## Proving the Existence of Irrational Numbers

The ancient Greeks first proved the existence of irrational numbers by proving that $\sqrt{2}$ is irrational.  The proof is, as modern proofs of irrationality go, fairly simple.  It is often the first example of a proof of irrationality that students see in, say, a course in real analysis.

Yesterday, though, I saw an even simpler proof of irrationality in Morgan’s Real Analysis [1] that I thought worth showing to my students in advanced calculus.

Theorem: The number $\log_{10} 5$ is irrational.

Proof: First, $\log_{10} 5 > 0$ because 5 and 10 are both greater than 1.  Suppose that $\log_{10} 5$ is rational.  Then there exist positive integers p, q such that $10^{p/q} = 5$.  Thus $\displaystyle 10^p = 5^q.$  However, $10^p$ ends in 0, while $5^q$ ends in 5.  This is a contradiction, and so $\log_{10} 5$ is irrational.

References

1. Frank Morgan, Real Analysis, American Mathematical Society, 2005.
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### 3 Responses to Proving the Existence of Irrational Numbers

1. Valerio De Angelis says:

Thanks, that is a nice proof.
Here is a very short proof that sqrt(2) is irrational. I believe it is due to John Conway, but I am not sure:
Suppose sqrt(2) is rational. Then there is a smallest positive integer n with the property that
n sqrt(2) is an integer. But then m=n sqrt(2) -n has the same property, and it’s smaller, contradiction.

• mzspivey says:

Nice!

(Note for other readers: This proof uses the property that $\sqrt{2}$ must satisfy $\sqrt{2} \leq 2$, which is easy to prove.)