall y \in Y,\left( {y = f\left( x\right) \Leftrightarrow x = g\left( y\right) }\right) . \]\n\nThen the following statements are also true:\n\n(a) \( g\left( {f\left( x\right) }\right) = x \) for all \( x \in X \) . and\n\n(b) \( f\left( {g\left( y\right) }\right) = y \) for all \( y \in Y \) , | Proof. The proof of (a) runs as follows. Suppose that \( y = f\left( x\right) \Leftrightarrow x = g\left( y\right) \) for all \( x, y \) in the respective domains of \( f \) and \( g \) . Then for any \( x \in X \), we may define \( z \) as \( z = f\left( x\right) \) . By the \( \Leftrightarrow \) statement it follows that \( x = g\left( z\right) \) . But then we may substitute the first equation into the second and obtain \( g\left( {f\left( x\right) }\right) = x \) . Since \( x \) was an arbitrary element of \( X \), it follows that \( g\left( {f\left( x\right) }\right) = x \) for all \( x \in X \) . | Yes |
Example 6.7.8. The husband of the wife of any married man is the man himself - in other words,\n\n\[ \n\\text{husband}\\left( {\\operatorname{wife}\\left( y\\right) }\\right) = y\\text{.}\n\]\n\nAlso, the wife of the husband of any married woman is the woman herself, so that\n\n\[ \n\\text{wife(husband(x))} = x\\text{.}\n\] | It follows that the wife function is an inverse of the husband function. In fact, it's pretty clear that husband is the only inverse of wife. | No |
Proposition 6.7.11. Suppose \( f : X \rightarrow Y \) . Then \( f \) has an inverse \( g : Y \rightarrow X \) if and only if \( f \) is a bijection. | Proof. (forward direction) Assume there is a function \( g : Y \rightarrow X \) that is an inverse of \( f \) . Then by the definition of an inverse function,\n\n(a) \( f\left( {g\left( y\right) }\right) = y \) for all \( y \in Y \), and\n\n(b) \( g\left( {f\left( x\right) }\right) = x \) for all \( x \in X \) .\n\nSuppose then that \( f\left( {x}_{1}\right) = f\left( {x}_{2}\right) \) for some \( {x}_{1},{x}_{2} \in X \) . Then since \( g \) is a function we have\n\n\[ g\left( {f\left( {x}_{1}\right) }\right) = g\left( {f\left( {x}_{2}\right) }\right) \]\n\nTherefore by (b), \( {x}_{1} = {x}_{2} \) . Hence \( f \) is one-to-one.\n\nNow suppose \( y \in Y \) . Then since \( g \) is a function, there exists a unique \( x \in X \) such that \( g\left( y\right) = x \) . Substituting into (a) we get\n\n\[ f\left( x\right) = y. \]\n\nTherefore \( \forall y \in Y,\exists x \in X \) s.t. \( f\left( x\right) = y \) . Hence \( f \) is onto. So \( f \) is both one-to-one and onto: thus \( f \) is a bijection. | Yes |
Example 7.2.1. One of the first and most famous private key cryptosystems was the shift code used by Julius Caesar. We first represent the alphabet numerically by letting \( \mathrm{A} = 0,\mathrm{\;B} = 1,\ldots ,\mathrm{Y} = {24},\mathrm{Z} = {25} \) . This means for example that the word BAY would be represented numerically as:\n\n\[ 1,0,{24}\text{.} \]\n\nAn example of a shift encoding function is\n\n\[ f\left( n\right) = {\;\operatorname{mod}\;\left( {n + 3,{26}}\right) }. \]\n\nwhich can also be written as\n\n\[ f\left( n\right) = n \oplus 3 \]\n\nwith the understanding that \( n \) refers to the numerical value assigned to each letter, and \( \oplus \) refers to addition in \( {\mathbb{Z}}_{26} \) . This encoding function takes\n\n\[ 0 \rightarrow 3,1 \rightarrow 4,\ldots ,{25} \rightarrow 1,{26} \rightarrow 2, \]\n\nso that our numerical representation of BAY is changed to: \( 4,3,1 \), which is the numerical representation of EDB.\n\nThe decoding function is the inverse of the function \( f \), which we can find in the usual way by solving the equation \( m = n \oplus 3 \) for \( n \) . The result is \( n = m \ominus 3 \), so that\n\n\[ {f}^{-1}\left( m\right) = m \ominus 3\;\text{ or }\;{f}^{-1}\left( m\right) = m \oplus {23}. \] | Suppose we receive the encoded message DOJHEUD. To decode this message, we first represent it numerically:\n\n\[ 3,{14},9,7,4,{20},3\text{.} \]\n\nNext we apply the decryption function to get\n\n\[ 0,{11},6,4,1,{17},0, \]\n\nwhich is the numerical representation of ALGEBRA. Notice here that there is nothing special about either of the numbers 3 or 26 . We could have used a larger alphabet or a different shift. | Yes |
Example 7.2.5. Suppose we receive a message that we know was encrypted by using a shift transformation on single letters of the 26-letter alphabet. To find out exactly what the shift transformation was, we must compute \( b \) in the equation \( f\left( n\right) = n + b{\;\operatorname{mod}\;{26}} \) . We can do this using frequency analysis. The letter \( \mathrm{E} = {04} \) is the most commonly occurring letter in the English language. Suppose that \( \mathrm{S} = {18} \) is the most commonly occurring letter in the ciphertext. Then we have good reason to suspect that \( {18} = 4 \oplus b \), or \( b = {14} \) . Therefore, the most likely encoding function is | \[ f\left( n\right) = n \oplus {14} \]\n\nThe corresponding decoding function is\n\n\[ {f}^{-1}\left( m\right) = m \oplus {12}. \]\n\nIt is now easy to determine whether or not our guess is correct. | Yes |
Example 7.2.12. Let's consider the affine cryptosystem encoding function \( f\left( n\right) = \left( {a \odot n}\right) \oplus b \), where \( \odot \) and \( \oplus \) are multiplication and addition mod 26 respectively. For this cryptosystem to work we must choose an \( a \in {\mathbb{Z}}_{26} \) that is invertible. This is only possible if \( \gcd \left( {a,{26}}\right) = 1 \) . Recognizing this fact, we will let \( a = 5 \) since \( \gcd \left( {5,{26}}\right) = 1 \) . The reader may check that \( {a}^{-1} = {21} \) . Therefore, we can take our encryption function to be \( f\left( n\right) = \) \( \left( {5 \odot n}\right) \oplus 3 \) . Thus, ALGEBRA is encoded as \( 3,6,7,{23},8,{10},3 \), or DGHXIKD. | The decryption function will be\n\n\[ \n{f}^{-1}\left( n\right) = \left( {{21} \odot n}\right) \ominus \left( {{21} \odot 3}\right) = \left( {{21} \odot n}\right) \oplus {15}. \n\] | Yes |
Example 7.3.1. Before exploring the theory behind the RSA cryptosystem or attempting to use large integers, we will use some small integers just to see that the system does indeed work. Suppose that we wish to send some message, which when digitized is 395 . Let \( p = {23} \) and \( q = {29} \) . Then\n\n\[ \n n = {pq} = {667}\;\text{ and }\;m = \left( {p - 1}\right) \left( {q - 1}\right) = {616}. \]\n\nWe can let \( E = {487} \), since \( \gcd \left( {{616},{487}}\right) = 1 \) . The encoded message is computed to be\n\n\[ \n{\;\operatorname{mod}\;\left( {{395}^{487},{667}}\right) } = {570}. \]\n\n(This may seem like a very long computation, but there are fast ways of doing this: see Exercise 7.3.3 below.) Using the Euclidean algorithm, we determine that \( {191E} = 1 + {151m} \) ; therefore, the decrypting key is \( \left( {n, D}\right) = \left( {{667},{191}}\right) \) . We can recover the original message by calculating\n\n\[ \n{\;\operatorname{mod}\;\left( {{570}^{191},{667}}\right) } = {395}. \]\n\nThis really seems like magic. How in the world does it work? | First of all, we know that \( {DE} \equiv 1{\;\operatorname{mod}\;m} \) ; so there exists a \( k \) such that\n\n\[ \n{DE} = {km} + 1\text{.} \]\n\nThis means that\n\n\[ \n{y}^{D} = {\left( {x}^{E}\right) }^{D} = {x}^{DE} = {x}^{{km} + 1} = {\left( {x}^{m}\right) }^{k}x. \]\n\nAt this point we need Euler's theorem from Chapter 15, which states the following. Suppose \( m \) is the number of positive integers less than \( n \) that are relatively prime to \( n \) . Then it is true that:\n\n\[ \n{x}^{m} \equiv 1\;\left( {\;\operatorname{mod}\;n}\right) \]\nfor any \( x \) that is relatively prime to \( n \) .\n\nWe can use this to simplify our previous expression for \( {y}^{D} \) :\n\n\[ \n{y}^{D} = {\left( {x}^{m}\right) }^{k}x \equiv {\left( 1\right) }^{k}x \equiv x{\;\operatorname{mod}\;n}, \]\n\nand presto! We have our result. | No |
