The proof that frightened Pythagoras

Adrian McMenamin Uncategorized June 21, 2013June 23, 2013 1 Minute

I have been meaning to do this for a while …

Pythagoras was said to be shocked by the proof that $\sqrt 2$ is an irrational number (i.e., that it cannot be represented as a ratio of two whole numbers). The proof (by contradiction) is a simple one.

If $\sqrt 2$ is rational then $\sqrt 2 = \frac{a}{b}$ and $2 = \frac{a^2}{b^2}$ where $a$ and $b$ are the smallest nominator and denominator possible, i.e, in the lowest terms.

Hence $a^2 = 2b^2$ and so $a$ must be even (as two odd numbers multiplied always give an odd number – let $l = 2x$ and $m = 2y$ be even numbers then $(l + 1)(m + 1) = lm + l + m + 1$ ). Hence $a = 2k$ .

But we also have $4k^2 = a^2$ and thus $4k^2 = 2b^2$ and $b^2 = 2k^2$ and so $b$ is even.

So we have $a$ and $b$ sharing a common factor of 2, so they cannot be the nominator and denominator of the fraction in the lowest terms.

But if $a$ and $b$ are both even then they share a common factor of 2 and $\frac{a}{b} = \frac{2k}{2p} = \frac{k}{p}$ : implying that $a = k$ and $a = 2k$ , an obvious contradiction: hence $\sqrt 2$ cannot be a rational.

Update: I have made the final step of this shorter and clearer.

Further update: I have been told (see comments below) I would have been better sticking with a clearer version of the original ending ie., we state that $\frac{a}{b}$ are the lowest terms. Then, plainly as they have a factor in common (2) they cannot be the lowest terms and so we have a contradiction. Would be great if someone could explain why we cannot use the $a = 2k = k$ contradiction.

And another update: Should have stuck with the original explanation – which I have now restored in a hopefully clearer way. The comments below are really interesting and from serious mathematicians, so please have a look!

An interesting inequality (sirjackshen.wordpress.com)
Heuristic ideas about bounded prime gaps (motls.blogspot.com)
Online reading seminar for Zhang’s “bounded gaps between primes” (terrytao.wordpress.com)

Published by Adrian McMenamin

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Published June 21, 2013June 23, 2013

9 thoughts on “The proof that frightened Pythagoras”

castover says:

June 22, 2013 at 3:34 am

In your last step, you deduce that $a=k$ from $\frac{a}{b}=\frac{k}{p}$ , which isn’t necessarily correct. The only deduction that holds is that $ap=bk$ .

The part of this proof that usually lends the contradiction is from the fact that $\frac{a}{b}$ would have a 2 in common. In particular, an earlier step in the proof is to generally assume that $\sqrt{2}=\frac{a}{b}$ where $a$ and $b$ are in lowest terms (and hence have no common factors).

This was one of my favorite proofs as an undergraduate: Thanks for taking the time to share it with others. 🙂
1. Adrian McMenamin says:
  
  June 22, 2013 at 7:01 am
  
  Thanks. Not a maths graduate, so happy to accept I have got it wrong, but would like to know more 🙂 I did, essentially, have a version of the fact they were in lowest terms before and then changed it to current version – why doesn’t that necessarily hold?
  1. castover says:
    
    June 22, 2013 at 9:00 pm
    
    Sorry for not responding sooner; I haven’t been around much today.
    
    The short answer as to why that doesn’t necessarily hold can be shown by an example: $\frac{1}{2}=\frac{3}{6}$ though clearly $1\neq 3$ . As mentioned above, however, $6(1)=3(2)$ .
    
    The longer answer is from abstract algebra: Given a commutative ring with identity $R$ , its field of fractions $D^{-1}R$ is constructed by “attaching” to $R$ the set $D^{-1}$ of multiplicative inverses (namely, reciprocals) of nonzero elements of $R$ . This definition is well-defined precisely when the construction is made as an equivalence class in which elements $\frac{a}{b}$ and $\frac{c}{d}$ are considered equal whenever $ad=bc$ .
    
    Given this construction, the answer for your specific proof lies in the fact that $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$ , and so uniqueness of “fractions” (elements of $\mathbb{Q}$ ) exists only up to the equivalence mentioned above.
    
    If you ever want to talk about this – or anything mathematical, really – further, don’t hesitate to let me know! 🙂
  2. Adrian McMenamin says:
    
    June 22, 2013 at 11:02 pm
    
    Err, yes. I’ve been a bit silly really. Thanks
  3. castover says:
    
    June 22, 2013 at 9:04 pm
    
    Also, in my explanation of $D^{-1}R$ , I also apparently neglected to mention that if $R$ has zero divisors, they also can’t be given reciprocals. To be precise, the definition I gave sort of implies that $R$ is assumed to have been an integral domain, though the more general construction for $D^{-1}R$ is done similarly. For details, check here:
    
    http://en.wikipedia.org/wiki/Field_of_fractions
Lubos Motl says:

June 22, 2013 at 1:14 pm

One can also quickly complete the proof by saying that a^2 = 2b^2 is impossible because in the factorization of the left hand side to primes, each prime is exponentiated to an even exponent, while the decomposition of the right hand side to a product of primes has an odd power of 2 (because of the extra power of two), so they can’t be equal. One needs to use that the factorization exists for each positive integer and is unique.
1. Adrian McMenamin says:
  
  June 22, 2013 at 11:03 pm
  
  Yes, thanks.
prubin73 says:

June 22, 2013 at 8:15 pm

“Would be great if someone could explain why we cannot use the a = 2k = k contradiction.” If a/b = c/d and a/b = e/f, it does not imply that c = e and d = f UNLESS both c/d and e/f are in lowest terms. So going from a/b = k/p to a = k without the lowest terms bit is an unjustified leap.
1. Adrian McMenamin says:
  
  June 22, 2013 at 11:02 pm
  
  I suppose that is pretty obvious really. Not sure why i let it pass me by. Thanks