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Section 5.3 Practice Problems for Chapter 5

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Exercises

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1.

Consider the following proposition:

Proposition. For each integer a, if 3 divides a2, then 3 divides a.

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(a)

Write the contrapositive of this proposition.

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(b)

Prove the proposition by proving its contrapositive.

Hint.

Consider using cases based on the Division Algorithm using the remainder for “division by 3.” There will be two cases since the hypothesis of the contrapositive is, “3 does not divide a.”

Solution.
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Proof.

We will prove the contrapositive of this proposition, which is:

For each integer a, if 3 does not divide a, then 3 does not divide a2
So we let a be an integer, assume that 3 does not divide a, and will prove that 3 does not divide a2. Since 3 does not divide a, we can use the Division Algorithm to conclude that there exists an integer q such that a=3q+1 or a=3q+2.

For the case where a=3q+1, we obtain

a2=(3q+1)2=9q2+6q+1=3(3q2+2q)+1

By the closure properties of the integers (3q2+2q) is an integer, and so the last equation means that a2 has a remainder of 1 when divided by 3 and so 3 does not divide a2.

For the case where a=3q+2, we obtain

a2=(3q+2)2=9q2+12q+4=3(3q2+4q+1)+1

By the closure properties of the integers (3q2+4q+1) is an integer, and so the last equation means that a2 has a remainder of 1 when divided by 3 and so 3 does not divide a2.

Since we have proved that 3 does not divide a2 in both cases, we have proved the contrapositive of the proposition, and hence, we have proved that for each integer a, if 3 divides a2 , then 3 divides a.

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2.

Complete the details for the proof of Case 3 of Proposition 5.2.

Solution.

For the third case, n=3q+2 and r=2. When we substitute this into (n3−n), we obtain

n3−n=(3q+2)3−(3q+2)=(27q3+54q2+36q+8)−(3q+2)=27q3+54q2+33q+6=3(9q3+18q2+11q+2).

Since 9q3+18q2+11q+2 is an integer, the last equation proves that 3 | (n3−n).

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3.

Is the following proposition true or false? Justify your conclusion with a counterexample or proof.

For each integer n, if n is odd, then 8 divides n2−1.

Solution.
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Proof.

We let n be an integer, assume that n is odd, and will prove that 8 divides n 21. Since n is odd, there exists an integer k such that n=2k+1. We then see that

n2−1=(2k+1)2−1=4k2+4k(5)=4k(k+1)

We also know since k is an integer, either k or k+1 is even. In either case, the product k(k+1) must be even and so there exists an integer q such that

k(k+1)=2q

Substituting this into the right side of equation (5), we obtain n2−1=8q and so 8 divides n2−1. This proves that for each integer n, if n is odd, then 8 divides n2−1.