Consider:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{{10}}\left( {i + 2}\right) \] | In this case, the \( \sum \) symbol lets us know that this is a sum. The \( i = 1 \) serves two functions. It tells us that the index variable is \( i \), and that \( i \) has a starting value of 1 . The 10 is the final value, and the \( \left( {i + 2}\right) \) to the right of the \( \sum \) is the formula. The \( i \) in the formula, takes each integer value from the starting value (1) to the final value (10). Therefore we have:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{{10}}\left( {i + 2}\right) = 3 + 4 + 5 + 6 + 7 + 8 + 9 + {10} + {11} + {12} = {75}. \] | Yes |
Consider the product of the sums \( \mathop{\sum }\limits_{{i = 0}}^{2}{3}^{i} \) and \( \mathop{\sum }\limits_{{j = 0}}^{2}{3}^{-j} \) . | By the distributive law and additive commutivity we have:\n\n\[ \left( {\mathop{\sum }\limits_{{i = 0}}^{2}{3}^{i}}\right) \left( {\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{-j}}\right) = \mathop{\sum }\limits_{{i = 0}}^{2}\left( {{3}^{i}\left( {\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{-j}}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{2}\left( {\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{i - j}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{2}\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{i - j} \]\n\nThis sum has 9 terms, where each term corresponds to a pair \( \left( {i, j}\right) \) as shown in Figure 8.3.1. These terms can be arranged along diagonal lines (as shown in the figure) so that all terms on each diagonal have the same value. So we can add the terms diagonal-by-diagonal as follows:\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{2}\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{i - j} = {3}^{-2} + 2 \cdot {3}^{-1} + 3 \cdot {3}^{0} + 2 \cdot {3}^{1} + {3}^{2} \]\n\n\[ = 1/9 + 2/3 + 3 + 6 + 9 \]\n\n\[ = {18}\frac{5}{9}\text{.} \] | Yes |
Let \( \mathbf{v} \) be the \( {10} \times 1 \) matrix given by\n\n\[ \n{v}_{j,1} = j, j = 1\ldots {10}.\n\]\n\n(Note that \( \mathbf{v} \) is essentially a column vector.) Let us compute \( C\mathbf{v} \), where the matrix \( C \) defined by\n\n\[ \n{c}_{i, j} \mathrel{\text{:=}} \frac{1}{2}\left( {-{\delta }_{i - 1, j} + {\delta }_{i + 1, j}}\right) ,\;1 \leq i, j \leq {10}.\n\] | The summation notation expression for the product is:\n\n\[ \n{\left\lbrack C\mathbf{v}\right\rbrack }_{ij} = \mathop{\sum }\limits_{{k = 1}}^{{10}}{c}_{i, k}{v}_{k, j}\n\]\n\nThe first thing to notice is that the second index \( j \) must be 1 since \( \mathbf{v} \) is a \( {10} \times 1 \) matrix. We may also substitute the expressions for \( {d}_{i, k} \) and \( {v}_{k, j} \) and simplify:\n\n\[ \n{\left\lbrack C\mathbf{v}\right\rbrack }_{i1} = \mathop{\sum }\limits_{{k = 1}}^{{10}}\frac{1}{2}\left( {-{\delta }_{i - 1, j} + {\delta }_{i + 1, j}}\right) k\;\left\lbrack {\text{ Definitions of }{c}_{i, k}\text{ and }{v}_{k,1}}\right\rbrack\n\]\n\n\[ \n= - \frac{1}{2}\mathop{\sum }\limits_{{k = 1}}^{{10}}{\delta }_{i - 1, k}k + \frac{1}{2}\mathop{\sum }\limits_{{k = 1}}^{{10}}{\delta }_{i + 1, k}k\;\text{ [Summation rules] }\n\]\n\nAt this point, we need to think about how the \( \delta \) ’s function within these two sums. Consider the first sum, namely:\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{{10}}{\delta }_{i - 1, k}k\n\]\n\nFor each value of \( i = 1,\ldots ,{10} \), this sum will give a different result:\n\n- When \( i = 1 \), all terms in the sum are 0, so the result is zero.\n\n- When \( i = 2 \), the only term that contributes is the \( k = 1 \) term, since \( {\delta }_{1, k} = 0 \) unless \( k = 1 \) . So for \( i = 2 \), the sum gives 1 .\n\n- Similarly when \( i = 3,\ldots ,{10} \), the only term that contributes is the \( k = i - 1 \) term, since \( {\delta }_{3, k} = 0 \) unless \( k = i - 1 \) . So the sum gives \( i - 1 \) for \( 2 \leq i \leq {10} \) .\n\nWe may summarize these findings as follows:\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{{10}}{\delta }_{i - 1, k}k = \left\{ \begin{array}{ll} 0 & \text{ if }i = 1 \\ i - 1 & \text{ if }2 \leq i \leq {10}. \end{array}\right.\n\]\n\nThe second sum may be evaluated similarly: this time, \( i = {10} \) is the exceptional case:\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{{10}}{\delta }_{i + 1, k}k = \left\{ \begin{array}{ll} i + 1 & \text{ if }1 \leq i \leq 9 \\ 0 & \text{ if }i = {10} \end{array}\right.\n\]\n\nSubstituting these expressions into our matrix product formula gives:\n\n\[ \n{\left\lbrack C\mathbf{v}\right\rbrack }_{i1} = \left\{ \begin{array}{ll} - \frac{0}{2} + \frac{2}{2} = 1 & \text{ if }i = 1 \\ - \frac{i - 1}{2} + \frac{i + 1}{2} = 1 & \text{ if }2 \leq i \leq 9 \\ - \frac{9}{2} + \frac{0}{2} = - {4.5} & \text{ if }i = {10} \end{array}\right.\n\]\n\nThe result is a \( {10} \times 1 \) column vector with entries all 1, except for a -4.5 in the \( {10}^{\text{th }} \) entry. \( \blacklozenge \) | Yes |
Example 8.5.10. This time we'll compute the entries of the matrix product \( {FV} \), where the entries \( {f}_{ij} \) of \( F \) and \( {v}_{ij} \) of \( V \) are given by:\n\n\[ \n{f}_{ij} \mathrel{\text{:=}} {\delta }_{i + 1, j} - {\delta }_{i, j};\;{v}_{i, j} \mathrel{\text{:=}} {2}^{i + j},\;1 \leq i, j \leq {20}.\n\] | We may begin once again with the matrix product formula:\n\n\[ \n\left\lbrack {F{V}_{ij} = \mathop{\sum }\limits_{{k = 1}}^{{20}}{f}_{ik}{v}_{kj}\;}\right. \text{[Matrix mulitplication formula ]} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{{20}}\left( {{\delta }_{i + 1, k} - {\delta }_{i, k}}\right) {2}^{k + j}\;\text{ [Substitution] } \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{{20}}{\delta }_{i + 1, k}{2}^{k + j} - {\delta }_{i, k}{2}^{k + j}\;\text{ [Substitution] } \n\]\n\n\[ \n= {2}^{i + 1 + j} - {2}^{i + j}\;\text{[Select nonzero term in each summation]} \n\]\n\n\[ \n= {2}^{i + j}\left( {2 - 1}\right) \n\]\n[Factor out common factor]\n\n\[ \n= {2}^{i + j}\text{.} \n\]\n[Exponent rules] | Yes |
