Proof that $\sqrt{3/2}$ is irrational I'm taking a course in Real Analysis for the first time and I'm being asked to prove that $$\sqrt\frac{3}{2}$$ is irrational.
Here's what I have so far:
Assume $\sqrt\frac{3}{2}$ IS rational. 
This means that $\exists$ $m\in Z, n\in Z,  n \neq 0$ such that $\frac{m}{n}$ = $\sqrt\frac{3}{2}$ and gcd(m, n) = 1.
$\frac{m}{n}$ = $\frac{\sqrt3}{\sqrt2}$ $\Rightarrow$ $\frac{m^2}{n^2}$ = $\frac{3}{2}$ $\Rightarrow$ $2m^2$ = $3n^2$ $\Rightarrow$ $3n^2$ is an even number $\Rightarrow$ $\sqrt3n$ is an even number
However, $\sqrt3n$ cannot be an even number since that would imply that $\exists k \in Z$ such that $2k = \sqrt3n$, which implies that $\frac{2}{\sqrt3}k = n$.
It is intuitively obvious that no such k exists, but how would I rigorously prove such a thing? 
If I could say that no such k exists, I could then say that $\sqrt3n$ is not an even number and then therefore by contradiction $$\sqrt\frac{3}{2}$$ is irrational right?
 A: In your proof,concluding $\sqrt{2}n$ as even is not correct. That can be a non-integer.
Thus rational root theorem for $2x^2-3$ will help.
A: $$\frac{m}{n} = \sqrt{\frac{3}{2}}$$
means $$2 m^2 = 3 n^2$$.
The left side has an odd number of factors of the prime $2$ while the right side as an even number of factors of the prime $2$ (or possibly zero such factors).
By the fundamental theorem of arithmetic (unique prime factorization) this can never occur.
QED.
A: Do you know that every number has a unique prime factorization?  Do you know Euclid's lemma that if $p$ is  prime and $p|ab$ then either $p|a$ or $p|b$ or both?
If $2m^2 =3n^2$ then $2|3$ or $2|n$ or $2|n$.  As $2\not \mid 3$ we have $2|n$.  ANd $3|2m^2$ so $3|2$ or $3|m$ or $3|m$ so $3|m$.  So $3|m$.  Can you see how that leads to a contradiction.
If you don't know Euclid's lemma you can sit work with if $a$ is even then $a^2$ is even and if $a$ is odd then $a^2$ is odd to get a contradiction.
A: The fundamental theorem of arithmetics of positive rational numbers states:
THEOREM For every positive rational number $\ x\ $ there is exactly one finite set $\ F_x\ $ of primes, and exactly one integer function $\ f_x:F_x\rightarrow\mathbb Z\setminus\{0\}\ $ such that
$$ x\ =\ \prod_{p\in F_x}p^{f_x(p)} $$
However,
$$  y=x^n\quad \Leftrightarrow\quad f_y\ =\ n*f_x $$
Now you, by interpreting $\ \frac 32\ $ as $\ y\ $ above,
we see that $\ x:=\sqrt\frac 32\ $ cannot be a rational number.
A: If $2m^2 = 3n^2$
then
considering the prime factorizations
of $m$ and $n$,
3 occurs an even number of times
(which may be zero)
 on the left
and an odd number of times on the right,
and similarly for 2.
Since this is a contradiction (twice!),
the equation can not hold.
This generalizes nicely to show that if
$um^2=vn^2$
with $(u, v) = 1$
then $u$ and $v$
must both be squares.
