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I would like critiques on correctness, conciseness, and clarity. Thanks!

Proposition: There is no rational number whose square is 12

Proof: Suppose there were such a number, $a = \in \mathbb{Q}$ s.t. $a^2 = 12$.

This implies that $\exists$ $m, n \in \mathbb{Z}$ s.t. $\frac{m^2}{n^2} = 12.$ Assume without loss of generality that $m,~ n$ have no factors in common.

$\Rightarrow m^2 = 12n^2$.

This implies that $m^2$ is even, and therefore that $m$ is even; it can thus be written $2k = m$ for some $k \in \mathbb{Z}$.

Thus $m^2 = 12n^2 $

$\Rightarrow 4k^2 = 12n^2 $

$\Rightarrow \frac{k^2}{3} = n^2$

Because $n^2$ is an integer, it is clear that $3$ divides $k^2$ which imples that $k$ has $3$ or $\frac{k}{n}$ has a factor (because $\frac{k^2}{n^2}= 3$)

Suppose that the former is true, and $3$ is a factor of $k$. Then $k = 3j$ for some integer j, which implies that $(3j)^2 = 3n^2$

$\Rightarrow 9j^2 = 3n^2 $

$\Rightarrow n^2 = 3j^2 $

$\Rightarrow n^2 = \frac{k^2}{n^2}j^2$

$\Rightarrow k = \frac{n^2}{j}$ but this implies that $j$ divides $n^2$, but $j$ divides $m$, and by initial assumption $n$ and $m$ have no factors in common, so this is a contradiction.

Suppose now that $\frac{k}{n}$ is a factor of k. Then $k = \frac{k}{n}j$ for some integer $j$. Then $(\frac{k}{n}j)^2 = 3n^2$ which implies that $3j^2 = 3n^2 \Rightarrow j^2 = n^2 \Rightarrow j = n$. But this means that $n$ divides $m$, which again is a contradiction. Thus any rational representation of the number whose square equals $12$ leads to a contradiction and this number must therefore have no rational representation.

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    $\begingroup$ Looks fine to me. A shorter version: $\sqrt{12}=2\sqrt{3}$ belongs to $\mathbb{Q}$ iff $\sqrt{3}$ belongs to $\mathbb{Q}$, but that is impossible by the unique factorization theorem and the primality of $3$. $\endgroup$ Jan 26, 2017 at 18:45
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    $\begingroup$ Correct thinking, but you can get away with a lot less work and writing.:) To begin with, write: $$\sqrt{12} = 2 \sqrt{3}.$$ This means, all you need is to show that $\sqrt{3}$ is irrational. $\endgroup$
    – avs
    Jan 26, 2017 at 18:46
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    $\begingroup$ Looks good. An alternate approach is to look at the polynomial $x^2-12$ and use the rational root theorem to show that any square root of $12$ would have to be a divisor of $12$. $\endgroup$
    – lulu
    Jan 26, 2017 at 18:47
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    $\begingroup$ 3 divides k^2 which imples that k has 3 or k/n has a factor I don't understand the or part. Since $3 \mid k^2$ it follows that $3 \mid k\,$ and there is no or case. $\endgroup$
    – dxiv
    Jan 26, 2017 at 18:52
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    $\begingroup$ @BenL For any integers $a=bc \implies b \mid a$ and $c \mid a\,$. That's by definition, and requires no additional assumptions. You have $k^2 = 3 n^2\,$, therefore $3 \mid k^2\,$. $\endgroup$
    – dxiv
    Jan 26, 2017 at 18:59

3 Answers 3

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Proof.   Assume $\sqrt{12} \in \mathbb{Q}$ is rational, then it can be written as $\sqrt{12}=\cfrac{m}{n}$ with $m,n \in \mathbb{Z}$ coprime.

Squaring the equality gives $m^2 = 12 n^2 = 3 \cdot 4 \cdot n^2\,$. Therefore $3 \mid m^2 = m \cdot m$ and, since $3$ is a prime, it follows by Euclid's Lemma that $3 \mid m\,$.

Then $m = 3k$ for some $k \in \mathbb{Z}$ and substituting back gives $9 k^2 = 12 n^2 \iff 3 k^2 = 4 n^2\,$. Therefore $3 \mid 4 n^2$ and, since $3 \not \mid 4$ it follows that $3 \mid n^2$ then, again by Euclid's Lemma, $3 \mid n\,$.

But $3 \mid m$ and $3 \mid n$ contradicts the assumption that $m,n$ are coprime, so the premise that $\sqrt{12} \in \mathbb{Q}$ must be false, therefore $\sqrt{12}$ is irrational.


Critique of the posted proof.

Proof: Suppose there were such a number, $a = \in \mathbb{Q}$ s.t. $a^2 = 12$.

This implies that $\exists$ $m, n \in \mathbb{Z}$ s.t. $\frac{m^2}{n^2} = 12.$ Assume without loss of generality that $m,~ n$ have no factors in common.

$\Rightarrow m^2 = 12n^2$.

So far so good.

This implies that $m^2$ is even, and therefore that $m$ is even;

The fact that $2 \mid m^2 \implies 2 \mid m$ may sound obvious, but still needs some justification. You could argue by contradiction, or use Euclid's Lemma.

it can thus be written $2k = m$ for some $k \in \mathbb{Z}$.

Thus $m^2 = 12n^2 $

$\Rightarrow 4k^2 = 12n^2 $

Correct. As an observation, $k^2 = 3 n^2$ just eliminated the perfect square factor of $4$ and reduced the problem to proving that $\sqrt{3}$ is irrational.

