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Prove $\sqrt{3}$ is irrational. (Proof by contradiction).

Let $\sqrt{3}$ be a rational number in simplest form $\frac pq$.

So squaring both sides of $\sqrt{3}=\frac pq$ we get $3=(\frac {p}{q})^2$ which translates to $3=\frac{p^2}{q^2}$.

Multiply both sides of the equation by $q^2$ yields $3q^2=p^2$. Now $p^2$ is taken to be divisible by 3 and thus an odd number, $p$ is also odd because any odd number squared is also odd.

So let $p=3s$ where s is an integer. Then $3q^2=(3s)^2 = 3q^2=9s^2$. Dividing both sides of the equation by 3 leaves us with $q^2=3s^2$.

Here is is taken that $q^2$ is divisible by 3 and is odd and so is $q$.

Therefore both $q \text{ and}\; p$ have a common factor of being odd and divisible by 3, proving that the $\sqrt{3}$ is irrational.

Are there any gaps that I could improve on?

amWhy
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    The part "$p^2$ is divisible by $3$ and therefore an odd number" is not true, $6^2$ is divisible by $3$ but not odd. – André Nicolas Nov 11 '12 at 17:05
  • You might also want to simplify matters by assuming, for the sake of contradiction, that there exist $p,q \in \mathbb{Z}$ such that $\sqrt{3} = \frac{p}{q} \text{ and}; gcd(p,q) = 1.$ – amWhy Nov 11 '12 at 17:08
  • @AndréNicolas - so does $6$ itself :) – Belgi Nov 11 '12 at 17:08
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    True! But I wanted to quote the OP exactly. – André Nicolas Nov 11 '12 at 17:14
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    @Jake +1 for showing your work! – amWhy Nov 11 '12 at 17:18
  • possible duplicate of Critiques on proof showing $\sqrt{12}$ is irrational. – Dominic Michaelis May 17 '13 at 14:43
  • @DominicMichaelis Surely critiques are localised to the specific question, no? Jake cannot gain a critique of his proof by looking at critiques of user21154's proof... – user1729 May 17 '13 at 15:11
  • @user1729 their proof nearly looks quiet the same. I voted to close mainly to shorten the unanswered questions. As it seems here is no work left to done I thought linking to a pretty similiar question – Dominic Michaelis May 17 '13 at 15:20
  • @DominicMichaelis Why not upvote an answer then?! This is a perfectly good question which does not deserve to be closed! (Although I do understand your point, as it is an old question. But still, a simple upvote is much neater and more elegant!) – user1729 May 17 '13 at 15:22
  • @user1729 closing a question doesn't mean that it is bad. I think those check my prof questions are most times to localized. I did not found any answer extremely nice and wanted to clean up more so I need to save some upvotes :) – Dominic Michaelis May 17 '13 at 15:31
  • @DominicMichaelis Maybe I am a bit loose with my upvotes... but anyway, I do agree that "check my proof" questions are too localised, but then of course we could close it as such, no? (And perhaps say that this question is similar to the linked one, in that they make the same error.) – user1729 May 17 '13 at 15:45
  • @User1729 yeah maybe that have been the better way to do so – Dominic Michaelis May 17 '13 at 15:47

4 Answers4

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"Now $p^2$ is taken to be divisible by 3 and thus an odd number, $p$ is also odd because any odd number squared is also odd." That's a non-sequitur. The fact that any odd number squared is odd doesn't rule out other numbers being odd.

"So let $p=3s$ where s is an integer." That's illegitimate. $p$'s being odd doesn't make it multiple of 3.

So you need to repair the argument from $p^2$ being taken to be divisible by 3 to $p$ being of the form $p=3s$.

Peter Smith
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  • I'm a little confused on what to do to repair the argument, could you give me a nudge in the right direction?

    Could I say the LHS is odd and so is the RHS, so p^2 must be odd. Then go on to replace p with 3s?

     – 1101 Nov 11 '12 at 17:12
  • @Jake To repair your proof you need to supply a valid proof that $\rm:3\mid n^2:\Rightarrow:3\mid n\ \ $ You seem to be mimicking one well-known proof of the irrationality of $\sqrt{2},:$ which uses the parity of $\rm,n.:$ That idea can be generalized to give an analogous proof here by working mod $3$ instead of mod $2.\ \ $ – Bill Dubuque Nov 11 '12 at 17:17
  • @BillDubuque I'm doing an exercise that follows the proof of the irrationality of sqroot{2}. So I'm trying to use that as a base to start from. From the step p^2=3q^2, that p^2 is a multiple of 3, and so is p? Now setting p=3s. Would this satisfy the proof?

    Is the difference between the sqroot[2] and sqroot[3] reasoning be that in the case of 3 it won't necessarily be odd but rather a multiple of 3?

     – 1101 Nov 11 '12 at 19:32
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Yes.

Firstly, the analogoues of even number when proving $\sqrt2$ is irrational, is not the odd numbers for $\sqrt3$, but the numbers 'divisible by $3$' (and these are not necessarily odd, for example $12$).

Secondly, it is not finished yet. You have to divide the $3$'s for an infinite time, contradicting the fundamental thm of number theory, or, the easiest, is that $p$ and $q$ are assumed to be relatively primes (else $\displaystyle\frac pq$ would be simplifiable).

Berci
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1) I would change the first "Let" to be "assume by negation"

2) 2) $p^{2}$ is not taken to be divisible by $3$, we concluded this

3) "and thus an odd number" - this is wrong since, for example, $3\mid6$ but $6$ is not odd

Belgi
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You must assume that $gcd(p,q)=1$, then you get contradiction in the last part of your proof.

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    I guess this is what the OP meant by "in simplest form" – Belgi Nov 11 '12 at 17:04
  • When I do these proofs, rather than assuming coprimality, I prefer to assume that $p$ and $q$ are finite integers and point out that you can keep factoring out a 3 indefinetely. Thus contradiction. – Arthur Nov 11 '12 at 18:36
  • @Belgi: Yes: in simplest form is a fairly common English synonym of in lowest terms and reduced, especially in grade school. – Brian M. Scott Nov 11 '12 at 18:40