If true prove it If false give a counterexample If ab is odd

If true, prove it. If false, give a counterexample. If ab is odd, then a and b are odd. If n^2 is even, then n is even

Solution

Solution : ( a )

This is a proof by contradiction which means that we have to assume the opposite consequent, which is the say that a and b are not both odd. So either a is even or b is even.

If a is even, then a = 2k for some integer k, and b*2k will always be even.

If b is even, then b=2m for some integer m, and a*2m will always be even.

So either case leads to a contradiction that a*b is odd, and therefore if a*b is odd, then both a and b are odd.

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Solution : ( b )

If n² is even by the definition.

Since it is 2 times 2k², which is an integer

It follows that n² = 4k² = 2(2k²).

Thus, n = 2k for some k Z .

Hence, n is even.