Prove $\sqrt{5} \leq 3$ I was just thinking about how to prove  $\sqrt{5} \leq 3$. I believe I can prove by contradiction by saying suppose  $\sqrt{5} > 3$, then it must be true also that $5 > 9$ by squaring both sides, but this is absurd so it must be that  $\sqrt{5} \leq 3$. However, I was thinking that this doesn't seem valid since we can suppose $\sqrt{5} > -3$ then squaring both sides gives $5 > 9$, which is also absurd! By we know  $\sqrt{5} > -3$, so is something I'm doing not valid? I know this is a simple thing, but I was just trying to prove an irrational number is less than a certain rational number and am stumped.
 A: Hint: $\;\;3-\sqrt{5} \,=\, \cfrac{4}{3+\sqrt{5}}$
A: Here is another way to look at this problem.
The square-root function, $f(x)=\sqrt x$, is increasing on its domain because $f'(x)>0$ on $(0, +\infty)$. Since $3=\sqrt 9$,
$3=f(9)$.
And $\sqrt 5=f(5)$.
But since $f(x)=\sqrt x$ is increasing on its domain, by the definition of increasing (and because $5<9$):
$f(5) <f(9)$. 
Substitute the meanings of $f(5)$ and $f(9)$ and get:
$\sqrt 5 <3$.
Q.E.D.
A: Squaring is not an equivalence transformation.
For example $x=3$ squared becomes $x^2=9$ which has solutions $\pm 3$ rather than $3$ alone.
But you can argue as follows
If you assume $\sqrt{5}>3$ , you can conclude $5=\sqrt{5}\cdot \sqrt{5}>\sqrt{5}\cdot 3>3\cdot 3=9$
because it is clear that $\sqrt{5}$ and $3$ are positive.
This way you get the desired contradiction.
A: Squaring both sides of an inequality does not preserve the inequality unless both sides were originally positive.
The long way: suppose $\sqrt{5} > 3$. Multiplying both sides by $3$, we have $3\sqrt{5} > 9$. Since $3 < \sqrt{5}$, $3\sqrt{5} < \sqrt{5}^2 = 5$. So $5 > 3\sqrt{5} > 9$, a contradiction.
A: For positive numbers the function $f(x)=x^2$ is an increasing function. This follows from computing the derivative.
A: $\sqrt5<\sqrt9=\sqrt{3^2}=3\;.$
Hence, $\;\sqrt5<3\;.$
Alternative proof without using the fact that the function $\;f(x)=\sqrt x:[0,+\infty)\to\mathbb R\;$ is monotonically increasing.
By letting $\;\alpha=\sqrt5-3\;,\;$ we get that
$\sqrt5=3+\alpha\;\;,$
$5=9+6\alpha+\alpha^2\;\;,$
$\alpha=-\dfrac{4+\alpha^2}6<0\;.$
Since $\;\alpha<0\;,\;$ it follows that
$\sqrt5=3+\alpha<3\;\;,$
hence, $\;\sqrt5<3\;.$
Another way to write the previous proof :
$\sqrt5=3-\dfrac{18-6\sqrt5}6=3-\dfrac{4+\left(\sqrt5-3\right)^2}6<3\;\;,$
hence, $\;\sqrt5<3\;.$
