We know that the $\sqrt 2$ is an irrational number.
Doing $\sqrt {.2} \approx .4472135955...$ in Wolfram Alpha gives the decimal approximation but it doesn't specify if it is accurately equivalent as a fraction of two integer numbers (a rational number).
If it is rational, how can I find its fraction equivalent.
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$\begingroup$Since $\sqrt{5}$ is irrational and $$\sqrt{0.2} = \sqrt{\frac{2}{10}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}\;,$$ $\sqrt{0.2}$ is irrational.
$\endgroup$ $\begingroup$To complete Mike's answer, let's prove that $\sqrt 5$ is irrational. This is a standard proof and worth remembering. You can apply it to other square roots too.
We shall proceed by assuming that $\sqrt 5$ is rational and demonstrate that this leads to a contradiction.
Let $$ \sqrt 5 = a/b $$ where $a$ and $b$ are positive integers. Let us further assume that we are using the minimum such $a$ and $b$. That is, the fraction is in its lowest terms and $a$ and $b$ have no factor in common.
Then, squaring both sides we obtain $$ 5 = a^2/b^2 $$ Note that $a^2$ and $b^2$ also have no factors in common. Therefore the expression above is in its lowest terms and we may write $$ 5 = a^2, 1 = b^2 $$ However, 5 is not the square of an integer, which contradicts the original hypothesis.
An alternate way to get to the contradiction is to note that $5|a^2$, and so $5|a$. So let $$ a = 5c $$ where $c<a$. Substituing we obtain $$ 5 = (5c)^2 / b^2 $$ and so $$ 1/5 = c^2 / b^2 $$ or, equivalently $$ 5 = b^2 / c^2 $$ Now we go through the argument again, setting $b = 5d$, $d<b$ to obtain $$ 5 = c^2 / d^2 $$ and $$ \sqrt 5 = c/d $$ This is a contradiction, as we originally set $a/b$ to be the lowest terms, but now we have $d<b$
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