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

Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.

I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.

share|cite|improve this question
How do you use the inverse of $3/7$ to solve the equation $$\frac37\,x=\frac{11}{23}?$$ And no! The inverse is not the conjugate, but using the conjugate helps! – Jyrki Lahtonen Sep 21 '12 at 7:42
So should I just multiply by the conjugate? In the same way I do to rationalize complex denominators? then I would have 1+3sqrt(2) * (1+3sqrt(2)* 1-3sqrt(2))/ 1-3sqrt(2) – user40105 Sep 21 '12 at 7:49
up vote 4 down vote accepted

Let $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ be the inverse of $1+3\sqrt{2}$, i.e. $(a+b\sqrt{2})(1+3\sqrt{2})=1$. Then $$ 1=a+6b+(3a+b)\sqrt{2}, $$ i.e. $$ 3a+b=0,\ a+6b=1. $$ It follows that $$ a=-\frac{1}{17},\ b=\frac{3}{17}. $$ Now $$ (1+3\sqrt{2})x=1-5\sqrt{2} \iff x=(1+3\sqrt{2})^{-1}(1-5\sqrt{2}), $$ i.e. $$ x=\frac{1}{17}(-1+3\sqrt{2})(1-5\sqrt{2})=-\frac{31}{17}+\frac{8}{17}\sqrt{2}. $$

share|cite|improve this answer
Thank you for the help! – user40105 Sep 21 '12 at 8:06

The inverse is almost never the conjugate. However, it does end up being related to the conjugate. (Why and how?) We can also use the conjugate instead, and avoid having to determine the inverse explicitly. Multiplying both sides of $$(1+3\sqrt{2})x=1-5\sqrt{2}$$ by $1-3\sqrt{2}$ gives us $$-17x=31-8\sqrt{2},$$ from which we see that $$x=-\frac{31}{17}+\frac8{17}\sqrt{2}.$$

share|cite|improve this answer

Hint $\ $ If $\rm\: 0\ne\alpha\bar\alpha = n\in \Bbb Z\:$ then $\rm\: \alpha\, x = \beta\:$ times $\,\bar \alpha\,$ yields $\rm\: n\, x = \bar\alpha\beta, \:$ i.e. $\rm\: x = \dfrac{\beta}\alpha = \dfrac{\bar\alpha\beta}{\bar\alpha\alpha} = \dfrac{\bar\alpha\beta}n$

This is known as rationalizing the denominator. It exploits the fact that every irrational has a rational multiple (its norm), to reduce division by an irrational to division by a rational.

share|cite|improve this answer

Instead of finding inverse, you can directly find(less computation) $x$

Let $x=a+b\sqrt 2$

Then, $(1+3\sqrt 2)(a+b\sqrt 2)=1-5\sqrt 2\implies (a+6b)+(3a+b)\sqrt 2=1-5\sqrt 2$

$\implies a+6b=1$ and $3a+b=-5$.

Solving these equations, we get

$a=-\frac{31}{17}$ and $b=\frac{8}{17}\implies x=-\frac{31}{17}+\frac{8}{17}\sqrt 2$

share|cite|improve this answer

We try to find it of the form $a+b\sqrt 2$ where $a$ and $b$ are two rational numbers. Then we should have $a+6b=1$ and $3a+b=0$, as $\sqrt 2$ is irrational. Then we just have a system to solve.

share|cite|improve this answer

Your Answer


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