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I just started to learn how to do proofs, and I'm trying to solve this exercise, I have to do it by direct proof and by contradiction. The proof of my book is:

Direct proof:
Let $x,y \in \mathbb{R}^+$ such that $x \leq y$. Multiplying both sides by $x$ and $y$ respectively, we obtain $x^2 \leq xy$ and $xy \leq y^2$. Therefore $x^2 \leq xy \leq y^2$ and so $x^2 \leq y^2$

Proof by contradiction
Assume, to the contrary, that there exist positive numbers $x$ and $y$ such that $x \leq y$ and $x^2>y^2$. Since $x \leq y$, it follows that $x^2 \leq xy$ and $xy \leq y^2$. Thus, $x^2 \leq y^2$ producing a contradiction.

My proofs are not like this though, I did them in this way:

Direct proof:
Let $x,y \in \mathbb{R}^+$ such that $x \leq y$. Then $(x-y)(x+y)\leq0$, which means that $x^2-y^2\leq0$, and therefore $x^2 \leq y^2$

Proof by contradiction (exact same thing):
Assume $x\leq y$ and $x^2>y^2$. Then $(x-y)(x+y)\leq0$, which means $x^2-y^2 \leq0$. Therefore $x^2\leq y^2$ producing a contradiction.

Do my proofs make sense even though I did them differently from the book?

pipey
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  • Yes, that's a perfectly valid proof as well! Of course there's always the question of how much detail to include (personally I might say separately that $x-y\le0$ and $x+y\ge0$, so that the inequality $(x-y)(x+y)\le0$ doesn't seem as sudden). – Greg Martin Oct 04 '20 at 20:29
  • Well, your proofs are certainly okay, but I would consider the second one to be a bit cheat-y. Any direct proof of the implication $X \to Y$ can be turned into a proof by contradiction by going "Assume $X$ but suppose $Y$ is wrong, (insert direct proof of $X \to Y$), hence $Y$ must be right, contradiction to $Y$ being wrong. – Watercrystal Oct 04 '20 at 20:29
  • @Watercrystal oh yeah you're right! i've never thought about it, thank you! – pipey Oct 04 '20 at 20:32
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    However, in this case I also find it hard to give a proof by contradiction which does not essentially work like this, it was more of a general remark. But you will get a feeling for whether or not a proof feels right in that sense or not soon :) – Watercrystal Oct 04 '20 at 20:36
  • Which book are you referring to? – Shaun Oct 04 '20 at 20:40
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    @Shaun Mathematical proofs a transition to advanced mathematics – pipey Oct 04 '20 at 21:38

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