We will soon be looking at fields though we won't go too deeply into the topic. The study of fields can easily take up a full-year course. For the purpose of this course, we only need to know some terminology associated with fields and a few examples of fields.

Before we look at the properties of a field, it helps to ignore
subtraction and division. In other words, we restrict our attention
to addition and multiplication. To make up for the loss, we use
the notions of **additive inverse** and **multiplicative inverse**.
Knowing these notions is crucial to understanding how we can have
fields that involve objects that are not numbers.

Loosely speaking, an additive inverse of a field element \(a\) is an element \(b\) such that \(a + b = b + a = 0\). It turns out that only one additive inverse exists for each element. For example, over the rational numbers, the additive inverse of \(\frac{2}{5}\) is \(-\frac{2}{5}\) and the additive inverse of \(-5\) is \(5\).

For a field element \(a\) not equal to 0, a multiplicative inverse of \(a\) is an element \(b\) such that \(a \cdot b = b \cdot a = 1\). It turns out that only one multiplicative inverse exists for each element. For example, over the rational numbers, the multplicative inverse of \(\frac{2}{5}\) is \(\frac{5}{2}\) and the multiplicative inverse of \(-1\) is \(-1\).

Perhaps you now see why there is no need for subtraction or division. We can rewrite \(a - b\) as \(a\) plus the additive inverse of \(b\), normally denoted by \(-b\), and we can rewrite \(a/b\) as \(a\) times the multiplicative inverse of \(b\), normally denoted by \(b^{-1}\).

In modelling Lights Out, we use only 0 and 1 with altered arithmetic rules: Multiplication of numbers is as usual but addition is almost the usual addition with the exception that \(1 + 1 = 0.\) We refer to the set \(\{0,1\}\) with the altered arithmetic rule as \(GF(2).\) Note that over \(GF(2)\), the additive inverse of 1 is 1 because \(1+1 = 0\) and the multiplicative inverse of 1 is 1.

We show that the multiplicative inverse of \(3-\sqrt{7}\) can be written in the form \(a + b\sqrt{7}\) where \(a,b \in \mathbb{Q}\).

The requirement that the multiplicative inverse be written in the form \(a + b\sqrt{7}\) where \(a,b \in \mathbb{Q}\) makes our task more challenging. The reason is that as \(3-\sqrt{7}\) is a real number, its multiplicative inverse is simply \(\frac{1}{3-\sqrt{7}}\). However, \(\frac{1}{3-\sqrt{7}}\) is not in the required form.

Fortunately, we can employ the technique of rationalizing the denominator as follows: \begin{eqnarray*} \frac{1}{3-\sqrt{7}} & = & \frac{1}{3-\sqrt{7}} \cdot \frac{3+\sqrt{7}}{3+\sqrt{7}} \\ & = & \frac{3+\sqrt{7}}{3^2-\sqrt{7}^2} \\ & = & \frac{3+\sqrt{7}}{9-7} \\ & = & \frac{3+\sqrt{7}}{2} \\ & = & \frac{3}{2} +\frac{1}{2}\sqrt{7} \end{eqnarray*} Note that \(\frac{3}{2} +\frac{1}{2}\sqrt{7}\) is in the desired form.

**Remark. **
Using the same technique, one can in fact show more generally that for all
\(x,y \in \mathbb{Q}\) and primes \(p\),
if \(x + y\sqrt{p} \neq 0\), then
its multiplicative inverse can be written as
\(a + b \sqrt{p}\) for some \(a,b \in \mathbb{Q}\).

For each of the following, give its additive inverse and multiplicative inverse.

\(\frac{3}{2}\)

\(-5\)

\(-\sqrt{2}\)

Give the multiplicative inverse of \(-2+\sqrt{5}\) in the form \(a + b\sqrt{5}\) where \(a,b \in \mathbb{Q}\).