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Let's assume that $12 = (\frac{p}{q})$, where p,q $\in$ $\mathbb{R}$ and $p$ and $q$ are coprime.

Then we have,

$(\frac{p^2}{q^2})= 12^2 = 144.$

So,

$p^2 = 144*q^2$

and $p^2 = 2*(72)*q^2.$

This implies that p is even.

Then,

$p=(2k)$

So,

$(2k)^2=2*(72)*q^2$

$4k^2 = 2*(72)*q^2$

$2k^2 = 72q^2$

Thus,

$k^2 = 36q^2$

so,

$k=6q$

then $(\frac{p}{q}) = (\frac{12q}{q})$,

which contradicts p and q being coprime. Therefore 12 is irrational. QED

I was wondering if the proof I've provided is sound.

Skm
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    This is hard to follow. You want $\frac pq=\sqrt {12}$, yes? Thus $\frac {p^2}{q^2}=12$, not $144$. and surely you mean $p,q\in \mathbb N$. – lulu Sep 02 '17 at 20:44
  • Further, it amounts to proving the same assertion for $3$, since $12=2^2\cdot3$. – Bernard Sep 02 '17 at 20:46
  • Hmm... of course, one might have an urgent desire to prove the irrationality of the square-root of any rational not being a square, one by one (it's a countable set, after all), but... will there be enough spectators finding that thrilling? –  Sep 02 '17 at 20:49
  • A general case https://math.stackexchange.com/questions/2126586/for-which-natural-numbers-n-is-sqrt-n-irrational-how-would-you-prove-your/2126663#2126663 – rtybase Sep 02 '17 at 21:21
  • Um... "therefore 12 is irrational" "I was wondering if the proof I've provided is sound." Uh.... $12 = \frac {12}{1}$ is rational so... obviously not. – fleablood Sep 02 '17 at 23:07

3 Answers3

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Assume otherwise. then there exists $p,q\in \Bbb N$ such that $mcd\{p,q\}=1$ and $\frac{p}{q}=\sqrt{12}.$ From here $\frac{p^2}{q^2}=12$ which is equivalent to $p^2=12q^2=3(2q)^2.$ From here we deduce that $3|p,$ so $p=3k$ for some natural number $k.$ The equation is now $9k^2=3(2q)^2,$ or after simplify, $3k^2=(2q)^2.$ Now we deduce that $3|2q,$ and hence $3|q.$ But this is a contradiction because $3|p,$ so that $mcd\{p,q\}\geq 3>1.$ Hence $\sqrt{12}$ is irrational.

John D
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  • I really should have double checked my proof since the purpose was to get it critiqued. Thanks for understanding what I meant, though. – Skm Sep 02 '17 at 23:18
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Well, we know $\sqrt{12} = 2\sqrt 3$, and we know that $2$ is rational while $\sqrt 3$ is irrational (proof here). We also know that the product of an irrational number and a non-zero rational number is irrational (proof here).

Thus $\sqrt{12}$ is irrational.

gen-ℤ ready to perish
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Let's assume that $12=\frac {p}{q}$, where $p,q ∈ R$ and $p$ and $q$ are coprime.

Okay, that would have to be either $p = 12;q=1$ or $p=-12; q=-1$.

Then we have, $(p^2q^2)=12^2=144.$

Yep $(\pm 12)^2(\pm 1)^2 = 144$.

That is true.

So,$p^2=144∗q^2$ and $p^2=2∗(72)∗q^2$. This implies that $p$ is even.

Yes, that is all true. $p=\pm 12$ is even and $(\pm 12)^2 = 2*(72)*(\pm)^2$.

Then, $p=(2k)$

Yes, $k = \pm 6$ and $p = \pm 12 = 2*(\pm 6)$.

So $(2k)^2=2∗(72)∗q^2$

$4k^22=2∗(72)∗q2 > > $2k^2=72q^2$ > > Thus, > > $k^2=36q^2$ > > so, > > $k=6q$

Yes, $k =\pm 6$ is equal to $6q$ as $q = \pm 1$.

then $\frac pq= \frac {12q}{q}$, which contradicts $p$ and $q$ being coprime.

No, it doesn't $\frac {12q}q$ may not be in lowest terms. Even if you did not know that $q =\pm 1$ this wouldn't conclude anything.

What you want to say is $q|p = 12q$ so $p$ and $q$ are not relatively prime. But you would be wrong as $q = \pm 1$.

What you have proven is that $p = 12q$ (which you knew at the very beginning. so $q|p$ and $\gcd(p,q)= q$. So $p$ and $q$ are relatively prime if and only if $q = \pm 1$. Which is not only possible, but inevetible.

Therefore $12$ is irrational. QED

You do realize that $12$ is rational (it's an integer and all integers are rational) and that that was never in question.

I was wondering if the proof I've provided is sound.

Since it proved that $12$ was irrational, it obviously is not.

Not you need to prove that if $r^2 =(\frac pq)^2 = 12$ then $r$ is impossible to be rational. Not if $r = \frac pq =12$ that $r$ is impossible to be rational.

You attempted to prove the entirely wrong thing. Proved the trivial result that $12$ was rational, and declared it to be irrational.

fleablood
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