# proving transcendental numbers are irrational

I don't understand how every transcendental number is irrational, is there a way to prove that? I know it just means it's a non-algebraic number, but how does that correlate to irrationality?

• How about proving that every rational is algebraic? – Lord Shark the Unknown Oct 26 '18 at 15:58
• Well, can you write $\pi$ or $e$ as an exact fraction? If you could, wouldn't they be a root of an algebraic equation with rational coefficients. Therefore all transcendental numbers are irrational. – Mohammad Zuhair Khan Oct 26 '18 at 15:58
• \$p,q\in\mathbb Z, q\neq0$$qx-p=0\to x=\frac pq$$ – Don Thousand Oct 26 '18 at 16:00
• by definition transcendental number is: 1) irrational and 2) is not a root of polynomial with integer coefficients – Vasya Oct 26 '18 at 16:03

If $$x$$ is transcendental but not irrational, then $$x = a/b$$, with $$a,b$$ integers, and so $$x$$ solves the rational equation $$b t - a = 0$$, but then $$x$$ is algebraic and hence not transcendental.
Summing up the comments...,there are two types of numbers in $$\Bbb{R}$$ in the sense , one type is algebraic and the other one is transcendental.
In particular every rational $$x=\frac{p}{q}$$ is algebraic, since $$x$$ satisfies $$qx-p$$, which is a non zero integer polynomial. Therefore if any $$x$$ is not algebraic ,it cannot be a rational!