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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?

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    $\begingroup$ How about proving that every rational is algebraic? $\endgroup$ – Lord Shark the Unknown Oct 26 '18 at 15:58
  • $\begingroup$ 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. $\endgroup$ – Mohammad Zuhair Khan Oct 26 '18 at 15:58
  • $\begingroup$ $p,q\in\mathbb Z, q\neq0$$$qx-p=0\to x=\frac pq$$ $\endgroup$ – Don Thousand Oct 26 '18 at 16:00
  • $\begingroup$ by definition transcendental number is: 1) irrational and 2) is not a root of polynomial with integer coefficients $\endgroup$ – Vasya Oct 26 '18 at 16:03
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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.

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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!

So every transcendental number is irrational not every irrational is transcendental!

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