Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$R$ 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
up vote 4 down vote accepted

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 $-1 + \sqrt{2} \in \mathbb{Q}[\sqrt{2}]$.

share|cite|improve this answer
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 $\, 2 = (a/b)^2\Rightarrow\!\Leftarrow$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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