How can I find The Multiplicative Inverse of $1+\sqrt{2}$?

Question

I am doing contemporary abstract algebra and am working in an integral domain. I have found it necessary to compute the multiplicative inverse of $1+\sqrt{2}$; I know such the definition of a multiplicative inverse and that one should exist in $\mathbb Q[\sqrt{2}]$, but do not know how I would go about expressing it in any form simpler than $\frac{1}{1+\sqrt{2}}$. How can I determine a simpler way to express the value of $\frac{1}{1+\sqrt{2}}$?

Answer 1

You want to find an element $x = a+b\sqrt{2}, a, b \in \mathbb{Z}$ such that: $x(1+\sqrt{2})=1 \to x = \dfrac{1}{1+\sqrt{2}}=-1+\sqrt{2}$

Answer 2

The complex numbers which have finite multiplicative order are called roots of unity. The same is true in any ring. Roots of unity in a subring of $\Bbb C$ have absolute value $1$ (note absolute value is defined for all complex numbers), but $1+\sqrt{2}$ does not have absolute value $1$, hence it cannot be a root of unity and have finite order.

But $1+\sqrt{2}$ is a unit in $\Bbb Z[\sqrt{2}]$ (the smallest unital subring that contains $1+\sqrt{2}$) and hence in any subring of $\Bbb C$ that contains $1+\sqrt{2}$. One computes the following:

$$\frac{1}{1+\sqrt{2}}=\frac{1}{1+\sqrt{2}}\cdot\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{1-\sqrt{2}}{1-\sqrt{2}^2}=\frac{1-\sqrt{2}}{-1}=-1+\sqrt{2}. $$

This procedure is known as "rationalizing the denominator." Given a quadratic integer $a+b\sqrt{d}$, its so-called conjugate is $a-b\sqrt{d}$. Then the product of conjugates $(a+b\sqrt{d})(a-b\sqrt{d})$ is a difference of squares $a^2-(b\sqrt{d})^2=a^2-db^2$, which is an integer! So given $a+b\sqrt{d}$ in the denominator of a fraction, one may multiply numerator and denominator by its conjugate, thereby making the denominator an integer. From there it should be straightforward how to manipulate the quantity into the form $x+y\sqrt{d}$ with $x,y\in\Bbb Q$.

A much more general phenomena happens in Galois theory. The Galois group is essentially the "symmetry group" of a field extension $L/K$ (the set of all functions $L\to L$ which fix $K$ pointwise, and preserves the set of all true equations involving elements of $L$ upon relabelling of elements - also certain conditions must be met for $L/K$ to be called "Galois"). The orbit of an element of $L$ under the action of the Galois group - the elements of $L$ that "look the same" so to speak - is called that element's conjugates. The product of all conjugates will be in $K$. One can use this to rationalize much more complicated denominators with algebraic integers generally.

Answer 3

Another way to think of Ganymede's answer:

Let $x = a + b\sqrt{2}$ (where $a,b \in \mathbb{Z}$), and suppose that $x(1+\sqrt{2}) = 1$.

Then

\begin{aligned}(a + b\sqrt{2})(1+\sqrt{2}) = (a+2b)+(a+b)\sqrt{2} \\ \\ = 1 + (0)\sqrt{2} \\ \\ \\ \Longrightarrow \begin{cases} a+2b = 1 \\ a+b = 0 \end{cases} \\ \\ \Longrightarrow a = -1, b = 1 \end{aligned}

All we've done is to express both $x(1+\sqrt{2})$ and $1$ in the "standard form" of an element of $\mathbb{Z}[\sqrt{2}]$ (that is, as a linear combination of $1$ and $\sqrt{2}$ with coefficients in $\mathbb{Z}$), equate the respective coefficients, and solve the resulting system of linear equations.

Answer 4

In $\Bbb Z[\sqrt{2}]$, $1 + \sqrt{2}$ is a unit; that is $(1 + \sqrt{2}) \mid 1$; its inverse is $\sqrt{2} -1$:

$(\sqrt{2} - 1)(1 + \sqrt{2}) = -(1 - \sqrt{2})(1 + \sqrt{2})$ $= -(1^2 -(\sqrt{2})^2) = -(1 - 2) = -(-1) = 1; \tag{1}$

that $\sqrt{2} -1$ is in fact the multiplicative inverse of $1 + \sqrt{2}$ may be found by writing

$(1 + \sqrt{2})^{-1} = \dfrac{1}{1 + \sqrt{2}} \tag{2}$

and multiplying the right-hand side through by

$1 = \dfrac{1 - \sqrt{2}}{1 - \sqrt{2}}: \tag{3}$

$(1 + \sqrt{2})^{-1} = \dfrac{1}{1 + \sqrt{2}} = 1\dfrac{1}{1 + \sqrt{2}}$ $= \dfrac{1 - \sqrt{2}}{1 - \sqrt{2}}\dfrac{1}{1 + \sqrt{2}~}$ $= \dfrac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1. \tag{4}$