Question #c74dd

Nov 3, 2016

$\frac{\sqrt{2}}{2}$ can not be rationalized.

(see below)

Explanation:

Defintion
Rationalization is the expression of a number in as a ratio (or "fraction") of the form: $\frac{a}{b}$ where $a$ and $b$ are integers and have no common factors greater than $1$.

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$\textcolor{b l a c k}{\frac{\sqrt{2}}{2}}$ is an irrational number.

Proof:
Suppose that $\frac{\sqrt{2}}{2}$ were a rational number.
Then, by definition of rational numbers, we could write
$\textcolor{w h i t e}{\text{XXX}} \frac{\sqrt{2}}{2} = \frac{a}{b}$
where $a$ and $b$ were integer values with no common factors.

This would mean
$\textcolor{w h i t e}{\text{XXX}} \sqrt{2} b = 2 a$
Squaring both sides:
$\textcolor{w h i t e}{\text{XXX}} 2 {b}^{2} = 4 {a}^{2}$

$\textcolor{w h i t e}{\text{XXX}} {b}^{2} = 2 {a}^{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {b}^{2}$ must be even
and since squared odd numbers are odd
$\textcolor{w h i t e}{\text{XXX}} \rightarrow b$ must be even.

So we could replace $b$ with $2 c$ for some integer $c$
$\textcolor{w h i t e}{\text{XXX}} \frac{\sqrt{2}}{2} = \frac{a}{2 c}$

$\textcolor{w h i t e}{\text{XXX}} \cancel{2} \sqrt{2} c = \cancel{2} a$

$\textcolor{w h i t e}{\text{XXX}} \cancel{2} {c}^{2} = {\cancel{4}}^{2} {a}^{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {a}^{2}$ is even

$\textcolor{w h i t e}{\text{XXX}} \rightarrow a$ is even.

Therefore both $a$ and $b$ would need to be even;
that is both $a$ and $b$ would have a common factor of $2$

...but this is contrary to the initial declaration that $a$ and $b$ have no common factors.