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I'm reading the book 'How to prove it' from Daniel Velleman which he presents a proof for the following; if $0 < a < b$ then $a^2 < b^2$ as;

Proof. Suppose $0 < a < b$. Multiplying the inequality $a < b$ by the positive number $a$ we can conclude that $a^2 < ab$, and similarly multiplying by $b$ we get $ab < b^2$. Therfore $a^2 < ab < b^2$, as required. Thus if $0 < a < b$ then $a^2 < b^2$.

However, I was also wondering if the statement could be proved using the following method.

Proof. Suppose that $0 < a < b$. Taking the square root of both sides of the inequality $\sqrt{a^2} < \sqrt{b^2}$ we get our original hypothesis $a < b$ . Thus if $0 < a < b$ then $a^2 < b^2$.

Hanul Jeon
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redbandit
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1 Answers1

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In the proof from the book you presented, you've assumed $0<a<b$ and deduced $a^2<b^2$.

In your presented proof, you've essentially assumed $a^2<b^2$ and deduced $a<b$. Why? The hypothesis $0<a<b$ is never used and by taking the square root of $a^2<b^2$, you implicitly assume that statement, deducing $a<b$ from it.

Thus, conceptually you have shown $B\Rightarrow A$ for the corresponding statements $A,B$, i.e. you have established $A\Leftrightarrow B$ using $A\Rightarrow B$ from the previous proof.

EDIT: As discussed with Ennar in the comments, there are additionally some issues with deducing $a<b$ from $a^2<b^2$ as it requires the square root function to be monotone, a property that follows from the fact the the square function is monotone on $[0,\infty)$. Then of course, you can not establish $B\Rightarrow A$ without first establishing $A\Rightarrow B$, i.e. there is another circularity of reasoning.

As you can see, there are even more implicit assumptions than I had initially pointed out.

blub
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  • Actually, I'm not convinced that the proof of $a^2<b^2$ implies $a<b$ isn't circular. How does one know that square root is monotone without knowing that squaring is monotone (on positive numbers)? – Ennar Aug 29 '18 at 11:08
  • @Ennar This is another problem, yes. I just wanted to elaborate the concept assuming something implicitly and that the hypothesis is not just what you state it is(necessarily). – blub Aug 29 '18 at 11:11
  • I understand and it's a good answer, but your last row can be misleading to OP who is probably first time learning how to write proofs. – Ennar Aug 29 '18 at 11:14
  • @Ennar I think you're right, I'll add this in. – blub Aug 29 '18 at 11:18
  • @Ennar Feel free to also edit my answer if you have other improvements. I've tried to convey this point as good(and short) as possible. – blub Aug 29 '18 at 11:25
  • I think it's ok. Cheers. – Ennar Aug 29 '18 at 11:28