There is no rational number r with the property $r^2 =3$ 
There is no rational number r with the property $r^2 =3$

This is what I did. Proof by contradiction.
Assume that there exist a rational number $r=a/b$ where $a$ and $b$ are integers. 
This implies that:
\begin{align*}
r^2 & = 3\\
(a/b)^2 &=3 \\ 
a^2/b^2 &= 3 \\ 
a^2&=3b^2
\end{align*}
I'm stuck here, I don't know how to go from here... 
 A: You are almost done. From here $a^2=3b^2$, note that by the Fundamental theorem of arithmetic, the largest power of $3$ which divides the integer on  the left-side, $a^2$, is EVEN, whereas the largest
 power of $3$ which divides the integer on the right-side, $3b^2$, is ODD. But those integers, $a^2$ and $3b^2$, are equal...
A: Note: Choose $r=a/b$, with $\gcd(a,b)=1.$
$a^2=3b^2$ $ \rightarrow$  $3|a^2.$
Theorem of Number Theory:
If a prime $p$ divides $cd$, then $p$ divides $c$ or $p$ divides $d$.
($c,d$ integers).
Hence $3|a$, or $a = 3r$.
Then: $a^2= 3^2r^2;$  and $b^2= 3r^2.$
Hence $3|b$, or $b= 3s$.
Contradiction to $\gcd(a,b) =1.$
A: Another way to attack such problems is using polynomials with integer coefficients and Rational root theorem (RTT). In this case the polynomial is 
$$P(x)=x^2-3$$
Now, let's assume $r \in \mathbb{Q}$. From RTT, $r$ as a solution of $P(x)$, can only be an integer (or irrational, but we skip this part since we assumed $r \in \mathbb{Q}$), since the leading coefficient of $P(x)$ is $1$, so $r \in \mathbb{Z}$. Now, we have an integer $r$ s.t. 
$$r^2=3 \tag{1}$$
or $3 \mid r\cdot r$ which (from Euclid's lemma and since $3$ is prime) means $3 \mid r$. In another words, $\exists q\in \mathbb{Z}, q\ne0: r=3\cdot q$. Applying this to (1) we have
$$3\cdot q \cdot 3\cdot q =3 \Rightarrow 3\cdot q^2=1$$
which means $3 \mid 1$ - contradiction. Thus $r \notin \mathbb{Q}$.

On another note, related to your attempt, you need to assume $\gcd(a,b)=1$. Then apply Euclid's lemma against $a^2=3\cdot b^2$ twice, conclude that $\gcd(a,b)\geq 3$ and obtain a contradiction.
