Question

In this problem, we will prove for all integers n, n is even if and only if n² is even. This requires proving two implications, if n is even then n² is even and if n² is even then n is even (using the logical equivalence p ? q ? (p ? q) ? (q ? p) ). a. Proof: If n is even then n² is even. (Direct Proof) Assume n is even. Then we may write n = 2k for some integer k. We may now compute n² which is equivalent (in terms of k) to . To show that n² is even, we will next factor out a 2 from n² so that n² = 2. . Since the last answer is a product of integers (which results in another integer), we see that n² is 2 times an integer and is therefore even. b. Proof: If n² is even then n is even. (Indirect Proof) The contrapositive, if n is odd then n² is odd, will be proven directly. Assume n is odd. Then we may write n = 2t + 1 for some integer t. We may now compute n² which is equivalent (in terms of t) to . To show that n² is odd, we will write n² as one more than an even number so that n² = 2. + 1. Since the last answer is a product or sum of integers (which results in another integer), we see that n² is 2 times an integer plus one and is therefore odd. c. Since both implications have been proven, the biconditional (if and only if) is also true.

          In this problem, we will prove for all integers n, n is even if and only if n² is even.
This requires proving two implications, if n is even then n² is even and if n² is even
then n is even (using the logical equivalence p ? q ? (p ? q) ? (q ? p) ).

a. Proof: If n is even then n² is even.
(Direct Proof) Assume n is even. Then we may write n = 2k for some integer
k. We may now compute n² which is equivalent (in terms of k) to . To
show that n² is even, we will next factor out a 2 from n² so that n² = 2.
. Since the last answer is a product of integers (which results in
another integer), we see that n² is 2 times an integer and is therefore even.

b. Proof: If n² is even then n is even.
(Indirect Proof) The contrapositive, if n is odd then n² is odd, will be proven
directly. Assume n is odd. Then we may write n = 2t + 1 for some integer
t. We may now compute n² which is equivalent (in terms of t) to
. To show that n² is odd, we will write n² as one more than
an even number so that
n² = 2. + 1. Since the last answer is a product or sum of
integers (which results in another integer), we see that n² is 2 times an
integer plus one and is therefore odd.

c. Since both implications have been proven, the biconditional (if and only if) is
also true.

In this problem, we will prove for all integers n, n is even if and only if n² is even.
This requires proving two implications, if n is even then n² is even and if n² is even
then n is even (using the logical equivalence p ? q ? (p ? q) ? (q ? p) ).

a. Proof: If n is even then n² is even.
(Direct Proof) Assume n is even. Then we may write n = 2k for some integer
k. We may now compute n² which is equivalent (in terms of k) to . To
show that n² is even, we will next factor out a 2 from n² so that n² = 2.
. Since the last answer is a product of integers (which results in
another integer), we see that n² is 2 times an integer and is therefore even.

b. Proof: If n² is even then n is even.
(Indirect Proof) The contrapositive, if n is odd then n² is odd, will be proven
directly. Assume n is odd. Then we may write n = 2t + 1 for some integer
t. We may now compute n² which is equivalent (in terms of t) to
. To show that n² is odd, we will write n² as one more than
an even number so that
n² = 2. + 1. Since the last answer is a product or sum of
integers (which results in another integer), we see that n² is 2 times an
integer plus one and is therefore odd.

c. Since both implications have been proven, the biconditional (if and only if) is
also true.

Added by Adam G.

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