How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?

Question

How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?

I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.

Answer 1

One thing we tend to like to do when dealing with this stuff is have the denominator of fractionish things as a plain number, as much as possible. To do this, we'll multiply both top and bottom by something that will cancel out any radicals in the bottom. $$\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$ This even works for more complicated stuff. $$\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{1}{\sqrt{5}+\sqrt{3}}\cdot\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{5-3}=\frac{\sqrt{5}-\sqrt{3}}{2}$$

Answer 2

Since $2=\sqrt{2}\cdot\sqrt{2}$ you have that $$\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{1}{\sqrt{2}}$$

Answer 3

Let's square both of them: \begin{align} \sqrt{\frac{1}{2}}^2 &= \frac{1}{2}, \text{ while} \\ \left(\frac{\sqrt{2}}{2}\right)^2 &=\frac{\sqrt{2}^2}{2^2} \\ &=\frac{2}{4} \\ &= \frac{1}{2} \end{align}

So they're both positive numbers, and their squares are the same, so they must be the same.

Answer 4

The number $\frac{1}{\sqrt{2}}$ is defined to be the number such that, when you multiply it by $\sqrt{2}$, you get $1$. Symbolically, it is the solution to $x\cdot \sqrt{2}=1$. That's how division works - it's the inverse of multiplication. What is $\frac{\sqrt{2}}2\cdot \sqrt{2}$? Why, it's $\frac{2}2=1$ - so it must be $\frac{1}{\sqrt{2}}$, as it satisfies the definition of division for $\frac{1}{\sqrt{2}}$.

Answer 5

The sqrt(2)/2 turns into sqrt(2/4) when the two is squared and put under the square root which simplified is sqrt(1/2).

Answer 6

$\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}}$

$\frac{\sqrt{2}}{2}=\frac{2^{\frac{1}{2}}}{2^1}=2^{\frac{1}{2}-1}=2^{-\frac{1}{2}}$