# $1+\sqrt{2}$ is a unit in $\mathbb{Q}[\sqrt{2}]$. True or False

I believe that the statement is True, and this is my argument:

Since there exists an element $((1-\sqrt{2})/(1-(\sqrt{2})^2)\in\mathbb{Q}[\sqrt{2}]$ such that their products gives $1$ (multiplicative identity), therefore ( $1+\sqrt{2}$ ) is a unit.

A $unit$ is an element $a\in R|ab=1$ where as $b\in R$. Where $b$ is known as the multiplicative inverse of $a$.

Is my argument reasonable enough? And if there is/are any other possible argument, It would be really helpful to know.

-
$R$ is a ring. Sorry – Ozahm May 21 '14 at 1:57
Your ring is in fact a field. – Mariano Suárez-Alvarez May 21 '14 at 1:59
That's the best argument to make. – blue May 21 '14 at 2:00
Ah... That's a weird way of writing 2 ;-) – Mariano Suárez-Alvarez May 21 '14 at 2:02
Note that $1+\sqrt 2$ is a unit not only in ${\bf Q}[\sqrt 2]$, but also in ${\bf Z}[\sqrt 2]$. – tomasz May 21 '14 at 3:14

As noted in the comments, you have a field, so all non-zero elements are units.

But since you asked: to show that $1+\sqrt{2}$ is a unit, you just have to find an element $b$ such that $(1+\sqrt{2})b = 1$. And you can do $$(1 + \sqrt{2})(-1+\sqrt{2}) = -1 - \sqrt{2} + \sqrt{2} + 2 = 1.$$ And that is it. You don't have to say anything else. It just has to be clear that the $b$ is indeed an element of $\mathbb{Q}[\sqrt{2}]$. And since $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2}: a,b\in \mathbb{Q}\}$, it is very clear that $-1 + \sqrt{2} \in \mathbb{Q}[\sqrt{2}]$.

-
Now why didn't I think of $(-1+\sqrt{2})$ , Thanks @Thomas – Ozahm May 21 '14 at 2:17
Just one question, since $a,b \in Q$, doesn't imply that it has to be in the form of fraction? – Ozahm May 21 '14 at 11:15
@Ozahm: Remember that all integers are rational numbers. $1 = \frac{1}{1} \in \mathbb{Q}$. So no, the elements of $\mathbb{Q}$ don't have to be written as fractions. – Thomas May 21 '14 at 12:31
@Ozahm: It is correct to say that "Any integer is a rational number". In the same way we also say that a rational number is a real number. This is perfectly fine. That said, if you define your rational numbers as equivalence classes of integers, then you could make the argument that you should distinguish. – Thomas May 22 '14 at 23:32
@Ozahm: Right, not every rational number is an integer. Neither is every real number a rational number and so on. – Thomas May 23 '14 at 0:11

Probably this exercise is a warmup for the proof that $\,\Bbb Q[\sqrt{2}]\,$ is a field. The key idea is simple: to invert $\,w = 1+\sqrt{2}\,$ simply rationalize the denominator of $\,\dfrac{1}{1+\sqrt 2},\,$ i.e. multiply the numerator and denominator by the conjugate $\,w' = 1-\sqrt{2}\,$ to force the denominator to be rational $\,ww' = r.\,$ Being a nonzero rational, $\,r\,$ is invertible, which yields the sought inverse $\, \dfrac{1}w = \dfrac{w'}{ww'}= r^{-1} w'$

Thus, by taking norms $\,w\to ww',$ this method transforms the problem of inverting a quadratic irrational to the simpler problem of inverting a rational. Since the same method works for any irrational $\,w = a + b\sqrt{2} \in \Bbb Q[\sqrt{2}]\,$ we infer that $\Bbb Q[\sqrt{2}]$ is a field. Note that in this general proof it is crucial to show that the denominator $\,ww'= a^2-2b^2 \ne 0.\,$ But if it $=0\,$ then $\, 2 = (a/b)^2\Rightarrow\!\Leftarrow$

-