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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.

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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
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5 Answers

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. $$ x=\frac{1}{17}(-1+3\sqrt{2})(1-5\sqrt{2})=-\frac{31}{17}+\frac{8}{17}\sqrt{2}. $$

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Thank you for the help! –  user40105 Sep 21 '12 at 8:06
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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 $$x=-\frac{31}{17}+\frac8{17}\sqrt{2}.$$

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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.

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

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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.

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