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 \(a,b \in \mathbb{Q}\).