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.


  1. Frank Morgan, Real Analysis, American Mathematical Society, 2005.
This entry was posted in irrational numbers, real analysis. Bookmark the permalink.

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.

  2. Pingback: Proof of the Irrationality of e | A Narrow Margin

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s