Proposition 8.5.15. If \( A \) and \( B \) are matrices such that the matrix product is defined, then\n\n\[{\left( AB\right) }^{\mathrm{T}} = {B}^{\mathrm{T}}{A}^{\mathrm{T}}\] | Proof. We’ll prove this by expressing the \( \left( {i, j}\right) \) entry of the left-hand side in summation notation, doing some algebraic hocus-pocus, and showing that it agrees with the \( \left( {i, j}\right) \) entry of the right side. First we make things clear by specifying that \( A \) has \( n \) columns and \( B \) has \( n \) rows (these dimensions have to agree, or the product is not defined). This gives us\n\n\[{\left\lbrack AB\right\rbrack }_{i, j} = \mathop{\sum }\limits_{k}{a}_{i, k}{b}_{k, j}\]\n\n(remember that we decided to use abbreviated notation, so we leave off the summation limits) so the \( \left( {i, j}\right) \) entry of the left-hand side is:\n\n\[{\left\lbrack {\left( AB\right) }^{T}\right\rbrack }_{i, j} = {\left\lbrack AB\right\rbrack }_{j, i} = \mathop{\sum }\limits_{k}{a}_{j, k}{b}_{k, i}.\]\n\nAt this point we can introduce \( A \) and \( B \) transpose because the \( j, k \) entry of any matrix is the \( k, j \) entry of its transpose:\n\n\[\mathop{\sum }\limits_{k}{a}_{j, k}{b}_{k, i} = \mathop{\sum }\limits_{k}{\left\lbrack {A}^{\mathrm{T}}\right\rbrack }_{k, j}{\left\lbrack {B}^{\mathrm{T}}\right\rbrack }_{i, k.}\]\n\nSince the terms of \( A \) and \( B \) are being expressed as a summation, they commute (i.e. order doesn't matter), which allows us to say (using our definition of matrix product):\n\n\[\mathop{\sum }\limits_{k}{\left\lbrack {A}^{\mathrm{T}}\right\rbrack }_{k, j}{\left\lbrack {B}^{\mathrm{T}}\right\rbrack }_{i, k} = \mathop{\sum }\limits_{k}{\left\lbrack {B}^{\mathrm{T}}\right\rbrack }_{i, k}{\left\lbrack {A}^{\mathrm{T}}\right\rbrack }_{k, j} = {\left\lbrack {B}^{\mathrm{T}}{A}^{\mathrm{T}}\right\rbrack }_{i, j},\]\n\nVoilà, we have the \( \left( {i, j}\right) \) entry of the right-hand side, and the proof is complete. | Yes |
Proposition 8.6.16. Given any \( 3 \times 3 \) matrix \( A \), then\n\n\[ \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{ij}{a}_{k\ell } = \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{ji}{a}_{k\ell } \]\n\n(Observe the minute difference between the two sides: there’s an \( {a}_{ij} \) on the left-hand side which becomes an \( {a}_{ji} \) on the right. Minute differences matter!) | Proof. Let us consider the case \( i = 1 \) :\n\n\[ \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{1j}{a}_{k\ell } = \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{j1}{a}_{k\ell } \]\n\nand we’ll leave the cases \( i = 2,3 \) as exercises.\n\nOn both right and left sides there are terms with \( j = 1, j = 2 \), and \( j = 3 \) . We'll consider these cases one by one.\n\n- \( j = 1 \) : these terms are equal on both sides, since in this case \( {a}_{1j} = \) \( {a}_{j1} = {a}_{11} \)\n\n- \( j = 2 \) : in view of the \( {\epsilon }_{{jk}\ell } \) on both sides, since \( j = 2 \) the only nonzero terms are \( k = 3,\ell = 1 \) or \( k = 1,\ell = 3 \) . On the left-hand side this gives \( {a}_{12}{a}_{31} - {a}_{12}{a}_{13} \), while on the right-hand side we get \( {a}_{21}{a}_{31} - {a}_{21}{a}_{13} \)\n\n- \( j = 3 \) : once again, in view of the \( {\epsilon }_{{jk}\ell } \) on both sides, since \( j = 3 \) the only nonzero terms are \( k = 1,\ell = 2 \) or \( k = 2,\ell = 1 \) . On the left-hand side this gives \( {a}_{13}{a}_{12} - {a}_{13}{a}_{21} \), while on the right-hand side we get \( {a}_{31}{a}_{12} - {a}_{31}{a}_{21} \)\n\nAdding all left-hand side terms gives\n\n\[ {a}_{12}{a}_{31} - {a}_{12}{a}_{13} + {a}_{13}{a}_{12} - {a}_{13}{a}_{21} = {a}_{12}{a}_{31} - {a}_{13}{a}_{21}, \]\n\nwhile adding all right-hand side terms gives\n\n\[ {a}_{21}{a}_{31} - {a}_{21}{a}_{13} + {a}_{31}{a}_{12} - {a}_{31}{a}_{21} = - {a}_{21}{a}_{13} + {a}_{31}{a}_{12}. \]\n\nMiraculously, these turn out to be equal. | No |
Example 8.7.1. Evaluate \( \mathop{\sum }\limits_{{k = 1}}^{n}{2}^{k - 1}k \) . | In order to use the summation by parts formula, we need to define \( {a}_{k} \) and \( {B}_{k} \) so that the summand \( {2}^{k - 1}k \) is the product of \( {a}_{k} \) and \( {B}_{k} \) . Just as in integration by parts, we want to make our choice based on what makes the calculations easiest. Note that \( {B}_{k} \) is a partial sum of \( k \) terms, and that \( k = 1 + \ldots + 1 \) . So it’s natural to choose \( {B}_{k} = k \), which means that \( {a}_{k} = {2}^{k - 1} \) .\n\nBased on our choice of \( {B}_{k} \) and \( {a}_{k} \), we can now figure out \( {b}_{k} \) and \( {A}_{k} \) . As we noted above, \( {B}_{k} \) is the sum of \( {k1} \) ’s, so that \( {b}_{k} = 1 \) . This leaves us with \( {a}_{k} = {2}^{k - 1} \), so that\n\n\[ \n{A}_{k} = \mathop{\sum }\limits_{{j = 1}}^{k}{a}_{j} \n\]\n\nAs we've seen before, we may rewrite this by shifting the starting value of the index, so that\n\n\[ \n{A}_{k} = \mathop{\sum }\limits_{{{j}^{\prime } = 0}}^{{k - 1}}{2}^{{j}^{\prime }} = \frac{{2}^{k} - 1}{2 - 1} = {2}^{k} - 1 \n\]\n\nwhere we've used our standard formula for the sum of geometric series.\n\nSummarizing our progress so far, we have:\n\n\[ \n{a}_{k} = {2}^{k - 1};\;{B}_{k} = k;\;{b}_{k} = 1;\;{A}_{k} = {2}^{k} - 1. \n\]\nPlugging the above values into the summation by parts formula, we find:\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}{2}^{k - 1}k = \left( {{2}^{n} - 1}\right) n - \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}\left( {{2}^{k} - 1}\right) . \n\]\n\nThe summation on the far right can be evaluated by breaking it into two separate parts:\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{{n - 1}}\left( {{2}^{k} - 1}\right) = \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}{2}^{k} - \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}1 = \left( {{2}^{n} - 1}\right) - \left( {n - 1}\right) . \n\]\n\nWe can put this into our equality and do some further algebraic puttering to obtain the final result:\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}{2}^{k - 1}k = \left( {{2}^{n} - 1}\right) n - \left( {{2}^{n} - 1}\right) + \left( {n - 1}\right) \n\]\n\n\[ \n= {2}^{n}n - n - \left( {{2}^{n} - 1}\right) + \left( {n - 1}\right) \n\]\n\n\[ \n= {2}^{n}n - {2}^{n} \n\]\n\n\[ \n= {2}^{n}\left( {n - 1}\right) \text{.} \n\] | Yes |
Example 9.3.3. Express the following polynomials in summation notation:\n\n(a) \( {p}_{1}\left( x\right) = 1 + x + {x}^{2} + {x}^{3} \) | (a) \( {p}_{1}\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{3}{x}^{j} \) | Yes |
