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

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    Related: [Showing $a^20](http://math.stack$e$xchange.com/questions/52877/showing-a2-b2-if-0-a-b)2012-03-02

7 Answers 7

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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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    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.2011-09-24
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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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    @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)2014-07-11
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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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    One definition of |a| is $|a|=\sqrt{a^2}$. As the square root is monotonic this statement is natural.2013-12-18
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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!