How to prove square of a number between $0$ and $1$ is less than the number?
If $a\in \Bbb{R}$, prove that $0<a<1$ $\implies$ $0<a^2<a$ and $a>1 \implies a^2>a$.
I found some solution for rational $a$'s but I'm stuck with real $a$'s.
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$\begingroup$If you use axiomatic approach to real numbers then there is an axiom (or something equivalent to it) that
$\forall a, b \in \mathbb{R}~~$ and $~~c > 0~~$ if $~~a > b~~$ then $~~ac > bc$
This axiom is part of the axioms which specify what kind of ordering real numbers have. From it we deduce that if $~~0 < a < 1~~$ then $~~0 = 0*a < a*a = a^2~~$ and $~~ a^2 = a*a < a*1 = a~~$.
And if $~~ a > 1 ~~$ then $a^2 = a*a > 1*a = a$.
If you use constructive approach (Cauchy sequences, Dedekind cut, decimal representation etc.) then you can prove that the above property holds in your construction and then use it.
$\endgroup$ $\begingroup$If $x>0$ and $y>0$, then $xy>0$. If $,0<a<1$ then $0<1-a$ and thus $0<a(1-a)$. You conclude $0<a-a^2$ which implies $a^2<a$.
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