Question 1 of 5
Which proof correctly proves n² is even when n is even?
If n is even, n = 2k. Then n² = 2k². Since 2k² is divisible by 2, n² is even.
If n is even, n = 2k for some integer k. Then n² = (2k)² = 4k² = 2(2k²). Since 2k² is an integer, n² = 2m where m = 2k². Therefore n² is even.
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