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Suppose $a$ and $b$ are real numbers. Prove that if $a < b < 0$ then $a^2 > b^2$.

My attempt:

  1. We know that if $x > y$, then $-x < -y$.

  2. We also know that if $x > y\ge 0$, then $x^2 > y^2$.

Now consider our example:

Given that $a < b < 0$

$$\tag1 a<b$$

$$\tag2 -a > -b$$

$$\tag3(-a)^2 > (-b)^2$$

$$\tag4 a^2 > b^2$$

Is it correct? Any suggestions for improvement would be welcome.

José Carlos Santos
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Stokolos Ilya
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2 Answers2

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The idea is fine, but look again at what you wrote; it is just a sequence of assertions, with no links between them. I suggest that you type it as follows:\begin{align}a<b&\implies-a>-b\\&\implies(-a)^2>(-b)^2\text{ (since $-a>-b\geqslant0$)}\\&\iff a^2>b^2.\end{align} Another possibility consists in noting that$$a^2-b^2=\overbrace{(a-b)}^{\phantom{0}<0}\,\overbrace{(a+b)}^{\phantom{0}<0}>0.$$

José Carlos Santos
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    Do you need to explicitly show that there is a link between them? I've numbered each step, assuming that it is evident that each step follows from the previous one. – Stokolos Ilya Jul 17 '19 at 12:54
  • Either you do show explicitely a link between them, as I did, or you assert explicitely that each assertion follows from the previous one. – José Carlos Santos Jul 17 '19 at 12:56
  • If I am to choose the second option, how am I supposed to do it properly? – Stokolos Ilya Jul 17 '19 at 12:56
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    One possibility would be eliminate the sentence “Given that $a<b<0$” and to add, after your sequence of assertions “Since we are assuming that $(1)$ holds and since $(1)\implies(2)\implies(3)\implies(4)$, this proves that $a^2>b^2$”. – José Carlos Santos Jul 17 '19 at 13:01
  • One more question: $ (1)⟺ (2) ⟺(3) ⟺(4)$ Would that be correct in our scenario? – Stokolos Ilya Jul 17 '19 at 13:07
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    Yes. I often tell to my students to simply write $\implies$ instead of $\iff$ whenever $\implies$ is enough, in order to decrease the possibility of an oversight. But it is correct in this case. – José Carlos Santos Jul 17 '19 at 13:09
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Your solution is right!

$f(x)=x^2$ increases on $[0,+\infty).$

By your work $-a>-b>0.$

Thus, indeed $(-a)^2>(-b)^2$ and we are done!

I think, it's better to use that $f$ decreases on $(-\infty,0].$

From this we'll obtain the statement immediately.

Michael Rozenberg
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