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The book first introduces a property: if a > b > 0, then a^2 > b^2

By assuming that if a > b > 0, then a^2 > b^2 is true. This is because both a,b>0 making aa>bb true as no negative is introduced (inequality sign does not flip).

The book then introduces another property: if a < b < 0 then a^2>b^2

I tried to approach this with a < b < 0 by first finding a,b<0 meaning that if I squared both sides, I would get a*-|a|< b*-|b|, the sign would have flipped twice as both sides were multiplied by a negative number, making it return to normal so I thought a^2 < b^2.

Although I know if examples were given/by common sense that the property is true, I just wondered why the sign flipped only once rather than twice when 2 negative were multiplied in the equation. Could you guys help me prove the unique concept of squaring both sides of inequalities? Do I lack basic understanding towards the concept of inequalities?

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4 Answers 4

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You've already proved:$\;\;(*)\;\;$If $a > b > 0$, then $a^2>b^2$.

Now you want to prove:$\;\;\;\;\;\;\;\;$If $a < b < 0$, then $a^2>b^2$.

Here's one way . . . \begin{align*} &a < b < 0\\[4pt] \implies\;&-a > - b > 0&&\text{[multiplying by $-1$ reverses}\\[-1pt] &&&\;\text{the directions of the inequalities]}\\[4pt] \implies\;&(-a)^2 > (-b)^2&&\text{[from the previous line, using $(*)$]}\\[4pt] \implies\;&a^2 > b^2\\[4pt] \end{align*} as was to be shown.

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  • $\begingroup$ Just a question: By squaring inequalities such as (-a)^2, how can you assume and prove that the sign will not change? $\endgroup$
    – nabu1227
    Aug 24, 2017 at 15:27
  • $\begingroup$ If $a$ is negative, then $-a$ is positive. Didn't you already prove that if $x > y > 0$, then $x^2 > y^2$? $\endgroup$
    – quasi
    Aug 24, 2017 at 15:31
  • $\begingroup$ oh right sorry about that $\endgroup$
    – nabu1227
    Aug 24, 2017 at 15:38
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HINT: $$a^2>b^2$$ is equivalent to $$(a-b)(a+b)>0$$

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  • $\begingroup$ After all, does that mean to solve inequalities it is best to move everything to LHS? $\endgroup$
    – nabu1227
    Aug 24, 2017 at 15:08
  • $\begingroup$ in most of the cases you can do this $\endgroup$ Aug 24, 2017 at 15:08
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hint:

if $a<b<0$ then $|a|>|b|$ and since negative times negative is positive ( both a and b are negative in this version) you get $|a|^2>|b|^2\rightarrow a^2>b^2$

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If $a<b<0$, then you obtain first, multiplying by $a$ and by $b$ (which are negative): $$a^2>ba>0\qquad\text{and}\qquad ba>b^2>0.$$ Then, by transitivity $\;a^2>b^2>0$.

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