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Given $a > b$, where $a,b ∈ ℝ$, is it always true that $a^2 > b^2$?

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8 Answers 8

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If $\: \color{#c00}{a > b}\: $ then $\: a^2\! -\! b^2 = (\color{#c00}{a\!-\!b})(a\!+\!b) > 0 \iff a\!+\!b >0 $

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    $\begingroup$ Now that is neat! $\endgroup$
    – anon
    Oct 30, 2010 at 16:26
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    $\begingroup$ -which can also be taken as an allusion to a certain scenario involving the First Law of Thermodynamics, in which the change in energy of the system is positive, and a and b are associated with the work and heat, not necessarily respectively, of the process. $\endgroup$
    – Mike Jones
    Sep 25, 2011 at 5:11
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    $\begingroup$ Bill, this has left quite an impression on me, apparently! See this recent question $\endgroup$ Mar 21, 2012 at 16:31
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    $\begingroup$ @TheChaz I'm not sure on whose shoulders we stand here, but, alas, if only they were broader, so that there might have been room for the OP to enjoy the view too! Thanks for the shoulder company. $\endgroup$ Mar 21, 2012 at 20:19
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    $\begingroup$ @Julian Equivalently it states $\rm\ (a-b)(a+b) \le 0 \iff a+b \le 0\ $ when $\,a>b,\,$ which is true when $\ b < -a < 0\,\ (\iff a+b < 0 < a)$ $\endgroup$ Jul 11, 2014 at 13:20
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no its not. When $a,b$ are positive it happens. Consider $a=-2$ and $b =-3$. notice that inequality reverses.

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The correct statement is,$$|a|>|b|\iff a^2 > b^2 $$A counterexample of your hypothesis is $a = 7, b = -8.$

Yes, $a >b $, but $b^2 > a^2$, i.e.:$$ (-8)^2 > 7^2\\64 > 49$$

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    $\begingroup$ One definition of |a| is $|a|=\sqrt{a^2}$. As the square root is monotonic this statement is natural. $\endgroup$
    – Squirtle
    Dec 18, 2013 at 22:36
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If $a > b > 0$ then $a^2 > b^2$.

$a > b$ means there is a positive $k$ such that $a = b + k$. Squaring this equation we have $a^2 = b^2 + (2bk + k^2)$ but $2bk + k^2$ is just another positive so $a^2 > b^2$.

The reason we know $2bk + k^2$ is positive is because of the ordered field axioms, one says if $x$ and $y$ are positive so is $xy$ and another says that $x+y$ is positive. That is why we need $b$ to be positive.

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The conclusion is correct on $[0, +\infty)$ because that is precisely the interval over which the function $f(x) = x^2$ is an increasing function. No algebra is necessary. Draw the parabola and LOOK!

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Yes, when $a$ and $b$ are positive real numbers. In this case, we can write:

$a>b \implies a-b>0 \implies (a+b)(a-b)>0 \implies (a^2)-(b^2)>0$

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    $\begingroup$ This leaves the false impression, for the layman, that it is a necessary condition that a and b be positive for the inequality to occur, but I will refrain from down-voting your answer. The perfect answer, above, was given by Bill Dubuque. $\endgroup$
    – Mike Jones
    Sep 24, 2011 at 22:01
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Given that $ a > b $, it is not always true that $ a^{2} > b^{2} $. One counterexample would be that one of $ a $ or $ b $ is negative, say $ a = 1 $, and $ b = -1 $. Then $$ a^2 = 1^{2} = 1 $$ and $$ b^{2} = (-1)^{2} = 1 $$ making $ a^2 = b^2 $, a contradiction of our assumption that $ a^{2} > b^{2} $.

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This answer is the same as Bill Dubuque's, but goes into a bit more detail. The question

Does $a>b$ imply that $a^2>b^2$?

is equivalent to

Does $a-b>0$ imply that $a^2-b^2>0$?

In other words, the answer to the bottom question is the same as the answer to the top question.

By factorising $a^2-b^2$, we see that no, $a-b>0$ does not imply that $a^2-b^2>0$: $$ a^2-b^2=(a+b)(a-b) > 0 \text{ if, and only if, $a+b>0$} \, . $$ We know that $a-b>0$, but that is not enough. We also require that $a+b>0$ so that $(a+b)(a-b)>0$. If we instead ask

Does $a>b$ and $a>-b$ imply that $a^2>b^2?$

then the answer is in the affirmative.

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