Exercise 2: Proving "n^2 Even Implies n Even" by Contrapositive — Possible Solution
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IDENTIFYING THE CONTRAPOSITIVE OF THIS SPECIFIC STATEMENT
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The claim is "if n^2 is even, then n is even." Written as P -> Q, here
P = "n^2 is even" and Q = "n is even." Per this chapter, the
contrapositive of P -> Q is ~Q -> ~P - negate both, and swap their
order.

~Q = "n is NOT even" = "n is odd."
~P = "n^2 is NOT even" = "n^2 is odd."

So the contrapositive of "n^2 even -> n even" is:

  "if n is odd, then n^2 is odd"

THIS IS EXACTLY EXERCISE 1's OWN RESULT
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Exercise 1 already proved, directly, that "if n is odd, then n^2 is
odd" - which is precisely the contrapositive statement identified
above. No new proof work is needed at all.

WHY PROVING THE CONTRAPOSITIVE IS ENOUGH
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Per this chapter, "p -> q is logically equivalent to ~q -> ~p ... so
proving one genuinely proves the other." Since Exercise 1 already
proved the contrapositive (n odd -> n^2 odd) directly, and that
contrapositive is logically equivalent to the original claim (n^2
even -> n even), the original claim is proven too - with zero
additional algebraic work required.

WHY THIS SPECIFIC CLAIM IS A GOOD CANDIDATE FOR CONTRAPOSITIVE PROOF
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A direct proof of "n^2 even -> n even" would require starting from
"n^2 = 2m for some integer m" and somehow extracting a specific factor
of 2 out of n itself - genuinely awkward algebra, since taking a
square root inside a proof isn't a clean integer operation. The
contrapositive direction, by contrast, starts from the clean, concrete
fact "n = 2k+1" and simply expands a square - exactly the easy
direction Exercise 1 already worked through.

WHY THIS WORKS AS AN ANSWER
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It derives the contrapositive of this specific statement carefully
(negating and swapping both P and Q), recognizes it as identical to
Exercise 1's own already-proven result, and explains both why citing
that result is a complete proof and why the contrapositive direction
was the easier one to prove in the first place.