Proposition 9.5.3. Given that \( R \) satisfies conditions (I)-(V) listed above, then addition in \( R\left\lbrack x\right\rbrack \) is both commutative:\n\n\[ p\left( x\right) + q\left( x\right) = q\left( x\right) + p\left( x\right) \]\n\nand associative:\n\n\[ \left( {p\left( x\right) + q\left( x\right) }\right) + r\left( x\right) = p\left( x\right) + \left( {q\left( x\right) + r\left( x\right) }\right) . \] | Proof. First, we show commutativity: Given two polynomials \( p\left( x\right) \) and \( q\left( x\right) \) where\n\n\[ p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i};\;q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{b}_{i}{x}^{i}, \]\n\nthen\n\n\[ p\left( x\right) + q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{{\max \left( {m, n}\right) }}\left( {{a}_{i} + {b}_{i}}\right) {x}^{i}, \]\n\nand\n\n\[ p\left( x\right) + q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{{\max \left( {m, n}\right) }}\left( {{b}_{i} + {a}_{i}}\right) {x}^{i}. \]\n\nSince the addition is commutative, we have \( {a}_{i} + {b}_{i} = {b}_{i} + {a}_{i} \) for all \( i \) . It follows that all coefficients of \( p\left( x\right) + q\left( x\right) \) are equal to the corresponding coefficients of \( q\left( x\right) + p\left( x\right) \) . By the definition of polynomial equality, this means that \( p\left( x\right) + q\left( x\right) = q\left( x\right) + p\left( x\right) \) . | Yes |
Proposition 9.5.4. Given that \( R \) satisfies properties (I)-(V), then the additive identity of \( R\left\lbrack x\right\rbrack \) is \( 0{x}^{0} \), where 0 denotes the additive identity of \( R \) . | Proof. The proof has two parts: (i) \( p\left( x\right) + 0{x}^{0} = p\left( x\right) \) and (ii) \( 0{x}^{0} + p\left( x\right) = \) \( p\left( x\right) ,\forall p\left( x\right) \in R\left\lbrack x\right\rbrack \) . We’ll prove (i) and leave (ii) as an exercise.\n\n(i) Given an arbitrary polynomial \( p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \in R\left\lbrack x\right\rbrack \) . Then,\n\n\[ p\left( x\right) + 0{x}^{0} = \left( {\mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i}}\right) + 0{x}^{0} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{m}\left( {{a}_{i} + 0}\right) {x}^{i} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \]\n\n\[ = p\left( x\right) \]\n\nSo part (i) of the proof is finished. | No |
Proposition 9.5.7. Let \( p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) be a polynomial in \( R\left\lbrack x\right\rbrack \), where \( R \) satisfies properties (I)-(V). Then the additive inverse of \( p\left( x\right) \) is \( q\left( x\right) = \) \( \mathop{\sum }\limits_{{i = 0}}^{n}\left( {-{a}_{i}}\right) {x}^{i} \), where \( - {a}_{i} \) is the additive inverse of \( {a}_{i} \) in \( R \) . | Exercise 9.5.8. Prove Proposition 9.5.7 by showing that \( p\left( x\right) + q\left( x\right) \) and \( q\left( x\right) + p\left( x\right) \) both sum to the additive identity of \( R\left\lbrack x\right\rbrack \) . | No |
Proposition 9.5.12. Multiplication in \( R\left\lbrack x\right\rbrack \) is associative: | Proof. We’ve seen that the product of two polynomials \( p\left( x\right) \) and \( q\left( x\right) \) may be written in summation notation as:\n\n\[ p\left( x\right) q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}{a}_{i}{b}_{j}{x}^{i + j} \]\n\nNow we multiply a third polynomial, \( r\left( x\right) \), to calculate its product with \( \left( {p\left( x\right) q\left( x\right) }\right) \):\n\n\[ \left( {p\left( x\right) q\left( x\right) }\right) r\left( x\right) = \left( {\mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}{a}_{i}{b}_{j}{x}^{i + j}}\right) \left( {\mathop{\sum }\limits_{{k = 0}}^{\ell }{c}_{k}{x}^{k}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}\left( {{a}_{i}{b}_{j}{x}^{i + j}\left( {\mathop{\sum }\limits_{{k = 0}}^{\ell }{c}_{k}{x}^{k}}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}\mathop{\sum }\limits_{{k = 0}}^{\ell }{a}_{i}{b}_{j}{x}^{i + j} \cdot {c}_{k}{x}^{k} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}\mathop{\sum }\limits_{{k = 0}}^{\ell }{a}_{i}{b}_{j}{c}_{k}{x}^{i + j + k}. \]\n\nIn the above calculation we have twice brought multiplicative terms inside of summations, using the distributive law. The last step uses a familiar exponent rule.\n\nTo complete the proof of associativity, we need to show that the summation expression for \( p\left( x\right) \left( {q\left( x\right) r\left( x\right) }\right) \) may be simplified into the same expression. The calculation is very similar, and we leave it as an exercise:\n\nExercise 9.5.13. Show \( p\left( x\right) \left( {q\left( x\right) r\left( x\right) }\right) \) also simplifies to \( \mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}\mathop{\sum }\limits_{{k = 0}}^{\ell }{a}_{i}{b}_{j}{c}_{k}{x}^{i + j + k} \).\n\nGive a justification for each step of your calculation.\n\nThe exercise shows that \( \left( {p\left( x\right) q\left( x\right) }\right) r\left( x\right) \) and \( p\left( x\right) \left( {q\left( x\right) r\left( x\right) }\right) \) both simplify to the same expression, so they are equal. This completes the proof. | No |
Proposition 9.5.14. Polynomials in \( R\left\lbrack x\right\rbrack \) have both right distributivity across addition:\n\n\[ \left( {q\left( x\right) + r\left( x\right) }\right) p\left( x\right) = q\left( x\right) p\left( x\right) + r\left( x\right) p\left( x\right) ,\]\n\nand left distributivity across addition:\n\n\[ p\left( x\right) \left( {q\left( x\right) + r\left( x\right) }\right) = p\left( x\right) q\left( x\right) + p\left( x\right) r\left( x\right) . \] | Proof. To show right distributivity, we have:\n\n\[ \left( {q\left( x\right) + r\left( x\right) }\right) p\left( x\right) = \left( {\mathop{\sum }\limits_{{j = 0}}^{n}{b}_{j}{x}^{j} + \mathop{\sum }\limits_{{j = 0}}^{\ell }{c}_{j}{x}^{j}}\right) \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \]\n\n\[ = \left( {\mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\left( {{b}_{j} + {c}_{j}}\right) {x}^{j}}\right) \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \]\n\n\[ = \left( {\mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\left( {{b}_{j}{x}^{j} + {c}_{j}{x}^{j}}\right) }\right) \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\left( {\left( {{b}_{j}{x}^{j} + {c}_{j}{x}^{j}}\right) \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\mathop{\sum }\limits_{{i = 0}}^{m}\left( {{b}_{j}{x}^{j} + {c}_{j}{x}^{j}}\right) {a}_{i}{x}^{i} \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\mathop{\sum }\limits_{{i = 0}}^{m}\left( {{b}_{j}{x}^{j}{a}_{i}{x}^{i} + {c}_{j}{x}^{j}{a}_{i}{x}^{i}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\mathop{\sum }\limits_{{i = 0}}^{m}{b}_{j}{x}^{j}{a}_{i}{x}^{i} + \mathop{\sum }\limits_{{j = 0}}^{{\max \left( {n,\ell }\right) }}\mathop{\sum }\limits_{{i = 0}}^{m}{c}_{j}{x}^{j}{a}_{i}{x}^{i} \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{n}\mathop{\sum }\limits_{{i = 0}}^{m}{b}_{j}{x}^{j}{a}_{i}{x}^{i} + \mathop{\sum }\limits_{{j = 0}}^{\ell }\mathop{\sum }\limits_{{i = 0}}^{m}{c}_{j}{x}^{j}{a}_{i}{x}^{i} \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{n}{b}_{j}{x}^{j}\mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} + \mathop{\sum }\limits_{{j = 0}}^{\ell }{c}_{j}{x}^{j}\mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \]\n\n\[ = q\left( x\right) p\left( x\right) + r\left( x\right) p\left( x\right) \]\n\nwhich gives us right distributivity. We'll leave left distributivity up to you: | No |
