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I was doing some geometric series stuff today and was thinking about how numbers that are in between $-1$ and $1$ are always ‘farther from $0$’ than their squared equivalents, but how is that proven? Is it just taken as something basic? I mean, obviously any number multiplied by a number who’s absolute value is less than one will be smaller than before, but is there anything more formal/rigorous to explain that?

I guess you could say that for a number $-1<\frac{a}{b}<1$ that $|a|<|b|$ and so $|b^2-b|$ is greater than $|a^2-a|$, and so there is an even bigger different between $b^2$ and $a^2$, therefore $|\frac{a^2}{b^2}|$ must be smaller than $|\frac{a}{b}|$. Is that valid?

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  • $\begingroup$ Multiply both sides by $|x|$ assuming $|x|>0$. Then mention what happens at $x=0$ $\endgroup$ – David Peterson Oct 7 '18 at 0:30
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Your idea is right. In formulas, if $x = 0$ it is trivial. Otherwise, take $|x| < 1$ and multiply both sides by $|x|$ to get $|x|^2 < |x|$.

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If $0<a<1,$ $$ a^{n+1}-a^n=a^n(a-1)<0$$

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To make your proof look nicer, we can do something like this. Let $x$ be a real number so that $x\in[0,1)$. Then we can write $x=1-y$ for $y\in(0,1]$. Thus, $$ x^2 = 1-2y+y^2 $$ And $$ x-x^2 = 2-3y+y^2 = (y-1)(y-3)>0 $$ when $y\in(0,1]$. Now, try proving the case when $x\in(-1,0)$ with the same idea; the only difference that there are now some negative signs around. After that, you can even go on to prove this for complex numbers.

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Consider $f(x) = |x| - |x|^2 = |x|(1-|x|)$

Now $f(x) \geq 0 \iff |x|(1-|x|) \geq 0 \underset{\mathrm{since} |x| \geq 0}\iff 1-|x| \geq 0 \iff |x| \leq 1$

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