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I have attempted to produce a proof of the intuitive rule for squaring inequalities, according to which, given any two numbers x and y and regardless of their sign,

1) if |x| < |y| then $x^2<y^2$;

2) if |x| > |y| then $x^2>y^2$.

Proof for Case 1:

Let x and y being any two real numbers such that |x| < |y|.

  • Multiplying both sides by |x| gives |x||x| < |y||x|. Hence, by the result |a||b| = |ab|, we have |x|$^2$<|xy|;

  • Multiplying both sides by |y| gives |y||x| < |y||y|. Hence, by the same result, we get |xy|<|y|$^2$.

Now, merging the two inequalities we get |x|$^2$<|xy|<|y|$^2$, to give |x|$^2$<|y|$^2$. By the result |a|$^2$ = _a_$^2$ we get _x_$^2$<_y_$^2$. This completes the proof in the first case.

Proof for Case 2:

Let x and y being any two real numbers such that |x| > |y|.

  • Multiplying both sides by |x| gives |x||x| > |y||x|. Hence, by the result |a||b| = |ab|, we have |x|$^2$>|xy|;

  • Multiplying both sides by |y| gives |y||x| > |y||y|. Hence, by the same result, we get |xy|>|y|$^2$.

Merging the two inequalities we get |x|$^2$>|xy|>|y|$^2$, to give |x|$^2$>|y|$^2$. By the result |a|$^2$ = _a_$^2$ we get _x_$^2$>_y_$^2$. This completes the proof in the second case.

Substantially this proof (hopefully correct) follows the same path as the proof for squaring both sides of an inequality. The difference is that here we take into account only absolute values, which avoids us the tricky task of considering signs.

Any comment or suggestion to improve (or disprove!) this would be most welcome!

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    $\begingroup$ You don't have to prove case $2$: it's contained in case $1$ by exchanging the roles of $x$ and $y$. $\endgroup$
    – Bernard
    Jul 6, 2015 at 22:04
  • $\begingroup$ I've never seen such inefficient proof. $|x|<|y|$ means that $x^2=|x||x|<|y||x|<|y||y|=y^2$. That's it!. And you don't haev to prove 2) it is equivalent to 1) up to exchchange of symbols $x,y$. $\endgroup$ Jul 6, 2015 at 22:06
  • $\begingroup$ @AlexanderVigodner; I can't see any difference between your proof and the proof of the OP excep that the OP's proof is more wordy. Or do I miss something? $\endgroup$
    – miracle173
    Jul 6, 2015 at 22:28
  • $\begingroup$ @miracle173, well I am not merging any intermediate inequalities, but may be you are right, though for me it is really too many words and relations to prove the simple fact. $\endgroup$ Jul 6, 2015 at 22:35
  • $\begingroup$ Actually yes, probably too wordy and too obvious. Put this way however this may be a useful rule of thumb to decide the inequality sign after squaring simply by checking magnitudes. I have seen around many doubts about how doing this, so I hope this might provide a quick practical rule along with proof $\endgroup$ Jul 6, 2015 at 23:13

2 Answers 2

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Your proof seems correct, although a bit too complicated.

If you know that $x \mapsto x^2$ is a strictly increasing function on $\Bbb{R}_{\geq 0}$, then you immediately know that $|x| > |y|$ iff $x^2 > y^2$ for every $x,y \geq 0$ (by definition of strictly increasing). To prove that, the quickest way would be to observe that $\frac{d}{dx}(x^2) = 2x \geq 0$ for $x \geq 0$, with equality only for $x=0$.


To keep things elementary, you could instead observe that $|x| > |y|$ implies $$ |x||y| < |x||x| = x^2 \qquad \text{and} \qquad |x||y| > |y||y| = y^2 $$ Putting the two together gives $|x| > |y| \Rightarrow x^2 > y^2$. Reversing the roles of $x$ and $y$ gives (1), but note that the two cases are really just the same thing.

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Notice that both inequalities follow from proving that

$\;\;\;$If $0\le a<b,$ then $a^2<b^2$; and

$\;\;\;0\le a<b\implies a^2\le ab\;$ and $\;ab<b^2\implies a^2<b^2$.

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  • $\begingroup$ Absolutely correct! My aim however was to reflect about a general rule (perhaps obvious) to quickly determine the sign of the inequality after squaring along with proof of its truth. The proof given above considers only the case in which 'a' and 'b' are positive, while when considering magnitudes we are not concerned abut signs $\endgroup$ Jul 6, 2015 at 23:16

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