Dividing polynomials in \( \mathbb{R}\left\lbrack x\right\rbrack \) is very similar to long division of real numbers. For example, suppose that we divide \( {x}^{3} - {x}^{2} + {2x} - 3 \) by \( x - 2 \) . | In the example, we need to take the leading power term of \( x \) in the divisor and multiply by something that will make it equal to the the leading power term in the dividend. In this case it’s \( {x}^{2} \) . This gives \( {x}^{2} \cdot \left( {x - 2}\right) = {x}^{3} - 2{x}^{2} \) Subtract from the dividend to yield a remainder of \( {x}^{2} + {2x} - 3 \) and repeat until the remainder is of a degree less than the divisor. Hence, \( {x}^{3} - {x}^{2} + {2x} - 3 = \left( {x - 2}\right) \left( {{x}^{2} + x + 4}\right) + 5 \), which you may check by multiplying out the right-hand side. | Yes |
Example 9.6.3. Divide \( \left( {2{x}^{3} + 3{x}^{2} + x + 4}\right) \) by \( \left( {x + 2}\right) \) where both polynomials are in \( {\mathbb{Z}}_{5}\left\lbrack x\right\rbrack \) . |  | No |
Proposition 9.6.7. (Division algorithm for polynomials) Suppose that the set \( F \) has addition and multiplication operations that satisfy (I)-(VII). Let \( f\left( x\right) \) and \( g\left( x\right) \) be nonzero polynomials in \( F\left\lbrack x\right\rbrack \), where the degree of \( g\left( x\right) \) is greater than 0 . Then there exist unique polynomials \( q\left( x\right) \) and \( r\left( x\right) \) in \( F\left\lbrack x\right\rbrack \) such that \[ f\left( x\right) = g\left( x\right) q\left( x\right) + r\left( x\right) , \] where the degree of \( r\left( x\right) \) is less than the degree of \( g\left( x\right) \) . | Proof. We will first prove the existence of \( q\left( x\right) \) and \( r\left( x\right) \) . We define a set \( S \) as follows: \[ S = \{ f\left( x\right) - g\left( x\right) h\left( x\right) ,\text{ for all }h\left( x\right) \in F\left\lbrack x\right\rbrack \} . \] This set is nonempty since \( f\left( x\right) \in S \) . Let \( r\left( x\right) \) be a polynomial of smallest degree in \( S.{}^{2} \) This means that there must exist a \( q\left( x\right) \) such that \[ r\left( x\right) = f\left( x\right) - g\left( x\right) q\left( x\right) . \] We need to show that the degree of \( r\left( x\right) \) is less than the degree of \( g\left( x\right) \) . Let's prove this by contradiction. So we assume the contrary, namely that \( \deg g\left( x\right) \leq \deg r\left( x\right) \) . Let \( n, m \) be the degree of \( g\left( x\right), r\left( x\right) \) respectively, where \( n \leq m \) . Then we may write \[ g\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \] and \[ r\left( x\right) = {b}_{0} + {b}_{1}x + \cdots + {b}_{m}{x}^{m} \] where \( {a}_{n} \neq 0 \) and \( {b}_{m} \neq 0 \) . Taking a cue from the process of long division, we define a new polynomial \( {r}^{\prime }\left( x\right) \) by \[ {r}^{\prime }\left( x\right) \mathrel{\text{:=}} r\left( x\right) - {b}_{m}{a}_{n}^{-1}{x}^{m - n}g\left( x\right) \] \( {}^{2} \) At this point we can’t assume that there’s only one such polynomial, so we have to say \ | Yes |
Suppose that we would like to find the gcd of \( a\left( x\right) = {x}^{4} - 5{x}^{3} + 5{x}^{2} + {5x} - 6 \) and \( b\left( x\right) = {x}^{4} + 5{x}^{3} + 5{x}^{2} - {5x} - 6 \) . | We first divide \( a\left( x\right) \) by \( b\left( x\right) \) to determine the remainder, \( {r}_{1} \).\n\n\n\nSo \( {r}_{1} = - {10}{x}^{3} + {10x} \). We then divide \( b\left( x\right) \) by \( {r}_{1} \) to determine the second remainder, \( {r}_{2} \).\n\n\n\nSo \( {r}_{2} = 6{x}^{2} - 6 \). We then divide \( {r}_{1} \) by \( {r}_{2} \) to determine the third remainder, \( {r}_{3} \).\n\n\n\nNotice that \( {r}_{3} = 0 \). This means that \( 6{x}^{2} - 6 \) divides both \( a\left( x\right) \) and \( b\left( x\right) \). Furthermore, any real, nonzero multiple of \( 6{x}^{2} - 6 \) will divide both \( a\left( x\right) \) and \( b\left( x\right) \). For convenience, we choose the multiple with a leading coefficient of 1. This means that \( {x}^{2} - 1 \) is the gcd of \( a\left( x\right) \) and \( b\left( x\right) \). You should check that \( {x}^{2} - 1 \) divides both \( a\left( x\right) \) and \( b\left( x\right) \). | Yes |
Suppose that we would like to find the gcd of \( a\left( x\right) = {x}^{4} + 2{x}^{3} + 5{x}^{2} + {5x} + 1 \) and \( b\left( x\right) = {x}^{4} + 5{x}^{3} + 5{x}^{2} + {2x} + 1 \) in \( {\mathbb{Z}}_{7}\left\lbrack x\right\rbrack \) . | We first divide \( a\left( x\right) \) by \( b\left( x\right) \) to determine the remainder, \( {r}_{1} \) . \n\nSo \( {r}_{1} = 4{x}^{3} + {3x} \) . We then divide \( b\left( x\right) \) by \( {r}_{1} \) to determine the second remainder, \( {r}_{2} \) .\n\n\n\nSo \( {r}_{2} = 6{x}^{2} + 1 \) . We then divide \( {r}_{1} \) by \( {r}_{2} \) to determine the third remainder, \( {r}_{3} \) .\n\n\n\nNotice that \( {r}_{3} = 0 \) . Therefore, \( 6{x}^{2} + 1 \) divides both \( a\left( x\right) \) and \( b\left( x\right) \) . We multiply \( 6{x}^{2} + 1 \) by the inverse of 6 to obtain the gcd for \( a\left( x\right) \) and \( b\left( x\right) \) . The result is \( {x}^{2} + 6 \) . We leave it to the reader to check that \( {x}^{2} + 6 \) divides both \( a\left( x\right) \) and \( b\left( x\right) \) . | Yes |
Proposition 9.6.12. Let \( F \) satisfy properties (I)-(VII), \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), and \( a \in F \) . When \( f\left( x\right) \) is divided by \( x - a \), the remainder is \( f\left( a\right) \) . | Proof. According to Proposition 9.6.7, if we divide \( f\left( x\right) \) by \( x - a \), it will produce two unique polynomials \( q\left( x\right) \) and \( r\left( x\right) \) such that \( f\left( x\right) = \left( {x - a}\right) q\left( x\right) + \) \( r\left( x\right) \) . Since the degree of \( x - a \) is 1, then according to the division algorithm, the degree of \( r\left( x\right) \) must be less than 1 . Therefore \( r\left( x\right) \) must be a constant \( r \), and we may write:\n\n\[ f\left( x\right) = \left( {x - a}\right) q\left( x\right) + r. \]\n\nIf we set \( x = a \) then we get:\n\n\[ f\left( a\right) = \left( {a - a}\right) q\left( x\right) + r \]\n\n\[ = 0 \cdot q\left( x\right) + r \]\n\n\[ = r\text{.} \] | Yes |
Proposition 9.6.14. Let \( F \) satisfy properties (I)-(VII), \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), and \( a \in F \) . Then \( x - a \) divides \( f\left( x\right) \) if and only if \( f\left( a\right) = 0 \) . | Proof. From Proposition 9.6.17 \( f\left( x\right) = \left( {x - a}\right) \cdot q\left( x\right) + f\left( a\right) \) . Therefore \( f\left( a\right) = 0 \) if and only if \( f\left( x\right) = \left( {x - a}\right) \cdot q\left( x\right) \), which is true if and only if \( x - a \) divides \( f\left( x\right) \) . | Yes |
