How can I prove that if $|x|<1$, then $|x| \geq |x|^2$?

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?

• Multiply both sides by $|x|$ assuming $|x|>0$. Then mention what happens at $x=0$ – David Peterson Oct 7 '18 at 0:30

Your idea is right. In formulas, if $$x = 0$$ it is trivial. Otherwise, take $$|x| < 1$$ and multiply both sides by $$|x|$$ to get $$|x|^2 < |x|$$.
If $$0 $$a^{n+1}-a^n=a^n(a-1)<0$$
To make your proof look nicer, we can do something like this. Let $$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)$$ with the same idea; the only difference that there are now some negative signs around. After that, you can even go on to prove this for complex numbers.
Consider $$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$$