$\Rightarrow \frac{k^2}{3} = n^2$

Because $n^2$ is an integer, it is clear that $3$ divides $k^2$ which imples that $k$ has $3$

You should generally avoid fractions where they are not necessary. The previous line gave $k^2 = 3 n^2\,$, which directly implies that $3 \mid k^2\,$.

or $\frac{k}{n}$ has a factor (because $\frac{k^2}{n^2}= 3$)

This makes no sense, and it is in fact not needed to complete the proof.

Suppose that the former is true, and $3$ is a factor of $k$. Then $k = 3j$ for some integer j, which implies that $(3j)^2 = 3n^2$

$\Rightarrow 9j^2 = 3n^2 $

$\Rightarrow n^2 = 3j^2 $

The proof is complete right here at this point, if you just note that the last equality implies that $3 \mid n^2\,$, and therefore $3 \mid n$ which contradicts the assumption that $m,n$ are coprime.

[ rest of post snipped ]

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  • $\begingroup$ My main concern is proving that $3 | n^2 \Rightarrow 3|n$. I'm familiar with Euclid's Lemma, but I'm not clear whether it's legit in the context of a real analysis course. Thanks though, this was very helpful! $\endgroup$
    – BenL
    Jan 26, 2017 at 20:01
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    $\begingroup$ @BenL Euclid's Lemma is of course true, regardless of context. Here, $n$ is an integer, and $n^2=n\cdot n$ is a product of two integers (which happen to be equal). $3$ is a prime which divides that product, so it must divide (at least) one of the factors. Ergo, $3 \mid n\,$. $\endgroup$
    – dxiv
    Jan 26, 2017 at 20:03
  • $\begingroup$ @BenL You don't need Euclid's Lemma for that. mod $3\!:\ n\not\equiv 0\,\Rightarrow\,n\equiv \pm1\,\Rightarrow\, n^2\equiv 1\,\Rightarrow\,3\nmid n^2.\ $ But you do need Euclid (or equivalent) if you want to prove that for *all* primes $p$ since then you cannot brute force check all possible residues as above. $\endgroup$ Jan 26, 2017 at 20:03
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"This implies that ∃ m,n∈Z s.t. m2n2=12. Assume without loss of generality that m, n have no factors in common."

I, personally, would not argue "without loss of generality" . $q \in \mathbb Q$ is defined as $q = \frac mn$ for some relatively prime integers. So we declare them to have no factors in common by fiat-- not merely by lack of loss of generality. (It's not that big of an issue.)

"This implies that $m^2$ is even, and therefore that m is even"

I'd accept this but dxiv very much has a point, that it should require some justification. I personally would simply put it in more definitive language. I'd say: "Therefore $2|m^2$ and, as $2$ is prime, $2|m$". This could require a little justification in that all numbers have a unique prime factorization so that for prime $p$ we know if $p|ab$ then $p|a$ or $p|b$ so if $p|m^2$ then $p|m$ or $p|m$.

"Because $n^2$ is an integer, it is clear that 3 divides $k^2$ which implies that k has 3 or $\frac kn$ has a factor ".

As $\frac {k^2}3$ is a integer, it implies $3|k^2$. Period. That always happens. That any thing else may happen doesn't matter. It may have $k/n$ as a factor or it may have $7$ as a factor. Or it may not. Those don't matter.

Also, if $\frac kn$ is an integer at all, then it is trivial that $\frac kn$ is a factor $k$ whether or not $k^2/3$ is an integer or not. And if $\frac kn$ is not an integer then the statement $\frac kn$ is a factor of $k$ is meaningless.

And if $k/n$ is an integer, then $n|m = 2k$ and as $n,m$ have no factor in common then $n = 1$. (Which would mean $\sqrt{12} = 2\sqrt{3}$ is an integer which is easy to verify is not the case).

"if the former ($3|k$) then .... "

All that is just fine and the rest is unneeded.

But the rest is a bit of a mess.

"Suppose now that $k/n$ is a factor of $k$" Again, this is trivial if $n|k$ and is meaningless if $n \not \mid k$.

And we can rule out $n|k$ as that would imply $\sqrt{12} = m/n = 2k/n$ is an integer. Which

But beware. This is true of all numbers and nothing relevant is likely to arise. And it doesn't:

" Then $(\frac knj)^2=3n^2$ which implies that $3j^2=3n^2$"

Actually, no, it implies $3j^2 = 3n^4$. And thus we get $j = n^2$ (we can assume $j$ is positive this time as we can assume $k$ and $n$ are positive).

"But this means that n divides m, which again is a contradiction."

Actually it's not a contradiction if $n = 1$.

But this isn't a contradiction that needed to be reached. $k/n$ is a factor of $k$ only makes sense if $n|k$ which would imply $n|m = 2k$.

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See this proof that if $n$ is not a perfect square then $\sqrt{n}$ is irrational:

Follow-up Question: Proof of Irrationality of $\sqrt{3}$

The proof starts by saying that if $n$ is not a perfect square then there is a $k$ such that $k^2 < n < (k+1)^2$. The proof breaks down if $k^2 = n$.

Note that this proof does not use divisibility.

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  • $\begingroup$ It is misleading to say the proof does not use divisibility. It essentially uses the division algorithm to achieve descent on denominators. I explain this further in this May 20, 2009 sci.math post, where I highlight the beautiful view of irrationality proofs in terms of Dedekind's conductor ideal. $\endgroup$ Jan 26, 2017 at 21:18
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    $\begingroup$ This has nothing to do with the question asked, so it should be a comment, not an answer. $\endgroup$ Jan 26, 2017 at 21:26

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