Proposition 9.6.15. Suppose that \( F \) satisfies properties (I)-(VII). Given \( a, b \in F \) and \( {ab} = 0 \), then either \( a = 0 \) or \( b = 0 \) . | Exercise 9.6.16. Prove Proposition 9.6.15. (Hint: you may follow the steps given in Exercise 2.2.12 | No |
Proposition 9.6.17. Suppose \( F \) satisfies Properties (I)-(VII), and suppose \( p\left( x\right), q\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( p\left( x\right) q\left( x\right) = 0 \) iff either \( p\left( x\right) = 0 \) or \( q\left( x\right) = 0 \) . | Proof. Since this is a \ | No |
Proposition 9.6.19. Suppose \( F \) satisfies properties (I)-(VII), and let \( c \) be any element \( F \) . Then \( c \) has at most \( n{n}^{\text{th }} \) roots. | Proof. Given \( C \in F \), then the polynomial \( {x}^{n} - c \) is an element of \( F\left\lbrack x\right\rbrack \) . By Proposition 9.6.18, the equation \( {x}^{n} - c = 0 \) has at most \( n \) solutions. This is exactly the same thing as saying that \( c \) has at most \( n{n}^{\text{th }} \) roots. | Yes |
Find the roots of \( p\left( x\right) = 2{x}^{2} + {2x} + 5 \) . | Since this is a quadratic polynomial we can use the famous quadratic formula:\n\n\[ x = \frac{-b \pm \sqrt{{b}^{2} - {4ac}}}{2a} \]\n\nIn \( p\left( x\right), a = 2, b = 2 \), and \( c = 5 \) . We substitute those values into the formula and obtain the following:\n\n\[ x = \frac{-2 \pm \sqrt{{2}^{2} - 4 \cdot 2 \cdot 5}}{2 \cdot 2} = \frac{-2 \pm \sqrt{-{36}}}{4} = \frac{-2 \pm {6i}}{4} \]\n\n\[ = \frac{-1 \pm {3i}}{2}\text{. } \]\n\nSo the roots of \( p\left( x\right) \) are \( x = - \frac{1}{2} + \frac{3}{2}i, - \frac{1}{2} - \frac{3}{2}i \) . Neither of these roots are elements of \( \mathbb{R} \) . As noted above this does not contradict FTOA, which only guarantees there won't be more than 2 roots. | Yes |
Find the roots of \( f\left( x\right) = 3{x}^{3} + {10}{x}^{2} + {11x} + 6 \) . | Since this is a cubic polynomial, we can't use the quadratic formula, at least not to begin with. The coefficients are integers, so we may use Proposition 9.6.22, which says that any rational roots of \( p\left( x\right) \) have numerators that are factors of \( {a}_{0} \) and denominators that are factors of \( {a}_{n} \) . This does not guarantee that there are rational roots: sometimes polynomials are irreducible, but we still try every method possible to find those roots unless we know that we can't reduce the polynomial. So we will proceed with trying to find the roots of \( f\left( x\right) \) using Proposition 9.6.22.\n\nIn \( f\left( x\right) \), possible numerators of any rational roots are: \( p = \pm 1, \pm 2, \pm 3, \pm 6 \) . The possible denominators are: \( q = \pm 1, \pm 3 \) . So we have as possible rational roots the following: \( p/q = \pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 6 \) . By Proposition 9.6.14, if \( f\left( {p/q}\right) = 0 \) then \( \left( {x - p/q}\right) \) is a factor of \( f\left( x\right) \) ; which would make \( p/q \) a root of \( f\left( x\right) \) . After testing all possibilities we find the following rational root: \( f\left( {-2}\right) = 3{\left( -2\right) }^{3} + {10}{\left( -2\right) }^{2} + {11}\left( {-2}\right) + 6 = 0 \) . Therefore, \( x = - 2 \) is a root of \( f\left( x\right) \) and \( \left( {x + 2}\right) \) is a factor of \( f\left( x\right) \) . We then use long division to factor \( f\left( x\right) \).\n\n\n\nSo now we have \( f\left( x\right) = \left( {x + 2}\right) \left( {3{x}^{2} + {4x} + 3}\right) \) . We use the quadratic formula to find the following roots for \( 3{x}^{2} + {4x} + 3.x = \frac{-2 \pm \sqrt{5}i}{3} \) . So there are two complex roots and one real root. They are \( x = \frac{-2 - \sqrt{5}i}{3}, - 2,\frac{-2 + \sqrt{5}i}{3} \) . | Yes |
Example 9.6.29. We begin with an example of a linear binomial in \( \mathbb{C}\left\lbrack x\right\rbrack \) . Let \( p\left( x\right) = \left( {3 + {2i}}\right) x + \left( {2 - i}\right) \) . Find the root of \( p\left( x\right) \) (since \( p\left( x\right) \) is linear, it will only have one root). | First we set \( p\left( x\right) \) equal to zero and then proceed to find the root as follows. Begining with \( \left( {3 + {2i}}\right) x + \left( {2 - i}\right) = 0 \), we may rearrange to obtain\n\n\[ x = \frac{-2 + i}{3 + {2i}} \]\n\nand multiplying numerator and denominator by \( 3 - {2i} \) and simplifying gives\n\n\[ x = \frac{4 + {7i}}{13} \in \mathbb{C}\left\lbrack x\right\rbrack \] | Yes |
Let \( p\left( x\right) = \left( {1 + i}\right) {x}^{2} + \left( {2 - i}\right) x + \left( {3 + {3i}}\right) \). Find the roots of \( p\left( x\right) \). | Since this is a quadratic polynomial we can use the quadratic formula and obtain the following:\n\n\[ x = \frac{-\left( {2 - i}\right) \pm \sqrt{{\left( 2 - i\right) }^{2} - 4\left( {1 + i}\right) \left( {3 + {3i}}\right) }}{2\left( {1 + i}\right) } = \frac{\left( {-2 + i}\right) \pm \sqrt{3 - {4i} - {24i}}}{\left( 2 + 2i\right) }\n\]\n\n\[ = \frac{\left( {-2 + i}\right) \pm \sqrt{3 - {28i}}}{\left( 2 + 2i\right) } = \frac{\left( {-2 + i}\right) \pm \sqrt{3 - {28i}}}{\left( 2 + 2i\right) } \cdot \frac{2 - {2i}}{2 - {2i}}\]\n\n\[ = \frac{6 + {6i} \pm \left( {2 - {2i}}\right) \sqrt{3 - {28i}}}{8} = \frac{3 + {3i} \pm \left( {1 - i}\right) \sqrt{3 - {28i}}}{4}. \]\n\nSo the two roots are \( x = \left\{ {\frac{3 + {3i} - \left( {1 - i}\right) \sqrt{3 - {28i}}}{4},\frac{3 + {3i} + \left( {1 - i}\right) \sqrt{3 - {28i}}}{4}}\right\} \). | Yes |
Proposition 9.6.32. Any polynomial \( p\\left( x\\right) \) of degree \( n \) in \( \\mathbb{C}\\left\\lbrack x\\right\\rbrack \) can be completely factored as a constant times a product of \( n \) linear terms, as follows:\n\n\[ p\\left( x\\right) = b\\left( {x - {a}_{1}}\\right) \\left( {x - {a}_{2}}\\right) \\ldots \\left( {x - {a}_{n}}\\right) . \]\n\nwhere \( b,{a}_{1},\\ldots ,{a}_{n} \\in \\mathbb{C} \) . | Proof. Let \( p\\left( x\\right) \) be an arbitrary polynomial of degree \( n \) in \( \\mathbb{C}\\left\\lbrack x\\right\\rbrack \) . By Proposition 9.6.28, \( p\\left( x\\right) \) has at least one complex root \( a \) . So \( \\left( {x - a}\\right) \) is a factor of \( p\\left( x\\right) \) and we can write \( p\\left( x\\right) = \\left( {x - {a}_{1}}\\right) {p}_{1}\\left( x\\right) \) ; where the degree of \( {p}_{1}\\left( x\\right) \) is \( n - 1 \) . If \( {p}_{1}\\left( x\\right) \) is linear, then we are done, but if \( {p}_{1}\\left( x\\right) \) is not linear, then by Proposition 9.6.28 it also has a complex root \( {a}_{2} \) . So \( \\left( {x - {a}_{2}}\\right) \) is a factor of \( {p}_{2}\\left( x\\right) \) and we can write \( p\\left( x\\right) = \\left( {x - {a}_{1}}\\right) \\left( {x - {a}_{2}}\\right) {p}_{2}\\left( x\\right) \) ; where the degree of \( {p}_{2}\\left( x\\right) \) is \( n - 2 \) . The same argument continues until we reach \( {p}_{n - 2}\\left( x\\right) \), which has degree 1 . But this means that \( {p}_{n - 2}\\left( x\\right) \) can be written as \( {bx} - c \), which we may rewrite as \( b\\left( {x - {a}_{n}}\\right) \), where \( {a}_{n} = c/b \) . It follows finally that \( p\\left( x\\right) = b\\left( {x - {a}_{1}}\\right) \\left( {x - {a}_{2}}\\right) \\ldots \\left( {x - {a}_{n}}\\right) \) . | Yes |
