SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I ...

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(1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso ... But the check digit here is 2, and 2 + 2 mod 10 = 4 = 0. So, this is not a ...

SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

(1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso minestrone soup? Show why the other numbers are not valid UPC’s. 0 41196 01012 1 0 52010 00121 2 0 05055 00505 3 Note: UPC stands for Universal Product Code, which is basically the product bar codes we had been discussing in lectures. Solution : For the first one, without the check digit, 3 · 0 + 4 + 3 · 1 + 1 + 3 · 9 + 6 + 3 · 0 + 1 + 3 · 0 + 1 + 3 · 2 mod 10 = 4 + 3 + 1 + 27 + 6 + 1 + 1 + 6 mod 10 = 49 mod 10 = 9. Clearly the check digit, 1, adds up to 9 to give 0 mod 10. Hence, this is a UPC indeed. For the second one, without the check digit, 3 · 0 + 5 + 3 · 2 + 0 + 3 · 1 + 0 + 3 · 0 + 0 + 3 · 1 + 2 + 3 · 1 mod 10 = 5 + 6 + 3 + 3 + 2 + 3 mod 10 = 22 mod 10 = 2. But the check digit here is 2, and 2 + 2 mod 10 = 4 6= 0. So, this is not a UPC. For the third one, without the check digit, 3 · 0 + 0 + 3 · 5 + 0 + 3 · 5 + 5 + 3 · 0 + 0 + 3 · 5 + 0 + 3 · 5 mod 10 = 15 + 15 + 5 + 15 + 15 mod 10 = 5

mod 10.

But the check digit is 3, and 5 + 3 mod 10 = 8 6= 0 So, this is not a UPC.

(2) (Exercise 13, Page 107) The following is the UPC for Hellman’s 8 oz. Real Mayonnaise. Find the missing digit. 0 48001 26 04 2 1

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SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

Solution : Let us call this missing digit x. Then, without the check digit, 3 · 0 + 4 + 3 · 8 + 0 + 3 · 0 + 1 + 3 · 2 + 6 + 3x + 0 + 3 · 4 mod 10 = 4 + 24 + 1 + 6 + 6 + 3x + 12 mod 10 = 53 + 3x mod 10 = 3 + 3x mod 10. Now, we must have that, 3 + 3x + 2 mod 10 = 0, i.e., 5 + 3x mod 10 = 0. This is possible only if x = 5 which gives us the missing digit.

(3) A bank identification number is a 9 digit number that occurs in the lower left hand corner of bank checks. Let the digits be n1 , n2 , n3 , n4 , n5 , n6 , n7 , n8 , n9 . For a 9 digit number to be a valid bank identification number, it must satisfy 7n1 + 3n2 + 9n3 + 7n4 + 3n5 + 9n6 + 7n7 + 3n8 + 9n9 ≡ 0

mod 10.

(Exercise 21, Page 108) Determine the check digits (i.e. last digits) of the following bank codes: 3 1 0 6 1 4 8 3 0 2 5 7 1 1 0 8 Solution : Let us call the check digit c. In the first code, we need, 7 · 3 + 3 · 1 + 9 · 0 + 7 · 6 + 3 · 1 + 9 · 4 + 7 · 8 + 3 · 3 + 9c ≡ 0 =⇒ 21 + 3 + 0 + 42 + 3 + 36 + 56 + 9 + 9c ≡ 0 =⇒ 0 + 9c ≡ 0 mod 10.

mod 10. mod 10.

So, the digit c needed to make 9c ≡ 0 mod 10 is 0. So, in the first code, the check digit is 0. In the second code, we need, 7 · 0 + 3 · 2 + 9 · 5 + 7 · 7 + 3 · 1 + 9 · 1 + 7 · 0 + 3 · 8 + 9c ≡ 0 mod 10. =⇒ 0 + 6 + 45 + 49 + 3 + 9 + 0 + 24 + 9c ≡ 0 mod 10. =⇒ 6 + 9c ≡ 0 mod 10. Now, the digit c needed to make 6 + 9c ≡ 0 mod 10 is 6. So, in the second code, the check digit is 6. (4) Show that the following statements are true: (a) 547 ≡ 6 × 17 mod 23. Solution : By Fermat’s Little Theorem, 523−1 ≡ 1 mod 23. That is, 522 ≡ 1 mod 23. We can now square both sides to get, 544 ≡ 1 mod 23. Multiplying both sides by 53 , we get, 544+3 ≡ 53 mod 23. That is, 547 ≡ 125 mod 23. Now, 125 = 115 + 10 = 3 × 23 + 10. So, 125 mod 23 is 10. Therefore, 547 mod 23 = 10.

SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

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Next, 6 × 17 = 102 = 92 + 10 = 23 × 4 + 10 and so, 6 × 17 mod 23 = 10. Thus, 547 ≡ 6 × 17 mod 23. (b) 42 × 33 ≡ 251 mod 13. Solution : By Fermat’s Little Theorem, 213−1 ≡ 1 mod 13, that is, 212 ≡ 1 mod 13. Taking a power of 4, 248 ≡ 1 mod 13. So, multiplying by 23 = 8 on both sides, 251 ≡ 8 mod 13. That is, 251 mod 13 = 8. Next, 42 × 33 mod 13 = 16 × 33 mod 13 = (13 + 3) × (26 + 7) mod 13 = 3 × 7 mod 13 = 21 mod 13 = 8. Thus, 42 × 33 ≡ 251 mod 13. (5) Carefully look at the following mathematical reasoning. Write down what is wrong with it. I believe that 6832 ≡ 1 mod 17. This is because, Fermat’s theorem says that if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. Solution : Fermat’s little theorem says that if p is prime and a is an integer N OT DIV ISIBLEBY p, then, ap−1 ≡ 1 mod p. Although, 17 is prime here, 68 is indeed divisible by 17 (17 × 4 = 68), and so Fermat’s Little Theorem does not apply here. The correct answer should be 6832 mod 17 = 0. This is because, as 17 divides 68, 17 will divide any power of 68, and so, 17 divides 6832 . (6) Prove the following statements using the principle of induction: (a) 1 + 3 + 5 + ... + (2n − 1) = n2 , for all natural numbers n. Solution : The statement S(n) is “1 + 3 + 5 + ... + (2n − 1) = n2 ”. Step 1: We check if this is true for n = 1. The left hand side is just 1, and the right hand side is 12 and hence is 1. So, S(1) is true. Step 2: We assume S(m) is true. That is, we assume, 1+3+...+(2m−1) = m2 . Step 3: Now we need to prove that S(m + 1) is true. That is, we need

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SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

to prove that 1 + 3 + ... + (2(m + 1) − 1) = (m + 1)2 . The left hand side is 1+3+...+(2m+1) = 1+3+5+...+(2m−1)+(2m+1) = m2 +2m+1 (from Step 2). But, m2 +2m+1 = (m+1)2 . Therefore, 1+3+...+(2(m+1)−1) = (m+1)2 , which completes our induction steps. Hence, by induction, the statement 1 + 3 + 5 + ... + (2n − 1) = n2 is true for all natural numbers n. (b) 32n−1 + 2n+1 is divisible by 7, for all natural numbers n. Solution : The statement S(n) is “32n−1 + 2n+1 is divisible by 7”. Step 1: We check if this is true for n = 1. 32×1−1 + 21+1 = 3 + 22 = 3 + 4 = 7, which of course is divisible by 7. So, S(1) is true. Step 2: We assume S(m) is true. That is, we assume, 32m−1 + 2m+1 is divisible by 7. So, there is an integer k for which, 32m−1 + 2m+1 = 7k. Then, 2m+1 = 7k − 32m−1 . Step 3: Now we need to prove that S(m + 1) is true. That is, we need to prove that 32(m+1)−1 + 2(m+1)+1 is divisible by 7. Now, 32(m+1)−1 + 2(m+1)+1 = 32m+1 + 2m+1 · 2 = 32m+1 + (7k − 32m−1 ) · 2, (from Step 2) = 14k + 32m+1 − 2 · 32m−1 , = 14k + 32m−1+2 − 2 · 32m−1 , = 14k + 32m−1 (32 − 2) = 14k + 32m−1 · 7, which, of course, is divisible by 7. This completes our induction steps. Hence, by induction, 32n−1 + 2n+1 is divisible by 7, for all natural numbers n. (c) Let Fn denote the nth Fibonacci number (so, F1 = 1, F2 = 1, F3 = 2, F4 = 3 and so on). Show that F12 + F22 + ... + Fn2 = Fn · Fn+1 , for all natural numbers n. Solution : The statement S(n) is “F12 + F22 + ... + Fn2 = Fn · Fn+1 ”. Step 1: We check if this is true for n = 1. The left hand side F12 , which is just 1, and the right hand side is F1 · F2 = 1 · 1 = 1. So, S(1) is true.

SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

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2 = Step 2: We assume S(m) is true. That is, we assume, F12 + F22 + ... + Fm Fm · Fm+1 .

Step 3: Now we need to prove that S(m + 1) is true. That is, we need 2 to prove that F12 + F22 + ... + Fm+1 = Fm+1 · Fm+2 . The left hand side is 2 2 2 2 2 F1 + F2 + ... + Fm + Fm+1 = Fm · Fm+1 + Fm+1 = Fm+1 (Fm + Fm+1 ). Now, because Fn is the nth Fibonacci number, Fm + Fm+1 = Fm+2 . Therefore, 2 F12 + F22 + ... + Fm+1 = Fm+1 · Fm+2 , which completes our induction steps. Hence, by induction, the statement F12 + F22 + ... + Fn2 = Fn · Fn+1 is true for all natural numbers n. (7) Prove the following statements using the method of contradiction: (a) The negative of an irrational number is irrational. Solution : Let x be an irrational number. Assume that −x is rational. So, we can write −x as pq where p, q are integers with q 6= 0. Thus, −x =

p q

=⇒ x = −

p −p = . q q

Since −p and q are still integers, with q 6= 0, x is a rational number, which is a contradiction. Hence, the negative of an irrational number is irrational. (b) There are no even primes that are bigger than 2. Solution : Suppose there are even primes bigger than 2. Let p be one such prime. Since p is even, this means that we can write p = 2k for some natural number k > 1. But, this implies that both 2 and k divide p, which is a contradiction, since p is prime. Hence, there are no even primes that are bigger than 2. (c)



6 is irrational. √ Solution : Assume 6 is rational. Then we can write, √ p 6= , q where p and q are integers that have no common factors, and q 6= 0. Squaring both sides, p2 , q2 6q 2 = p2 . 6=

6

SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

Thus, 6 divides p2 . This means that both 2 and 3 divide p2 . Now, from what we did in class, 2 divides p2 means 2 divides p and 3 divides p2 means 3 divides p. That is, both 2 and 3 divide p, which gives us that 6 divides p. So, we can write p = 6k, for some k. Then, plugging this in the previous equation we get, 6q 2 = (6k)2 = 36k 2 , =⇒ q 2 = 6k 2 .

So, 6 divides q 2 . By the same reasoning as before, this means that 6 divides q. Therefore √ p and q have a common factor 6, which is a contradiction. Therefore, 6 is irrational. (d)

√ 2 + 3 is irrational. √ √ Solution : Assume 2 + 3 is rational. Then we can write,





2+



p 3= , q

where p and q are integers that have no common factors, and q 6= 0. Squaring both sides, √ √ √ √ p2 ( 2 + 3 · ( 2 + 3) = 2 , q √ p2 =⇒ 2 + 3 + 2 6 = 2 , q √ 2 2 =⇒ 5q + 2q 6 = p2 , √ p2 − 5q 2 =⇒ 6 = . 2q 2 The right hand side above is clearly a rational number, with the numerator and denominator both being integers and √ the denominator being non-zero. Therefore, the above expression shows that 6 is a rational number, which is a contradiction by what we proved in part (c) of this problem. (8) Let A and B be two sets. Draw Venn diagrams and shade the regions for the following set expressions:

SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012)

(a) (A − B)c . Solution : The shaded area is the required region.

(b) (A4B) − A. Solution : The area shaded in dark grey is the required region.

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