Prove that if $a$ is a real number, and $0\lt a \lt 1$, then $\sqrt{a} \gt a$ I wanted to use this fact in a proof, it seems obvious, but I should probably prove this as well.  Any proof will do.  Also for the future is there a way to search for these proofs google does not work very well.
EDIT: Ah, it may also be good to show that the n-th root of any real number a s.t. $0\lt a\lt1$ is also $\lt1$
Thanks.
 A: If you know that whenever $a\lt b$ and $c\gt 0$ then $ac\lt bc$, then for any $x$ with $0\lt x\lt 1$, multiplying through by $x$ gives 
\begin{equation*}
0 \lt x^2 \lt x \lt 1.
\end{equation*}
(The last inequality is "inherited" from the fact that you already know that $x\lt 1$). Repeat to get
\begin{equation*}
0\lt x^3 \lt x^2 \lt x \lt 1.
\end{equation*}
And so on; you get $0\lt x^n \lt x^{n-1}\lt\cdots \lt x \lt 1$.
Working the other way, if $1\lt y$, then multiplying through by $y$ you get 
\begin{equation*}
1\lt y\lt y^2
\end{equation*}
(with the first inequality "inherited from your original assumption) and repeating this process leads to
\begin{equation*}
1\lt y\lt y^2 \lt\cdots \lt y^n
\end{equation*}
So: if $0\lt z\lt 1$, then $0\lt z^n\lt 1$. If $1\lt z$, then $1\lt z^n$. And of course, if $z=1$ then $z^n = 1$.
So, start with $0\lt a\lt 1$. Then $\sqrt[n]{a}$ cannot be greater than $1$ (then $a = \left(\sqrt[n]{a}\right)^n \gt 1$ by the above), so $\sqrt[n]{a}\lt 1$. 
Now replace $x$ with $\sqrt[n]{a}$ to get the desired inequality.
A: HINT $\rm \ \sqrt a - a \ =\ \sqrt a\ \:(1 - \sqrt a)\ > 0\ $ since both factors are $> 0$  on the interval $(0,1)$.
A: We argue by contradiction. Suppose $a \in (0,1)$ and $\sqrt{a} \leq a$. Then 
\begin{eqnarray}
\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a} \leq 1, 
\end{eqnarray}
which implies that $\sqrt{a} \geq 1$ and therefore $a \geq 1$, which is absurd by construction.
A: Depending on the level of the audience, it might be seen as obvious.  I would just argue that $\sqrt{x}$ is monotonic and $\gt 0$ on $(0,1)$, so $\sqrt{a}\lt1$ then multiply both sides by $\sqrt{a}$
A: In a simply way
we have $a<1$.
Multiply both side by $a$ 
then we get $a^2<a$
Taking the square root for both sides 
we get $a < \sqrt{a}$.
