So the question goes let $x = a\sqrt3 + b\sqrt 5$ where $a,b$ rational. Prove that $x$ is rational iff $a=b=0$. I think I can prove this but I'm not sure if my proof is correct or rigorous.
Well obviously if $a=b=0$ then $x=0$ which is rational.
Now proving it the other way. If $x$ is rational then either $x$ is $0$ or $x$ is not $0$. If $x$ is $0$ then $$b\sqrt5 = -a\sqrt3$$ $$\frac{b}{-a} = \frac{\sqrt3}{\sqrt5}$$ But the LHS is Rational while the RHS is not. You can see this by assuming $\frac{\sqrt3}{\sqrt5} = \frac mn$ for some integers $m,n$. Then $3n^2=5m^2$ but if you factorize $m, n$ into primes you'll see that the power of 3 is one greater on the LHS than it is on the RHS. This contradicts the Fundamental Theorem of Arithmetic. So it must be that $x$ is Rational but not $0$. Not $100\%$ sure that this step is correct though.
So then $$a\sqrt3 = x-b\sqrt5$$ $$3a^2=x^2-2bx\sqrt5+5b^2$$ $$\frac{3a^2-x^2-5b^2}{-2bx}=\sqrt5$$ If $a,b\neq0$ then the LHS is rational since the quotient of rationals is rational as long as the denominator is not 0. However the RHS is irrational since the sqrt of a prime is irrational. So we cant's have $a,b\neq0$
If $a=0$ but $b\neq0$ then same deal. LHS rational, RHS irrational. So we can't have $a=0$ but $b\neq0$.
If $b=0$ but $a\neq0$ then simply start with the equation $b\sqrt5 = x-a\sqrt3$ and you will get a similar contradiction.
So the only plausible case left is $a=b=0$
Is this proof correct?
Here is an simpler proof:
Since $x^2= 3a^2+5b^2+2ab\sqrt{15}$, if $ab\ne 0$ and $x$ is rational, then so would $\sqrt{15}$.
Thus, $a=0$ or $b=0$.
If $a=0$, $b\ne0$, and $x$ is rational, then so would $\sqrt{5}$.
If $b=0$, $a\ne0$, and $x$ is rational, then so would $\sqrt{3}$.
Therefore, $a=0$ and $b=0$.