# If 1 $\leq x$, then $\sqrt{x} \leq x$

This is a really simple problem but I am unsure if I have proved it properly.

Suppose that $x \geq 1$ and $x< \sqrt{x}$. Then $x\cdot x \geq x \cdot 1$ and $x^2 < x$ (squaring both sides), which is a contradiction.

-
I think you're only missing a note that $x\cdot x\ge x\cdot 1\to x^2\ge x$ contradicts the squaring of $x^2\lt (\sqrt x)^2\to x^2\lt x$... –  abiessu Jan 29 '14 at 20:15
So you're saying that it would be better to write out that step? Ok –  grayQuant Jan 29 '14 at 20:17
Yes; note that $x^2\gt 1$ does not necessarily contradict $x^2\lt x$, and is not necessarily guaranteed by the given conditions. –  abiessu Jan 29 '14 at 20:18
The first time I squared both expressions, I changed it to $x\cdot x$ for the first one then squaring the second. Now I have a valid proof? –  grayQuant Jan 29 '14 at 20:21
Close, just one more note that you extend $x\ge 1$ to $x^2\gt x$ (note the change from "greater than or equal" to "greater than") which is not necessarily true. –  abiessu Jan 29 '14 at 20:24

Here is a direct proof (without contradiction)

$x=1+r$ with $r\geq0$ . Then $x^2=(1+r)^2=1+2r+r^2\geq1+r=x\rightarrow x^2\geq x\rightarrow x\geq\sqrt x$

-
Nice thanks, just curious did you have a reference for this or just solved it. I would like to look at the reference if you used any –  grayQuant Jan 29 '14 at 20:23
Sorry, I don't have a refference, I just happened to think about this in the shower the other day. but transforming the variable can be useful sometimes. –  Gamamal Jan 29 '14 at 20:25
This is a neat way to do it. I've found that I sometimes rely on contradiction too much so seeing direct proofs of things I would otherwise use contradiction on is very refreshing. –  Cameron Williams Jan 29 '14 at 21:13

Assume $x \geq 1.$ Then $x - 1 \geq 0$ and $x > 0,$ and hence

$$x(x-1) \; = \; x^2 - x \; \geq \; 0$$

since the product of two non-nonegative expressions is non-negative.

Factoring $x^2 - x$ as a difference of squares gives

$$\left(x - \sqrt{x} \right) \left( x + \sqrt{x} \right) \; \geq \; 0$$

Since $x + \sqrt{x}$ is positive (both $x$ and $\sqrt{x}$ are positive), it follows that $x - \sqrt{x}$ is non-negative, which is easily seen to be equivalent to what you wanted to prove.

-
Neat :). But, the $0$ answer is neater :D –  Dror Jan 29 '14 at 21:21

As a contra-positive, assuming $x$ is not negative: $$\sqrt{x} > x \implies x>x^2 \land x \neq0 \implies 1>x$$ (1st implication is by squaring (Which is obviously an increasing function here), 2nd implication is by dividing by $x$)

So, equivalently: $$x \geq 1 \implies x \geq\sqrt{x}$$

Alternatively, for a more direct proof, you can rely on $\sqrt{\cdot}$ being an increasing function in $\mathbb R_{\ge 0}$ domain: $$x \geq 1 \implies x^2 \geq x \implies x \geq\sqrt{x}$$ (1st implication is by multiplying by $x$, 2nd implication is by taking the square root, relying on it being increasing)

-