Critiques on proof showing $\sqrt{12}$ is irrational. My only exposure to proofs was in a math logic class I took in University. I was wondering if my attempt at proving that $\sqrt{12}$ is irrational is OK.
$$\Big(\frac{m}{n}\Big)^2 = 12$$
$$\Big(\frac{m}{2n}\Big)^2 = 3$$
$$m^2=3*(2n)^2$$
This implies $m$ is even and so $n$ must be odd.
The problem can be reduced to:
$$\Big(\frac{p}{n}\Big)^2 = 3$$ 
Because $n$ is odd, $p^2$ is odd, so $p$ is odd.
This implies:
$$4a+1 = 3(4b+1)$$
$$4a - 12b = 2$$
$$2a - 6b = 1$$
I'm kind of stuck at this point. I know that this can't be true but I don't know how to state it. Any critiques or suggestions? Thanks!
 A: If you know that $\sqrt{3}$ is irrational then we have easier method as follows:
If $\sqrt{12}$ want to be rational so it should be at form $\frac{m}{n}$ but we know $\sqrt{12}=\sqrt{2^{2}.3}=2\sqrt{3}$ so $\sqrt{3}=\frac{m}{2n}$ and should be rational too which is contradiction. So $\sqrt{12}$ can not be rational.
And if you don't know $\sqrt{3}$ is irrational you can prove it as usual way that is described by others and then use this method to conclude $\sqrt{12}$ is irrational according to $\sqrt{3}$ is irrational.
A: Yet another simple way to show this!
The question is equivalent with the irrationality of $\sqrt3$, which is the same as showing that there is no rational solution to $x^2-3=0$. By Eisenstein's criterion, the polynomial is indeed irreducible over $\mathbb Q$. So this finishes the proof.
A: To do it directly ignore the prime 2 altogether, and go for the prime 3 which appears to an odd power in the equation $$m^2=12n^2$$ (assume lowest terms)
The right hand side is divisible by 3, so the left hand side must be divisible by 3, so we must have $m=3r$. Our equation becomes $$9r^2=12n^2 \text{ or }3r^2=4n^2$$
Now we see similarly that $n$ must be divisible by 3, contradicting our lowest terms assumption.

To show that $3\mid m^2\implies 3\mid m$ let $m=3r+d$ with $d\in \{-1,0,1\}$ then $m^2=9r^2+6dr+d^2=3(3r^2+2dr)+d^2$ and this is not divisible by $3$ if $d^2=1$ so we must have $d=0$.
This establishes the result for the prime $3$ using the division algorithm and cases (which is pretty much equivalent to using the concepts odd and even when dealing with the prime $2$).
This can be done by cases for small primes, but to prove the general theorem for all primes does require more complex machinery as Pete Clark notes in the comments.
A: If you’re willing to use the Fundamental Theorem of Arithmetic, which says that the decomposition of any nonzero integer as a product of primes is unique, then this proof, and all others for irrationality of $r$-th roots, drops right out.
Write $m^2=12n^2$. This contradicts FTA because there are evenly many $3$’s on the left but oddly many on the right.
A: You made it too complicated in my opinion.
First we show that a rational number (different from $0$) times an irrational is irrational.
Proof by contradiction:
Let $x\in \mathbb{R}\setminus\mathbb{Q}$, $a,c\in\mathbb{Z}\setminus\{0\}$, $b,d\in \mathbb{N}, d \neq 0$.
$$x\cdot \frac{a}{b}=\frac{c}{d} \iff x=\frac{bc}{ad},$$ so $x$ would be rational.
Use
$$\sqrt{12}=\sqrt{4\cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2 \cdot \sqrt{3}$$
So $\sqrt{12}$ irrational $\iff \sqrt{3}$ is irrational.
Now, we show  $3|p^2 \implies 3|p$: We know 3 is prime so with Euclid's lemma we have
$$3|p^2 \implies 3|p \lor 3|p \implies 3|p$$
To prove that $\sqrt{3}$ is irrational, you derive a contradiction:
$$\sqrt{3}=\frac{p}{q}\iff 3q^2=p^2 \implies \exists k \in \mathbb{N}: 3k =p$$
$$q^2=3k^2$$
So $q$ and $p$ both have the divisor $3$. Now there are two different ways to use this information. The first proceeds by contradiction, assuming that $p$ and $q$ don't have a common divisor, but as you can show they always have the divisor $3$, you can't write $\sqrt{3}$ as a fraction of numbers without common divisors. This is the more elegant way in my opinion.
The other way is to show that both $p$ and $q$ can't be finite, because
by repeating this argument we see $3^n|p$ for all $n \in \mathbb{N}$ and similarly for $q$.
But because of $p,q\in \mathbb{N}$, $3^n> 1+2n$ and the Archimedean principle we get a contradiction.
