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

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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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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 '10 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 '11 at 5:11
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    $\begingroup$ Bill, this has left quite an impression on me, apparently! See this recent question $\endgroup$ – The Chaz 2.0 Mar 21 '12 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. $\endgroup$ – Bill Dubuque Mar 21 '12 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 $\ b < -a < 0\,\ (\iff a+b < 0 < a)$ $\endgroup$ – Bill Dubuque Jul 11 '14 at 13:20
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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 '13 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 '11 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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