Complex numbers: Known numbers
Adjunction of a square root
Imagine that you know nothing of real numbers and only know rational numbers. With this you can add, subtract, multiply and divide. Solving an equation like \[2x+3=9\] causes no problem. But sooner or later you run against an equation like \[x^2=2\] and maybe you can even proof that there exists no rational number that is a solution of this equation. How can you go from here? How can you introduce \(\sqrt{2}\) and calculations with \(\sqrt{2}\)?
The solution is to create a new collection, say \(V\), that meets certain requirements.
Wish list of properties of numbers in V
Our wish list of properties is:
- the rational numbers must be part of it or at least one must be able to identify a subset with this;
- you must be able to add, subtract, multiply and divide 'numbers' in \(V\); the result of such an operation must be again an element of \(V\);
- the calculation rules of rational numbers should apply as much as possible;
- the equation \(x^2=2\) should have a solution in \(V\).
We are going to do such a construction and it will be a leading example for the introduction of complex numbers based on the set of real numbers.
The construction is called the adjunction of a square root (because you can also use this method to \(\sqrt{3}\) , \(\sqrt{5}\) , etc., to add to the number system). The idea behind this is that we sometimes imagine that a solution of the equation \(x^2=2\) , and just see what happens when adding this solution to the rational numbers \(\mathbb{Q}\) . If you know the real numbers, then of course you know that \(\sqrt{2}\) and \(-\sqrt{2}\) both solutions of the equation \(x^2=2\) and that you will \(\sqrt{2}\) to \(\mathbb{Q}\) can add. But in view of the future construction of the complex numbers, we pretend we do not know this format and we note the (symbolic) solution of the quadratic equation \(x^2=2\) simply with the letter \(\alpha\) . We do not make us even pressure on the choice of positive or negative root, but use the drop-in calculations only property \(\alpha^2=2\) .
What does it mean by the process of adjunction of \(\alpha\) to the rational numbers? We certainly want to be able to multiply a rational number with \(\alpha\) and this gives numbers of the form \(y\cdot \alpha\). We would also, of course,be able to add numbers and in this way we get numbers of the form \(x+y\cdot \alpha\) with \(x,y\in\mathbb{Q}\) . A welcome surprise is that this is sufficient, and that there are no more numbers needed to be able to do proper arithmetics. The usual notation for this collection of number is \(\mathbb{Q}(\sqrt{2})\). Because the calculation rules of rational numbers should apply as much as possible, only the following definitions of addition and multiplication are possible:
Addition and multiplication
- Addition is done component-wise, where we consider \(x\) and \(y\) as the components of the number \(x+y\cdot \alpha\): \[(x_1+y_1\cdot \alpha)+(x_2+y_2\cdot \alpha)= (x_1+x_2) + (y_1+y_2)\cdot \alpha\]
- Fro multiplication, we must 'simply' eliminate the parentheses and use \(\alpha^2=2\): \[\begin{aligned}(x_1+y_1\cdot \alpha)\cdot (x_2+y_2\cdot \alpha) &= (x_1\cdot x_2) + (x_1\cdot y_2+y_1\cdot x_2)\cdot \alpha + y_1\cdot y_2\cdot \alpha^2\\ &= (x_1\cdot x_2+2\cdot y_1\cdot y_2) + (x_1\cdot y_2+y_1\cdot x_2)\cdot \alpha\end{aligned}\]
After all:\[\begin{aligned} (-4+4\alpha)\cdot(2+3\alpha) &= (-4\cdot2)+(-4\cdot3\cdot\alpha) + (4\cdot2\cdot\alpha) + (4\cdot3\cdot\alpha^2)\\ & \phantom{abcxyz}\color{blue}{\text{expansion of brackets}}\\ &= -8 + (-12+8)\cdot\alpha+ 12\cdot 2\\ & \phantom{abcxyz}\color{blue}{\text{use of the property }\alpha^2=2}\\ &= 16-4\alpha \\ &\phantom{abcxyz}\color{blue}{\text{standard representation}}\end{aligned}\]
Moreover, we must be able in the new set f numbers to divide by a nonzero number. The formula is complicated and therefore we will first explain it on the basis of an example. We make use of the following calculation rule \[(x+y\cdot \alpha)\cdot (x-y\cdot \alpha) = x^2-y^2\cdot \alpha^2=x^2-2y^2\] The right-hand side is always a rational number. This property we will use in division.
The above example may illustrate the following general method for division. \[\frac{x_1+y_1\cdot \alpha}{x_2+y_2\cdot \alpha}=\frac{(x_1+y_1\cdot \alpha)\cdot(x_2-y_2\cdot \alpha)}{(x_2+y_2\cdot \alpha)\cdot(x_2-y_2\cdot \alpha)}\] The numerator expands, with the property that \(\alpha^2=2\), into \[x_1\cdot x_2 -2\cdot y_1\cdot y_2+ (-x_1\cdot y_2+y_1\cdot x_2)\cdot \alpha\] The denominator simplifies, with the property that \(\alpha^2=2\) , into \[x_2^2 -2\cdot y_2^2\] Thus we arrive the following definition:
Division
\[\frac{x_1+y_1\cdot \alpha}{x_2+y_2\cdot \alpha}= \frac{x_1\cdot x_2 -2\cdot y_1\cdot y_2}{x_2^2 -2\cdot y_2^2}+ \frac{(-x_1\cdot y_2+y_1\cdot x_2)\cdot \alpha}{x_2^2 -2\cdot y_2^2}\]