How to prove that $\sqrt{x} > x$ for any $0 < x < 1$

Question

I want to prove that, for any x between $0$ and $1$, the $\sqrt{x}$ is bigger than $x$. Maybe it's easy and I'm not getting the point of the proof, but I tried to prove it a lots of times and I never get to something that makes it look proved.

Answer 1

If $x<0,\sqrt x$ is imaginary. So, $\sqrt x,x$ are not comparable.

See

Can a complex number ever be considered 'bigger' or 'smaller' than a real number, or vice versa?

OR

Comparing complex numbers

OR

this

So $x\ge0,$

We have $\sqrt x> x\iff\sqrt x(\sqrt x-1)<0$

As $\sqrt x\ge0,$

if $\sqrt x=0,x=0$

else $\sqrt x-1<0\iff \sqrt x<1\iff x<1$

Answer 2

As $0<x<1$ $\sqrt{x}>0$ Now devide the inequality by $\sqrt{x}$ (note that it is positive!). You will get $1>\sqrt{x}$ which is true as $x<1$.

Answer 3

We want to show for $x\in(0,1)$ that $$ \sqrt{x}>x \text{ but this is equivalent to } x>x^2 $$ well, and this is again equivalent - since we have $x>0$ - to $$ x>x^2>0\Leftrightarrow (\text{deviding by } x):1>x>0 $$ which is a true statement.

Answer 4

Let $\sqrt x=y\implies x=y^2$

We need $y>y^2\iff y(y-1)<0$

If $y<0,$ we need $y-1>0\iff 0>y>1$ which is impossible

If $y>0,$ we need $y-1<0\iff y<1\implies 0<y<1$