Example 9.6.35. Let \( f\left( x\right) = {x}^{4} - 4{x}^{3} + {10}{x}^{2} - {24x} + {24} \) be a polynomial in \( \mathbb{C}\left\lbrack x\right\rbrack \) . Notice that the coefficients of \( f\left( x\right) \) are integers, so \( f\left( x\right) \) is also in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Therefore we can use Proposition 9.6.22 to factor out our first linear term. Since \( {a}_{0} = {24} \) and \( {a}_{n} = 1 \), possible rational roots are \[ \frac{p}{q} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm {12}. \] By Proposition 9.6.14, if \( f\left( {p/q}\right) = 0 \) then \( \left( {x - p/q}\right) \) is a factor of \( f\left( x\right) \) i.e. \( p/q \) a root of \( f\left( x\right) \) . | After testing all possibilities we find the following rational root: \( f\left( 2\right) = {2}^{4} - 4{\left( 2\right) }^{3} + {10}{\left( 2\right) }^{2} - {24}\left( 2\right) + {24} = 0 \) . Therefore, \( x = 2 \) is a root of \( f\left( x\right) \) and \( \left( {x - 2}\right) \) is a factor of \( f\left( x\right) \) . After dividing \( f\left( x\right) \) by \( \left( {x - 2}\right) \) we have the following: \[ f\left( x\right) = \left( {x - 2}\right) \left( {{x}^{3} - 2{x}^{2} + {6x} - {12}}\right) = \left( {x - 2}\right) \left( {{x}^{2}\left( {x - 2}\right) + 6\left( {x - 2}\right) }\right) \] \[ = \left( {x - 2}\right) \left( {x - 2}\right) \left( {{x}^{2} + 6}\right) = {\left( x - 2\right) }^{2}\left( {{x}^{2} + 6}\right) \text{.} \] Solving \( {x}^{2} + 6 = 0 \) for \( x \) gives us additional roots: \( x = - \sqrt{6}i,2,\sqrt{6}i \) . In summary, we have \[ f\left( x\right) = \left( {x - 2}\right) \left( {x - 2}\right) \left( {x + \sqrt{6}i}\right) \left( {x - \sqrt{6}i}\right) . \] So as Proposition 9.6.32 states, \( f\left( x\right) \) factors completely into a product of linear terms. | Yes |
Example 10.1.5. Figure 10.1.2 shows a \( {60}^{ \circ } \) counterclockwise rotation of a regular hexagon where the vertices of the hexagon are labeled \( A, B, C, D, E, F \) . (Notice how the letters run counterclockwise around the hexagon. We will consistently follow this pattern. The reason is that in mathematical convention, a counterclockwise rotation is considered as positive, while a clockwise rotation is considered as negatve.) | The rotation moves \( A \) to \( B, B \) to \( C \), and so on. Now of course there are other points on our figure, namely all the points on the line segments between the vertices. But notice that if we account for where the vertices are moved to, then the movement of the line segments is automatically accounted for. If we know where \( A \) and \( B \) are moved to, we know exactly where \( \overline{AB} \) is. Therefore, our 60 degree rotation can be defined by the movement of the vertices \( \{ A, B, C, D, E, F\} \) .\n\nNow if we input a point from \( \{ A, B, C, D, E, F\} \), our rotation outputs a point from \( \{ A, B, C, D, E, F\} \) . We have used this \ | No |
Proposition 10.1.7. If \( S \) is the set of points that represent a figure, all symmetries of the figure are bijections from \( S \rightarrow S \) . | Proof. Since the result of any symmetry acting on \( S \) must be all of \( S \), then every point of \( S \) must be in the range of \( S \). Thus any symmetry is onto.\n\nFurthermore, the symmetry must map two different points to two different points, since the distance between points must be left unchanged by the symmetry. Hence any symmetry is one-to-one. So since any symmetry is both onto and one-to-one, it follows that any symmetry is a bijection. | Yes |
Suppose we wanted to find \( {r}_{180} \circ {s}_{v} \) using the tableau forms for \( {r}_{180} \) and \( {s}_{v} \) above. That is\n\n\[ \n{r}_{180} \circ {s}_{v} = \left( \begin{array}{llll} A & B & C & D \\ C & D & A & B \end{array}\right) \circ \left( \begin{array}{llll} A & B & C & D \\ D & C & B & A \end{array}\right) = ? \n\] | To see how this works, let’s \ | No |
Proposition 10.2.5. Suppose \( f \) and \( g \) are both symmetries of a figure. Then \( f \circ g \) is itself a symmetry of the same figure. | Proof. Recall that composition works from right to left. Since \( g \) is a symmetry, \( g \) takes the points of the figure and rearranges them so that the angles and distances of points in the figure are preserved. The symmetry \( f \) then takes the points of this preserved figure and moves them in such a way that the angles, and distances of points in the figure are preserved. Hence the net result of \( f \circ g \) preserves angles and distances between points in the figure. Therefore by definition, \( f \circ g \) is a symmetry of the figure. | Yes |
Proposition 10.3.5. The set of symmetries \( S \) of any figure under composition is associative. | Proof. By definition, we know any symmetry of a figure is a function. From the Functions chapter, we know | No |
Proposition 10.3.6. The set of symmetries \( S \) of any figure has an identity. | Proof. By the definition of a symmetry, the \ | No |
Proposition 10.3.7. All elements of the set \( S \) of symmetries of any figure have inverses. | Proof. Given a symmetry \( s \in S \), by definition \( s \) is a bijection. In the Functions chapter, we showed that every bijection has an inverse \( {s}^{-1} \) . It remains to show that \( {s}^{-1} \) is itself a symmetry. This means that we have to show:\n\n(i) \( {s}^{-1} \) leaves distances unchanged between points in the figure;\n\n(ii) \( {s}^{-1} \) leaves angles unchanged between points in the figure;\n\nii) \( {s}^{-1} \) leaves the appearance of the figure unchanged.\n\nThese three items are proved as follows:\n\n(i) This proof is similar to (ii), and we leave it as an exercise.\n\n(ii) We show that \( {s}^{-1} \) leaves angles and distances between points unchanged as follows:\n\n- Choose any three points \( A, B, C \) in the figure, and let \( {A}^{\prime } = \) \( {s}^{-1}\left( A\right) ,{B}^{\prime } = {s}^{-1}\left( B\right) ,{C}^{\prime } = {s}^{-1}\left( C\right) \).\n\n- By the definition of inverse, it follows that \( s\left( {A}^{\prime }\right) = A, s\left( {B}^{\prime }\right) = \) \( B, s\left( {C}^{\prime }\right) = C \).\n\n- Since \( s \) is a symmetry, it follows that \( \angle {A}^{\prime }{B}^{\prime }{C}^{\prime } = \angle {ABC} \).\n\n- Since \( A, B, C \) were arbitrary points in the figure, we have shown that \( {s}^{-1} \) leaves angles between points unchanged.\n\n(iii) In the Functions chapter, we showed that \( {s}^{-1} \) is also a bijection. Hence it leaves the appearance of the figure unchanged. | No |
Proposition 10.3.9. The set \( S \) of symmetries of any figure forms a group. | Exercise 10.3.10. Prove Proposition 10.3.9 (make use of the propositions that we've proved previously.) | No |
