Show that sqrt(3/2) is irrational. What I have so far is (sqrt(3/2))^{2}=a^{2}/b^{2}, so than 3/2=a^{2/}b^{2}. So how does this prove irrational
$\dfrac{3}{2} = \dfrac{a^2}{b^2}$ is the same as saying:
$2a^2 = 3b^2$, for integers $a,b$.
Since $2$ divides $2a^2$, it must divide $3b^2$, and since $2$ does not divide $3$, it must be that $2$ divides $b^2$, and thus $2$ divides $b$.
So suppose $k > 0$ is the largest power of $2$ that divides $b$, so that $b = 2^km$ so that $b^2 = 2^{2k}m^2$.
Thus $2a^2 = 3(2^{2k}m^2)$, so that $a^2 = 2^{2k-1}(3m^2)$. Now $k > 0$, so $2k - 1 \geq 1 > 0$. Thus $2$ divides $a^2$, and thus $a$.
If $a = 2^ts$, where $t > 0$ is the largest power of $2$ that divides $a$, then:
$a^2 = 2^{2t}s^2 = 2^{2k-1}(3m^2)$.
Now $s^2$ is an integer, and $2$ does not divide $s^2$, so we conclude that:
$s^2 = 2^{2k - 2t - 1}(3m^2) = 3m^2$, that is: $2k - 2t - 1 = 0$. But this is the same as: $2(k - t) = 1$, and $1$ is not divisible by $2$.