Question #c74dd

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Question #c74dd

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Answer 1 · 2016-11-03T12:05:14

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

(see below)

Explanation:

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\frac{\sqrt{2}}{2}$ $color(black)(sqrt(2)/2)$ is an irrational number.

Proof:
Suppose that $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ were a rational number.
Then, by definition of rational numbers, we could write
$XXX \frac{\sqrt{2}}{2} = \frac{a}{b}$ $color(white)("XXX")sqrt(2)/2=a/b$
where $a$ $a$ and $b$ $b$ were integer values with no common factors.

This would mean
$XXX \sqrt{2} b = 2 a$ $color(white)("XXX")sqrt(2)b=2a$
Squaring both sides:
$XXX 2 b^{2} = 4 a^{2}$ $color(white)("XXX")2b^2=4a^2$

$XXX b^{2} = 2 a^{2}$ $color(white)("XXX")b^2=2a^2$

$XXX \to b^{2}$ $color(white)("XXX")rarr b^2$ must be even
and since squared odd numbers are odd
$XXX \to b$ $color(white)("XXX")rarr b$ must be even.

So we could replace $b$ $b$ with $2 c$ $2c$ for some integer $c$ $c$
$XXX \frac{\sqrt{2}}{2} = \frac{a}{2 c}$ $color(white)("XXX")sqrt(2)/2=a/(2c)$

$color(white)("XXX")cancel2sqrt(2)c=cancel2a$

$color(white)("XXX")cancel2c^2=cancel4^2a^2$

$color(white)("XXX")rarr a^2$ is even

$color(white)("XXX")rarr a$ 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.

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Question #c74dd

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