why does $\sqrt2 = \frac{2}{\sqrt2}$? I noticed just now that $\sqrt2 = \frac{2}{\sqrt2}$
I'm suprised because isn't this like saying $x = \frac{2}{x}$?
 A: I think I have simply stated the definition of the square root, since multiplying  $x = \frac{2}{x}$ by $x$ gives  $x^2 = 2$, so $x = \sqrt2$
A: $$\frac{2}{\sqrt{2}} = \frac{\sqrt{2}\sqrt{2}}{\sqrt{2}} = \sqrt{2}$$
A: Mr_CryptoPrime's answer is the right way to think about this, but here is another way to see it. Since $\sqrt{2}$ can be defined as a positive real $x$ number such that $x^2=2$, we can ask ourselves whether $2/\sqrt{2}$ satisfies these two properties:


*

*$2/\sqrt{2}$ is a positive real number (because it is a quotient of two positive real numbers), and

*$\displaystyle \left(\frac{2}{\sqrt{2}}\right)^2 = \frac{2^2}{(\sqrt{2})^2} = \frac{4}{2} = 2.$


Thus, $2/\sqrt{2}$ satisfies the properties that define $\sqrt{2}$, and the two numbers must be equal.
A: Yes, just make sure you make sure you understand the distinctions below:
$\frac{2}{\sqrt2}$ is obtained by multiplying the $\sqrt2$ by 1 or $\frac{\sqrt2}{\sqrt2}$.
$$\left(\frac{\sqrt2}{1}\right)\left(\frac{\sqrt2}{\sqrt2}\right) = \frac{2}{\sqrt2}$$
This happens clearly because the two square roots cancel each other out due to the fact that the squaring operator is the inverse of the square-root operator, and vice versa.
Now let's perform the same method on x.
$$\left(\frac{x}{1}\right)\left(\frac{x}{x}\right) = \frac{x^2}{x}$$
$x \neq \frac{2}{x}$ unless we first explicitly state that x = $\sqrt2$.
A: That is pretty easy when you know how to deal with quadratic irrationals in a denominator. Just multiply by $\sqrt{2}$ numerator and denominator: $2/\sqrt{2} = 2\sqrt{2}/2 = \sqrt{2}$.
