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The proof (taken from http://www.themathpage.com/aPreCalc/rational-irrational-numbers.htm#proof):

"To prove that there is no rational number whose square is 2, suppose there were. Then we could express it as a fraction m/n in lowest terms.

That is, suppose $$\frac{m}{n} \cdot \frac{m}{n} = \frac{m \cdot m}{n \cdot n} = 2$$
but that is impossible. Since $\frac{m}{n}$ is in lowest terms, then $m$ and $n$ have no common divisors except 1. Therefore $m \cdot m$ and $n \cdot n$ also have no common divisors - they are relatively prime - and it will be impossible to divide $n \cdot n$ into $m \cdot m$ and get 2."

My problem:

I'm confused about how the fact that $m \cdot m$ and $n \cdot n$ are relatively prime makes it impossible to divide $n \cdot n$ into $m \cdot m$ to get 2.

Since $n$ could be 1, it seems to me that it relies on the assumption that $m$ can't be $\sqrt{2}$, which is what we're trying to prove in the first place, so the argument seems circular to me.

ps: I Wasn't sure about the proof-verification tag; this proof comes from a math Professor so my assumption here is that I'm obviously missing something, I don't mean for the tag to be insulting.

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    $\begingroup$ Nice spotting of the gap! We do need to show that $n\ne 1$, that is, that $\sqrt{2}$ is not an integer. This is clear, since $1^2\lt 2$ and if $m\ge 2$ then $m^2\gt 2$. That's what goes 'wrong" if we try, for example, to prove that $\sqrt{25}$ is not an integer, using a proof along the same lines. There is an additional gap, showing that indeed if $m$ and $n$ are relatively prime so are $m^2$ and $n^2$. For that we need unique factorization, or Euclid's Lemma (if a prime divides $ab$ then it divides at least one of $a$ and $b$). $\endgroup$ – André Nicolas May 5 '15 at 23:14
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    $\begingroup$ No it is not circular, it just is missing the verification that $\sqrt{2}$ is not an integer, which is a different assertion than the fact it is not rational, and quite a bit easier to prove. $\endgroup$ – André Nicolas May 5 '15 at 23:31
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    $\begingroup$ @jeremy The "proof" is also missing the justification that $\,m,n\,$ coprime $\,\Rightarrow\,$ $m^2,n^2$ coprime, which is the key inference. This is a consequence of the fundamental theorem of arithmetic (existence and uniqueness of prime factorizations), or related properties such as Euclid's Lemma, Bezout's gcd identity, gcd distributive law, etc. $\endgroup$ – Bill Dubuque May 6 '15 at 1:45
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    $\begingroup$ Euclid's Lemma says that if $\,\gcd (\color{#c00}{n,m})=1\,$ then $\ \color{#c00}{n\mid m}\,k\ \,\Rightarrow\ n\mid k.\ $ Therefore, applying this Lemma in our case above we deduce that: $\ 2n^2 = m^2\,$ $\Rightarrow$ $\,\color{#c00}{n\mid m}\,m\,\Rightarrow\, n\mid m,\ $ so $\,m/n\,$ is an integer which squares to $\,2,\,$ contradiction. That's one rigorous way to complete the proof. See this thread for many other proofs. $\endgroup$ – Bill Dubuque May 6 '15 at 1:58
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    $\begingroup$ @jeremy Great. If anything remains unclear please feel welcome to ask questions. $\endgroup$ – Bill Dubuque May 6 '15 at 15:46
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If you assume your number is rational, this means you can write it as a fraction in it's lowest terms, i.e. it cannot be factorised any more. This is, if your $\sqrt{2}=\frac{m}{n} $, where $m,n \in \mathbb{Z} $ cannot be simplified. In other words, if you write $m$ and $n$ as a product of primes you will not have any common primes in expansions of $m$ and $n$. So when you take $m \cdot m=m^{2}$ and $n \cdot n=n^{2}$ and write them as product of primes, they will not have any common primes, and hence cannot be factorised any more. So the only time this can be $2$ is when $n=1$. This however is a contradiction with your earlier assumption $\frac{m}{n}=\sqrt{2}$ as this would imply $m=\sqrt{2} \Rightarrow \sqrt{2} \in \mathbb{Z}$, and here is your contradiction.

Hope it makes sense...

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  • $\begingroup$ But isn't the fact that $\sqrt{2}$ is not an integer what we're trying to prove in the first place? $\endgroup$ – jeremy radcliff May 5 '15 at 23:31
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    $\begingroup$ @jeremyradcliff No, you're trying to prove that $\sqrt 2$ is irrational. $\endgroup$ – Théophile May 5 '15 at 23:36
  • $\begingroup$ @Théophile, yes my bad, but still the proof assumes that $\sqrt{2}$ is not an integer by not specifying that $n$ cannot be 1. $\endgroup$ – jeremy radcliff May 5 '15 at 23:38
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    $\begingroup$ @jeremy The argument in this answer implicitly uses the fundamental theorem of arithmetic. This need to be made explicit for the argument to be rigorous. $\endgroup$ – Bill Dubuque May 6 '15 at 2:07

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