Sorry for the dumb question; something about this proof seems off and I was wondering what (if anything) is wrong with it.
Assume $p$, $q$ are integers. We prove by contradiction.
$\sqrt3 = p/q$
$(\sqrt3)^3 = (p/q)^3$
$3 = p^3/q^3$
$3q^3 = p^3$
$\sqrt[3]{3q^3} = \sqrt[3]{p^3}$
$3q = p$
Then we substitute into the original equation:
$\sqrt 3 = 3q/q$
$\sqrt 3 = 3$
But it $3\cdot 3$ is clearly not $3$; so we have a contradiction.
Thanks in advance for any help!
On
It's unclear whether you meant $\sqrt{3}$ or $\sqrt[3]{3}$. As $(\sqrt{3})^3 \ne 3$ your proof is off to a wrong start. And then when you take the the cube root latter you essentially undo it ... exept you don't. You argument uses $(\sqrt 3)^3 = 3$ (wrong) and $\sqrt[3]{3} = 3$ (very wrong) and so $\sqrt 3 = 3$ (wrong). In actuality you would have done either $(\sqrt 3)^2=3$ and $\sqrt{3} = \sqrt {3}$ so $\sqrt 3 = \sqrt 3$ (true but useless) or $(\sqrt[3] 3)^3 = 3$ and $\sqrt[3]{3} = \sqrt[3]{3} $ so $\sqrt[3] 3= \sqrt[3] 3$ (ditto).
Your proof should mirror the proof that $\sqrt 2$ is irrational.
$\color{blue}{\text { Assume }\sqrt 2 = \frac ab\text{ where }a,b\text{ are integers with no common factors}}$
Okay so Assume $\sqrt 3 = \frac pq$ where $p,q$ are integers with no common factors.
$\color{blue}{2 = \frac {a^2}{b^2}\\ 2b^2 = a^2\\ \text {so }2\text{ divides }a^2}$
Okay... so
$3 =\frac {p^2}{q^2}\\ 3q^2 = p^2\\$ So $3$ divides $p^2$.
$\color{blue}{\text{Either }a\text{ is odd or }a\text{ is even.}\\\text{If }a\text{ is odd, then }a^2\text{ is odd, which is a contradiction.}\\\text{So }a\text{ is even.}}$
Okay... this is where $3$ is slightly different than $2$.
Either $p$ is divisible by $3$ or $p$ has remainder $1$ or $2$ when divided by $p$.
If $p$ has remainder $1$ then $p = 3k+1$ for some integer $k$ and $p^2 = (3k+1)^2 = 3k^2 + 6k + 1$ is not divisible by $3$. If $p$ has remainder $2$ then $p = 3k + 2$ for some integer $k$ and $p^2 = (3k + 2)^2 = 3k^2 + 12k + 4$ is not divisible by $3$. Those are contradictions so $p$ is divisible by $3$ is the only option.
$\color{blue}{\text{So }a = 2k\text{ for some integer }k\text{ and }2b^2 = a^2 = (2k)^2 = 4k^2\text{.}\\\text {So }b^2 = 2k^2}$.
So $q= 3k$ for some integer $k$ and $3p^2 = q^2 = (3q)^2=9q^2$ and $p^2 = 3q^2$.
$\color{blue}{\text{So }b^2\text{ is even and by the exact same argument, }b\text{ is even. }}$
So $p^2$ is divisible by $3$ and by the exact same argument $p$ is divisible by $3$.
$\color{blue}{\text{So both }a\text{ and }b\text{ are even and have factors of }2\text{. }\\\text{But that's a contradiction.}}$
So both $p$ and $q$ are divisible by $3$. But that's a contradiction.
Your proof is incorrect. You've mixed up cube roots and square roots; in fact, it's hard to ascertain what you have done. $\sqrt{3}^3$ is certainly not equal to $3$, and then you simply took the cube root of that, which would make the statement $\sqrt{3}q=p$. I would recommend you to read up on surds again.
The correct proof would be a derivative of the proof that the square root of $2$ is irrational.
Here's how:
Let $\sqrt3 = \frac{p}{q}$, where $\gcd(p,q)=1$, and $p,q\in \Bbb N$.
Thus $3=\frac{p^2}{q^2} \Rightarrow 3p^2=q^2$, which implies that $3|q$. Why is this true? I urge you to try it out on some numbers yourself: if a prime divides the product of two numbers, it must divide the numbers themselves. This is known as Euclid's Lemma. Here, the two numbers are $q$ and $q$. They multiply together to form $q^2$, and thus since $p$ cannot possibly divide $q$, we are left with $3$ dividing $q$.
Again, let $q=3a \Rightarrow p^2=9a^2$, which implies that $3|p$ (Why?) contradicting the fact that they are relatively prime. This proves that our assumption was wrong, and $\sqrt{3}$ cannot possibly be written as a rational.