is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$?

Question

I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:

$\sqrt{2} = 1.414213562$

$\frac {2}{\sqrt{2}} = 1.414213562$

It is confirmed by a hand-calculator.

I tried to proof this as follows:

$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$

$2^{\frac 12} = \frac {2}{2^{\frac 12}}$

$2 = 2^{\frac 12} \cdot 2^{\frac 12}$

$2 = 2^{\frac 12 + \frac 12} $

$ 2 = 2^1 $

$ 2 = 2$

It is also true for:

$-\frac {2}{\sqrt5} = \frac {-2}{5}\sqrt 5$

I didn't know this relationship beforehand and it was new for me, my question is: Is there a general rule for this?

Answer 1

It will be true for any positive real $x$ that $$\frac{x}{\sqrt{x}} = \frac{x}{x^{1/2}} = x^1\cdot x^{-1/2} = x^{1-1/2} = x^{1/2} = \sqrt{x}$$

Answer 2

The steps are: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$You do the same thing for any positive number $a>0$: $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a}.$$The trick is "multiplying by $1$ in a smart way".

P.s.: Congratulations on finding this out on your own! This is how we should always do math. Good studies.

Answer 3

Yes. $$\frac{x^a}{x^b} = x^{a-b}$$ In this case we have $$\frac{2^1}{\sqrt{2}} = \frac{2^1}{2^{1/2}} = 2^{1- 1/2} = 2^{1/2} = \sqrt{2}$$

Answer 4

What you want is:

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

Answer 5

It's more fundamental than the formula $x^a/x^b=x^{a-b}$. The definition of square root is $\sqrt{a}\sqrt{a}=a$. Just divide both sides by $\sqrt{a}$.