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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.

  • 0
    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)2012-09-21

5 Answers 5

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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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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. in fraction language $\rm\: x = \dfrac{\beta}\alpha = \dfrac{\bar\alpha\beta}{\bar\alpha\alpha} = \dfrac{\bar\alpha\beta}n$

This is known as rationalizing the denominator. It exploits the fact that every irrational algebraic $\alpha$ has a "simpler" (i.e. rational) multiple (its norm), to reduce division by an irrational to division by a rational. This is a special case of the general method of simpler multiples.

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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.