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?

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    $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 11 '12 at 17:05
  • $\begingroup$ You might also want to simplify matters by assuming, for the sake of contradiction, that there exist $p,q \in \mathbb{Z}$ such that $\sqrt{3} = \frac{p}{q} \text{ and}\; gcd(p,q) = 1.$ $\endgroup$ – amWhy Nov 11 '12 at 17:08
  • $\begingroup$ @AndréNicolas - so does $6$ itself :) $\endgroup$ – Belgi Nov 11 '12 at 17:08
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    $\begingroup$ True! But I wanted to quote the OP exactly. $\endgroup$ – André Nicolas Nov 11 '12 at 17:14
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    $\begingroup$ @Jake +1 for showing your work! $\endgroup$ – amWhy Nov 11 '12 at 17:18

"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 $p=3s$.

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  • $\begingroup$ 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? $\endgroup$ – 1101 Nov 11 '12 at 17:12
  • $\begingroup$ @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.\ \ $ $\endgroup$ – Gone Nov 11 '12 at 17:17
  • $\begingroup$ @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? $\endgroup$ – 1101 Nov 11 '12 at 19:32


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).

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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

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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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    $\begingroup$ I guess this is what the OP meant by "in simplest form" $\endgroup$ – Belgi Nov 11 '12 at 17:04
  • $\begingroup$ 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. $\endgroup$ – Arthur Nov 11 '12 at 18:36
  • $\begingroup$ @Belgi: Yes: in simplest form is a fairly common English synonym of in lowest terms and reduced, especially in grade school. $\endgroup$ – Brian M. Scott Nov 11 '12 at 18:40

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