Standard exercise is to show $\sqrt{2} \notin \mathbb{Q}$ (e.g. Wikipedia). There are examples on Math.SE such as [1, 2, 3, 4].

If we adjoin an element to $\mathbb{Q}$ does the same proof by contradiction work? I'd like to show $\sqrt{2} \notin \mathbb{Q}(i)$ where $x = i$ is a solution to $x^2 + 1 = 0$. Perhaps I could begin the same way. We are solving:

$$ p^2 = 2 q^2 $$

where $p, q \in \mathbb{Z}[i]$. We certainly have $2 = (1+i)(1-i)$ so that $2$ does factor in this new ring, but these two numbers are relatively prime. We have:

$$ \sqrt{1 \pm i } \notin \mathbb{Q}(i)$$

And therefore, their product does not belong in that field as well. Does that look correct? Is has correct ideas, but I don't think the logic is presented correctly.

There are may solutions to this problem, so I've also indicated a certain line of proof I'm trying to follow, using infinite descent. I'm asking, Does the descent argument we typically use over $\mathbb{Z}$ carries over to $\mathbb{Z}[i]$ ?

Alternatively, we can write any element of $\mathbb{Q}(i)$ as $a + bi$ with $a,b \in \mathbb{Q}$. If we have $(a+bi)^2 = 2$ Then $a^2 - b^2 = 2$ and $2ab = 0$. Then necessarily $b = 0$ and $a^2 = 2$. This only uses descent over $\mathbb{Q}$, and we never use that $2$ factorizes into $(1+i) \times (1-i)$.

In contrast: Proving $\sqrt{2}\in\mathbb{Q_7}$?

  • $\begingroup$ Consider the field $\mathbb R$ which does not contain $\pm i$ but does contain $i\times -i$. So you have made an unjustified assumption in your proof. $\endgroup$ – Mark Bennet Dec 15 '17 at 15:49
  • $\begingroup$ perhaps I could argue that $1+i$ divides $p$ ? And also $1-i$ and therefore again $2$ divides $p$. $\endgroup$ – cactus314 Dec 15 '17 at 15:52
  • $\begingroup$ I suppose you can use the same proof. Suppose $a + bi \in \Bbb Q(i)$ has the property that $(a + bi)^2 = 2$. Then $a^2 + 2abi - b^2 = 2$. Comparing the coefficients, it must be that $a = 0$ or $b = 0$, so either $a^2 = 2$ or $-b^2 = 2$. Now use the usual proof that this can't exist in $\Bbb Q$. $\endgroup$ – ÍgjøgnumMeg Dec 15 '17 at 15:59

Suppose $\sqrt2=a+ib\in\Bbb{Q}[i].$ Then $a^2+b^2=2$ and clearing denominators we can consider the Diophantine equation $p^2+q^2=2r^2.$ Since RHS is even both $p$ and $q$ has to be same parity. Moreover


Then using primitive Pythagorean triples we have $p=(m+n)^2-2n^2,$ $q=(m-n)^2-2n^2$ and $r=m^2+n^2$ for some $m,n\in\Bbb{Z}.$

If $p=0$ or $q=0,$ then by we have a contradiction $\sqrt2\in\Bbb{Q}.$
Otherwise $b\neq 0,$ which is also contradiction as $\sqrt2\in\Bbb{R}.$


We want to show that $\mathbb Q(i)$ does not contain a root to $x^2+2=0$.

To do this ione possible route is to this we work in $\mathbb C$ and notice that $\mathbb Q(i)$ must contain $\{a+bi| a,b\in \mathbb Q\}$.

If we can show $\{a+bi| a,b\in \mathbb Q\}$ is a field then clearly we must have $\mathbb Q(i)=\{a+bi| a,b\in \mathbb Q\}$.

Clearly it is closed under addition,multiplication and additive inverses. Recall that $(a+bi)^{-1}=\frac{a-bi}{a^2+b^2}$ which is also of the desired form.

Thus all we have to prove is that $(a+bi)^2\neq -2$ for $a,b\in \mathbb Q$.

Clearly $\sqrt{2}i$ and $-\sqrt{2}i$ are not of this form. (One can also prove that $(a+bi)^2=-2$ has no solution by hand)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.