I'm trying to prove
There is a positive irrational number $x$ for which $x^2=3$.
I think this would be useful.. but I'm not too sure. $\sqrt{2}\in\mathbb{R}\setminus\mathbb{Q}$
Prove: there is a positive irrational number $x$ for which $x^2=3$.
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proof-explanation
1 Answers
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the range of the function $f(x)=x^2$ on $\mathbb R^+$ is $\mathbb R^+$ (this is proved with the intermediate value theorem). So there exists a positive real $\alpha$ with $\alpha^2=3$, we just have to prove it cannot be rational.
We can do it by contradiction, suppose a rational exists, of the form $\frac{p}{q}$ with $p$ and $q$ coprime such that $(\frac{p}{q})^2=3$, we have that $p^2=3q^2$, it follows that $p$ is a multiple of $3$ (by euclids theorem, since $3$ is prime). So let $p=3r$, notice that we have $3r^3=q^2$, so $3$ also divides $q$, a contradiction.