# Prove that the function: f ( x ) = { <mtable columnalign="left left"

Prove that the function:

is discontinuous at every irrational number using both the precise definition of a limit and the fact that every nonempty open interval of real numbers contains both irrational and rational numbers.
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Krish Finley
Let $a$ be irrational, so $f\left(a\right)=a\ne 0$. Let $ϵ=|a|$ and assume there is $\delta >0$ such that $|f\left(x\right)-f\left(a\right)|<ϵ$ for all $x$ with $|x-a|<\delta$. By the existence of rationals $x$ with $|x-a|<\delta$ (for example $x=\frac{1}{n}⌊na⌋$ for $x=\frac{1}{n}⌊na⌋$ with $n>\frac{1}{\delta }$) we arrive at a contradiction because for this $x$ we have$f\left(x\right)=0$ and hence $|f\left(x\right)-f\left(a\right)|\nless ϵ$.