Show that there is no rational number $x$ such that $x^2=12$.

Question

Show that there is no rational number $x$ such that $x^2=12$.

---

This is an exercise from Rudin, yes. I am posting this to ask you to verify my proof please.

---

Let $x=m/n$ and assume it is in its lowest form. Then $(m/n)^2=12 \,\,\,(*)\,$ and $m^2=12n^2$.

Now observe that $12n^2$ must be even, since any even number times another number is even again. Since the square of an uneven number is always uneven and the square of an even number is always even, we conclude that $m$ is even.

Now let us rewrite $m^2$ as another even number $12 q$ and substitute it into $(*)$: $12q=12n^2$. We can cancel the $12$ to get $q=n^2$. Thus $n^2$ is even and as such $n$ is even. Now we have that both $m$ and $n$ are even, which is in contradiction with $x$ being in its lowest form and as such the desired result is proven.

Answer 1

Your proof doesn't work. Yes, you can say that $m^2=12q$ (of course! Since $m^2=12n^2$ ...), but how do you know $q$ is even just because $m$ is even? All you know is that $12q$ is even, and that is also true if $q$ is odd.

In fact, by your logic, there is also no rational number such that $x^2=4$. Say it is $x=\frac{m}{n}$ in its lowest form. Then $m^2=4n^2$, so $m^2$ is even and thus $m$ is even. We can write $m=4q$ for some $q$ ... and if $q$ is even (again, here is your mistake!) then $q = n^2$ is even, and thus $n^2$ is even and thus $n$ is even, and thus both $m$ and $n$ are even, and thus $\frac{m}{n}$ is not in its lowest form, thus a contradiction is reached, and thus there is also no rational number such that $x^2=4$.

Answer 2

Take $x=m/n$ where $m$ and $n$ are co primes.
Now, $m^2=12n^2$
Now observe that RHS has a factor of $3$ which means n has $3$ as a factor(Since $12n^2$ is square no and $n$ is an integer)
Also that $m$ will have $9$ as a factor since RHS has a factor of $9$ which means $m$ and $n$ have common factor which is $3$.
Which is a contradiction.

Answer 3

Alternatively, note that $\sqrt{12} = 2 \sqrt{3}$ and that $\sqrt{3}$ is irrational. Now, use one of the first exercises in Rudin.

Answer 4

Given $x^2=12$ $$x=\frac mn$$ $$\frac{m^2}{n^2}=12$$ $$12n^2=m^2$$

Now, the prime factorization of $12$ is $2^2\cdot3^1$ and notice that $3$ has odd number of factors. Now lets divide the equation with $3$ on both sides and notice that $3$ divides the left side because it has a factor of $12$. So, $3$ must also divide right side of an equation, $m^2$. The important thing is to notice that if $3$ divides $m^2$ then it also divides $m$. Because if we factor $m^2$ into prime factors, but a square of an integer must have an even number of each factor, so that means $3^2$ must divide $m^2$ , and $3$ must divide $m$

Since we showed that $3^2$ divides the the right side, $3^2$ must divide the left side, but note that there is only one factor of $3$ in $12$, so that means $3$ divides $n^2$. Using the same method we can say that $3$ must divide $n$

Since we showed that $3$ divides both $m$ and $n$, which contradicts the fact that $m$ and $n$ to be relatively prime. Since our assumption leads to a contradiction, there is no rational number $x$ such that $x^2=12$

Answer 5

The fact that $12n^2$ is even doesn't imply $n$ is even: for $n=1$, $12n^2=12$ is surely even. You get tricked by $q=n^2$, but nowhere you find arguments to show that $q$ is even.

---

Suppose there is a rational number $x$ such that $x^2=12$. Then also $y=x/2$ is a rational number; since $x=2y$, we get $4y^2=12$ and therefore $$ y^2=3 $$ Can you go on?

Answer 6

Consider $\wp(x) = x^2 - 12$. By the rational root theorem, if $\wp$ has a rational root $p$, then $$p\in\Big\{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12\Big\}. $$ But, $\wp(p) \neq 0 $ for all possible $p$. Therefore, none of the roots are rationals (and they aren't complex too), so they must be irrationals. $$\therefore \sqrt{12} \in \mathbb{R}\backslash\mathbb{Q} . $$