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The proof of the irrationality of $\sqrt{2}$ starts with the supposition that $\sqrt{2} = \frac ab$ where $a$ and $b$ are integers. I understand that, but why is it important that $\frac ab$ is expressed in simplest terms? I see that it is a major part of the contradiction, but why?

EDIT: I see that having the fraction be irreducible means it is unique. Why does it have to be unique though?

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    $\begingroup$ Because the simplest term is unique. $\endgroup$ – Math Lover Nov 16 '17 at 22:32
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    $\begingroup$ It is worth to read the beautiful proofs people here have written down, e.g., here. $\endgroup$ – Dietrich Burde Nov 16 '17 at 22:34
  • $\begingroup$ Actually, it doesn't have to be in lowest terms. But the fact that you picked an arbitrary representation of two integers for this value and they CAN'T be in lowest terms despite that there was nothing in choosing them that they couldn't be is a contradiction. No rational number is such that it can not ever be in lowest terms. And Yet $\frac ab$ so that $(\frac ab)^2 = 2$ can not ever be in lowest terms. That is not normal. That is wrong. That is a contradiction. $\endgroup$ – fleablood Nov 17 '17 at 3:08
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The importance of $\gcd(a,b)=1$ depends on the proof you are following.


Assuming $\sqrt{2}=\frac{a}{b}$ with $a,b\in\mathbb{N}^+$ we get $$ a^2 = 2b^2 $$ but this is impossible since $\nu_2(a^2)$ is even while $\nu_2(2b^2)$ is odd.


Here $\gcd(a,b)=1$ is irrelevant, for instance.
For any $n\in\mathbb{N}^+$, $\nu_2(n)$ is defined as $\max\{m\in\mathbb{N}:2^m\mid n\}$.

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    $\begingroup$ @Joffan: fine, explanation added. $\endgroup$ – Jack D'Aurizio Nov 16 '17 at 22:38
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It's usually just a time saver and an assurence that an infinite regress is a contradiction.

If $\frac ab$ is such that $(\frac ab)^2 =2$ then $a^2 = 2b^2$ so $2|a^2$ and so $2|a$. Let $a = 2a_1$ so $2a_1^2 = b^2$ so $2|b$. Let $b = 2b_1$ and $a_1^2 =2b_1^2$ so $a_1$ is divisible by $2$. Let $a_1 = 2a_2$. and so on ad infinitim.

Well... so what is wrong with ad infinitim? And how do we know that just because we got the same result $3$ times we will get it forever? And why is that a contradiction?

For that matter why is that all rational numbers actually have "lowest terms"?

Those all actually have answers and the pertain to the well-ordering principle and they are not trivial (although they aren't hard either).

We have a series of $a > a_1 > a_2 > ...$ so that $a_i = 2a_{i+1}$ as we as $b > b_1 > b_2 > ....$ so that $b_i = 2b_{i+1}$. But the induction principle in natural numbers we know these $a_i, b_i$ exist for all $i \in \mathbb N$ so that $a_i^2 = 2b_i^2$ and $b_i^2 = 2a_{i+1}^2$.

But by the well-ordering principle, we know that for such collections of $a_i$ and of $b_i$ that they must each have a least element.

And that is a contradiction.

But that involved a bit of a diversion that we feel simply distracts the student. Surely it's easier to just avoid the issue by saying "Oh, we can assume $\frac ab$ is in lowest terms".

But actually knowing that if $a,b \in \mathbb Z, b\ne 0$ that there must be some $a',b'$ so so that $\gcd(a',b') = 1$ and $\frac ab = \frac {a'}{b'}$ requires a similar well-ordering principle argument.

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If they aren't in lowest terms, then it's possible that both $a^2$ and $b^2$ are even, while avoiding that possibility is the heart of the contradiction.

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  • $\begingroup$ I guess what I'm asking is why it's important for the contradiction -- couldn't 2a/2b still be rational? $\endgroup$ – Zoe Nelson Nov 16 '17 at 22:37
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    $\begingroup$ @ZoeNelson no, since $2a/2b$ is precisely equal to $a/b$. $\endgroup$ – jacer21 Nov 16 '17 at 22:51
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Depending how you want to do the proof, it doesn’t have to be in lowest terms. Just suppose $a$ and $b$ are both positive. If $\frac{a}{b}=\sqrt2$, then $a$ and $b$ are both even, by the usual argument. Say $a=2a’$ and $b=2b’$. Thus, $\sqrt2=\frac{a}{b}=\frac{a’}{b’}$, with $a’<a$ and $b’<b$.

Now, by the same argument, $a’$ and $b’$ are both even, so we reduce again, and obtain another, smaller pair. Continuing in this way, we obtain two infinitely descending sequences of positive integers, which is impossible. That’s your contradiction, and this method is called “infinite descent”.

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