A: You could have done this way too. \ 
Assume $\displaystyle\frac{m}{n}$ is written in its simplest form. Then
$m^2=2(6n^2)\Rightarrow m$ is even. Substituting $m=2k\Rightarrow n$ is even. Thus both $m$ and $n$ have a common factor of 2 contradicting you statement that $\displaystyle\frac{m}{n}$ is not in its simplest form.
A: Another angle on this is to prove that no integer is the square of a ratio. All squares are the squares of integers. Then, since $12$ does not have an integer square root, its square root cannot be rational, either.
To show that no integer is the square of a ratio, suppose $(\frac{n}{m})^2 = k$ where $m, n$ and $k$ are integers, $n/m$ is in lowest terms, $m\neq 1$, and all are integers. But that situation is impossible.
$$(\frac{n}{m})^2 = k$$ 
$$\frac{n\cdot n}{m\cdot m} = k$$
If $n/m$ is in lowest terms, as we assumed, that means that $m$ does not divide $n$. This implies that $m$ and $n$ have completely distinct prime factors. Which implies that no multiple of $m$ divides any multiple of $n$, because multiples of a number are just combinations of its prime factors.  Thus $\frac{n\cdot n}{m\cdot m}$ cannot be an integer, unless $m = 1$ which we ruled out.
Therefore, $\sqrt 12$ cannot be a ratio. It must either be an integer ($12$ is a square), or else irrational. Our goal is therefore to show that $12$ isn't a square.
Observe that $12$ factors into $2\cdot 2\cdot 3$. A square has prime factors that are all of even duplicity, so that these factors can be divided into two identical groups. There is no way to separate the factors $2\cdot 2\cdot 3$ into two identical groups because $3$ occurs only once. (By contrast, consider $36 = 2\cdot 2\cdot 3\cdot 3$ whose factors are each of duplicity 2, and so can be split into two groups $(2\cdot 3)(2\cdot 3) = 6\cdot 6$.)
Since $\sqrt 12$ isn't an integer, and no integer has a square root which is a ratio, $\sqrt 12$ must be irrational.
A: I'll assume you know how to show that $\sqrt 2$ is irrational, in the same way you can show that for any prime $p$ then  $\sqrt p$ is irrational, and we know that if $a$ is irrational and $b\ne 0$ is rational then $ab$ is irrational then we have$$\sqrt12=2\sqrt3$$now $2\ne 0$ is rational and 3 is prime which implies that $\sqrt3$ is irrational thus we have $\sqrt12=2\sqrt3$ is irrational.
A: Just call $\sqrt {12}$ a rational number. What can you say about it? $12$ must be a square of rational number.
$12= k^2$ where ($k =\dfrac{p}{q}$)
GCD$(p,q)=1$
$12= 2^2 \cdot 3=\dfrac{p^2}{q^2} \implies 3= \dfrac{p^2}{q^2 \cdot 2^2} \implies 3$ is a square. Which implies it has odd number of divisors, but it doesn't.
A: The following is different in style.  It avoids the use of the unique factorization property that arguments about the divisibility by the prime $3$ use.
Assume $\sqrt{12}$ is rational.
Choose among all equivalent fractions the one with the least positive denominator; let it be $m/n$. 
Thus $m^2 = 12 n^2$, and we have
$$9 n^2 < m^2 < 16 n^2$$
$$3n < m < 4 n$$
$$0 < m -3n < n$$
Now
$$\begin{align}
\left({12n - 3m \over m - 3n}\right)^2 &= { 9(16n^2 -8mn+m^2)\over m^2-6mn+9n^2}\\
 &= { 9(16n^2 - 8mn + 12n^2)\over 12n^2 - 6mn + 9n^2} \\
 &= { 36(7n-2m)n\over 3(7n-2m)n} = 12\,,
\end{align}$$
which is to say that
${12n - 3m \over m -3n}$ equals $\sqrt{12}$ 
and has a lesser denominator.
This is impossible since we chose $m/n$ to be in least terms.
QED

In case you're wondering how to find the fraction, it's from the continued fraction expansion of $\sqrt{12}$.  One has
$$\begin{align}
x=\sqrt{a^2+b}&=a+{b \over 2a + \displaystyle{b \over 2a + \displaystyle {b \over 2a + \cdots}}} \\
 &= a + {b \over a + x}
\end{align}$$
In this case $a=b=3$ and if $x = M/N = m/n$ are two square-roots of $12$, then
$$
{m\over n} = 3 + {3 \over 3 + {M \over N}} = {12 N + 3M \over 3N + M}
$$
Now solve the system
$$ m = 12N + 3M, \quad n = 3N + M$$
for $M$, $N$, and get the new fraction in terms of $m$, $n$:
$$
{M \over N} = {12n - 3m \over m -3n}
$$
You then have to check that it's still a square-root of $12$ and the denominator is positive and has decreased.
The procedure works in general as long as $a^2+b$ is not a square (and $a$ and $b$ are positive).