Proposition 10.4.1. The dihedral group, \( {D}_{n} \), is a group of order \( {2n} \) . | Let us try to characterize these \( {2n} \) elements of the dihedral group \( {D}_{n} \) .\n\n\n\nFigure 10.4.1. Rotations and reflections of a regular \( n \) -gon\n\nFirst, we know that the elements of the dihedral group includes \( n \) rotations:\n\[ \text{id,}\frac{{360}^{ \circ }}{n},2 \cdot \frac{{360}^{ \circ }}{n},\ldots ,\left( {n - 1}\right) \cdot \frac{{360}^{ \circ }}{n}\text{.} \]\n\nWe will denote the rotation \( {360}^{ \circ }/n \) by \( r \) . Notice that:\n\n- \( r \circ r = \) rotation by \( 2 \cdot \frac{{360}^{ \circ }}{n} \)\n\n- \( r \circ r \circ r = \) rotation by \( 3 \cdot \frac{{360}^{ \circ }}{n} \)\n\nWe can generalize this pattern by writing:\n\[ {r}^{k} = \text{ rotation by }k \cdot \frac{{360}^{ \circ }}{n}\;\left( {k = 1,2,3,\ldots }\right) , \]\n\nwhere the notation \( {r}^{k} \) means that we compose \( r \) with itself \( k \) times: \( r \circ r\ldots \circ r \) . We can also continue this pattern with \( k = 0 \) and write:\n\[ {r}^{0} = \text{rotation by}0 \cdot \frac{{360}^{ \circ }}{n} = \mathrm{{id}}\text{.} \]\n\nWe also have\n\[ {r}^{n} = \text{rotation by}n \cdot \frac{{360}^{ \circ }}{n} = \text{rotation by}{360}^{ \circ } = \mathrm{{id}}\text{,} \]\n\nsince rotation by 360 degrees is tantamount to not moving the figure at all. | No |
Example 10.4.16. In \( {D}_{5} \), to compute \( \left( {s \circ {r}^{3}}\right) \circ \left( {s \circ {r}^{4}}\right) \) we have (using Proposition 10.4.15 and associativity): | \[ \left( {s \circ {r}^{3}}\right) \circ \left( {s \circ {r}^{4}}\right) = s \circ \left( {{r}^{3} \circ s}\right) \circ {r}^{4}\text{by associativity} \] \[ = s \circ \left( {s \circ {r}^{2}}\right) \circ {r}^{4}\text{by Prop. 10.4.15(c)} \] \[ = \left( {s \circ s}\right) \circ r \circ {r}^{5}\text{by associativity} \] \[ = \mathrm{{id}} \circ r \circ \mathrm{{id}}\text{by Prop. 10.4.15(a) and (b)} \] \[ = r \] | Yes |
Example 11.1.2. Let us recall for a moment the equilateral triangle \( \bigtriangleup {ABC} \) from the Symmetries chapter. Let \( T \) be the set of vertices of \( \bigtriangleup {ABC} \) ; i.e. \( T = \{ A, B, C\} \) . We may list the permutations of \( T \) as follows. For input \( A \) , we have 3 possible outputs; then for \( B \) we would have two possible outputs (to keep the one-to-one property of each combination); and finally for \( C \) only one possible output. Therefore there are \( 3 \cdot 2 \cdot 1 = 6 \) permutations of \( T \) . Below are the six permutations in \( {S}_{T} \) : | \[ \left( \begin{array}{lll} A & B & C \\ A & B & C \end{array}\right) \;\left( \begin{array}{lll} A & B & C \\ C & A & B \end{array}\right) \;\left( \begin{array}{lll} A & B & C \\ B & C & A \end{array}\right) \] \[ \left( \begin{array}{lll} A & B & C \\ A & C & B \end{array}\right) \;\left( \begin{array}{lll} A & B & C \\ C & B & A \end{array}\right) \;\left( \begin{array}{lll} A & B & C \\ B & A & C \end{array}\right) \] Which of these permutations are symmetries of the equilateral triangle? In the Symmetries chapter we saw that they all are: so in this case the set of symmetries on \( T \) is equal to \( {S}_{T} \) . | Yes |
Proposition 11.2.1. Given any set \( X,{S}_{X} \) is a group under function composition. | Proof.\n\n- First then, if \( f, g \in {S}_{X} \), then \( f \circ g \) would be, by definition of composition, a function from \( X \rightarrow X \) . Further, since it is a composition of two bijections, \( f \circ g \) would be a bijection (proved in Functions chapter). Therefore by definition \( f \circ g \) is permutation from \( X \rightarrow X \) . In other words \( f \circ g \in {S}_{X} \) . So \( {S}_{X} \) is closed under function composition.\n\n- Second, the identity of \( {S}_{X} \) is just the permutation that sends every element of \( X \) to itself (We will call this permutation id, just like we did with symmetries.)\n\n- Third, if \( f \in {S}_{X} \), then by definition \( f \) is a bijection; hence from the Inverse section of the Functions chapter we know \( f \) has an inverse \( {f}^{-1} \) from \( X \rightarrow X \) that is also a bijection. Hence \( {f}^{-1} \in {S}_{X} \) . Therefore every permutation in \( {S}_{X} \) has an inverse.\n\n- Finally, composition of functions is associative, which makes the group operation associative.\n\nHence \( {S}_{X} \) is a group under function composition. | Yes |
Example 11.3.1. Suppose \( \rho \in {S}_{6} \) and \( \rho = \left( \begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 5 & 6 & 1 \end{array}\right) \) . Then | \[ \rho \left( 1\right) = 2,\rho \left( 2\right) = 3,\rho \left( 3\right) = 4,\rho \left( 4\right) = 5,\rho \left( 5\right) = 6\text{, and}\rho \left( 6\right) = 1\text{.} \] A shorter way to represent this is \[ 1 \rightarrow 2,2 \rightarrow 3,3 \rightarrow 4,4 \rightarrow 5,5 \rightarrow 6\text{and}6 \rightarrow 1\text{.} \] We can visualize this as a \ | Yes |
Example 11.3.9. Suppose \( \tau \in {S}_{6} \) and \( \tau = \left( \begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 4 & 2 & 3 & 5 & 6 \end{array}\right) \) . Then | - \( 1 \rightarrow 1 \), which means that 1 \ | No |
Suppose we want to form the product (that is, composition) \( {\sigma \tau } \), where \( \sigma ,\tau \in {S}_{6} \) and \( \sigma = \left( {1532}\right) ,\tau = \left( {126}\right) \) . | Figure 11.3.2 provides a visual representation of how the product \( {\sigma \tau } \) acts on 1. Remember that we operate from right to left, so the figure shows '1' coming in from the right. The action of \( \tau \) takes 1 to 2 . (For convenience we have \ | No |
Consider the product \( {\sigma \tau } \) where \( \sigma = \left( {AEDBF}\right) \) and \( \tau = \left( {ABDFE}\right) \) . | - \( \tau \) takes \( A \rightarrow B \) and \( \sigma \) takes \( B \rightarrow \mathrm{F} \) ; hence \( {\sigma \tau } \) takes \( A \rightarrow F \) .\n\n- \( \tau \) takes \( F \rightarrow E \), and \( \sigma \) takes \( E \rightarrow D \) ; hence \( {\sigma \tau } \) takes \( F \rightarrow D \) .\n\n- \( \tau \) takes \( D \rightarrow F \), and \( \sigma \) takes \( F \rightarrow A \) ; hence \( {\sigma \tau } \) takes \( D \rightarrow A \) .\n\nWe have finished a cycle: \( \left( {AFD}\right) \) . Let us check where the other letters \( B, C, E \) go:\n\n- \( \tau \) takes \( B \rightarrow D \), and \( \sigma \) takes \( D \rightarrow B \) ; hence \( {\sigma \tau } \) takes \( B \rightarrow B \) .\n\n- Neither \( \tau \) nor \( \sigma \) affects \( C \) ; hence \( {\sigma \tau } \) takes \( C \rightarrow C \) .\n\n- \( \tau \) takes \( E \rightarrow A \), and \( \sigma \) takes \( A \rightarrow E \) ; hence \( {\sigma \tau } \) takes \( E \rightarrow E \) .\n\nSince \( B, C, E \) are unaffected by \( {\sigma \tau } \), we conclude that \( {\sigma \tau } = \left( {AFD}\right) \) . | Yes |
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