Proof of the irrationality of $\sqrt{3}$ - logic question

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?

Solution 1:

"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 $p=3s$.

Solution 2:

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).

Solution 3:

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