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!

Solution 1:

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.

Solution 2:

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.

Solution 3:

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).

Solution 4:

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.