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**hmmmm** If we take the open interval $\displaystyle (a,b)$ and shift it $\displaystyle (ax,bx)$. Where $\displaystyle x=\sqrt(p)$ where p is a prime.

Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:

$\displaystyle ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b$.

Therefore as $\displaystyle \frac{q}{x}$ and we have $\displaystyle \frac{q}{x}\in(a,b)$

we have an irrational in any open interval.

Now as there are infinitely many primes we have infinitley many irrationals of the form $\displaystyle \sqrt(p)$, so we have infinitley many irrationals in the interval.