Irrational numbers and Complex numbers

More concepts about exponentials

In the previous article, I explained that the root process is long and difficult to do.

Calculating by hand

$$\sqrt{2}$$

is not only going to take a long time, but it is actually going to take an infinite amount of time. Why is that?

We know from normal "everyday maths", that $1*1=1$ so $1^2=1$ and $2^2=4$, so the number "has to be" kind of between 1 and 2. If we try $1.5*1.5$ (that is $1+\frac{5}{10}$ in normal fraction terms), we get... $2.25$. Clearly that's too high. Let's try to bisect our way.

$$1.25*1.25 = 1.5625$$

Too low. Has to be between $1.25 < x < 1.5$. Try $1.375$.

$$1.375*1.375=1.890625$$

Getting closer, but still too low. $1.375 < x < 1.5$. Try $1.4375$.

$$1.4375*1.4375=2.06640625$$

Now we're too high again.

As you can see, this process allows us to eventually get to a degree of "precision" to get pretty close, but not quite "2". If you were to type $\sqrt{2}$ on a calculator:

$$\sqrt{2} = 1.41421356237309504880$$

And if you were to fact check...

$$1.41421356237309504880*1.41421356237309504880=1.99999999999999999999$$

Somehow not exact. We're getting close, but never quite 2. The reason for this is that the square root of 2, like many other numbers is irrational. Before we expressed decimals all over the form of a fraction, but this number eludes that specific view, as we strictly "cannot" represent it as a fraction. It would have "infinite" numbers after our period ($1.4142\ldots$).

Historically, irrational numbers were deemed impossible, as they would be impossible to express in actual notation. Also, consider that, just the "infinite numbers" after the period is not a sufficient requirement for a number to be irrational. If you consider $\frac{1}{3}$, you would get on a calculator $0.3333333\ldots$ or in mathematical notation $0.\overline{3}$ (the line represents the part repeating forever). But we can easily express that same number as a fraction.

There are many proofs involving why square root of 2 cannot be a rational number, and they involve a number of logical steps that while not difficult to follow, would inflate this article a bit. But when did that stop us anyway.

Why is $\sqrt{2}$ irrational

This is going to be a proof by contradiction. We assume something is true, find a contradiction somewhere, therefore we know our assumption was false. (This is a similar reasoning to a trial by error, first we fail, therefore we find the correct way after. Since any statement is either true or false only, we try the "wrong way first", and we can deduce the other way was correct.)

• Assume that $\sqrt{2}=\frac{p}{q}$, $p$ and $q$ are our usual variables we want to find. We assume this fraction cannot be simplified (so there's no shared factors between the two numbers).

• Square up both sides

$$\sqrt{2}^2=\left(\frac{p}{q}\right)^2$$ $$2 = \frac{p^2}{q^2}$$

• We multiply both by $q^2$

$$2*q^2 = \frac{p^2}{q^2}*q^2$$ $$2*q^2 = p^2$$

Commutative property to turn it around:

$$p^2 = 2*q^2$$

• We now know that $p^2$ is double $q^2$. Here some issues arise already. Let us give the following definitions:

$$p = p_1*p_2*p_3*\ldots$$ $$q = q_1*q_2*q_3*\ldots$$

(These "subscripts" are to indicate these are all different variables, and of course I don't know how much there are. They are a sort of multiplication of all the prime factors that make up $p$ and $q$, just like $15=3*5$, or $16=2*2*2*2$.)

• We know any term $p_n$ is going to be different from all the terms of $q_n$. (That's because otherwise in the fraction I could simplify them, from the top and bottom parts).

• Since we know $p^2$ is double $q^2$, we know $p^2$ is even. If we consider an even square ($a*a$), we're basically repeating all the factors twice. So:

$$p^2=p_1*p_1*p_2*p_2*p_3*p_3*\ldots$$

We know 2 will be part of that factor list, so we include it.

$$p^2=2*2*p_1*p_1*p_2*p_2*p_3*p_3*\ldots$$

But wait, since we said $p^2=2*q^2$, it means (if we group it up):

$$2*2*p_1*p_1*p_2*p_2*p_3*p_3*\ldots = 2*q^2$$ $$2*(2*p_1*p_1*p_2*p_2*p_3*p_3*\ldots) = 2*q^2$$

• Divide both by 2:

$$2*p_1*p_1*p_2*p_2*p_3*p_3*\ldots = q^2$$

• Not only is $q^2$ even (and therefore also $q$ even), but as a square number, it now has an "odd" number of multiplications. (If $a=b*c*d \iff a^2=b*b*c*c*d*d$, the number of terms is even). So it means that both, $q^2$ can't be a square number, and also that since $p$ and $q$ are both even, the fraction we assumed "simplified and without shared factors" is now entirely wrong.

• Therefore $\sqrt{2}$ cannot be expressed as a fraction of two numbers. Q.E.D. (quod erat demonstrandum, "that which was to be demonstrated", usually a pretentious way to declare that we've proven what we've set up to do)

The Real numbers

Since we have to live with the conscience that irrational numbers exist, and can be inbetween all the roots, or declared in "infinite operations/expansions", we need to expand our concept of numbers. Again.

$$\mathbb{R}$$

That's it, I cannot declare it under the usual fancy curly braces as I did for $\mathbb{Q}$, because all these numbers are not always algebraically possible to express in a common manner. You can consider $\mathbb{Q} \subset \mathbb{R}$, as real numbers encompass all rationals, along with irrational ones. A lot of irrational numbers can be expressed algebraically, as we've seen with $\sqrt{2}$, but most of them won't (there are possibly ways to define some of them, but require a few extra definitions, do not panic, it's fun to go into the unknown!).

Another problem with roots

The previous article, I hinted that the following is still "up for unknown".

$$\sqrt{-1} = ???$$

As we've seen with the "irrational numbers" up above, mathematics is a tool where we just explore more and more questions, assuming that what previously we thought as impossible, we try to take as true and consider the implications. By this philosophy, it's up for debate whether mathematical structures are made up or discovered. In this case, we're going to assert that the square root of $-1$ is equal to a new quantity, a new type of number we're crafting specifically.

$$\sqrt{-1} = i$$

The properties of this $i$ are as follows:

$$i=\sqrt{-1}$$ $$i^2=\sqrt{-1}*\sqrt{-1}=-1$$ $$i^3=i^2*i=-1*i=-i$$ $$i^4=i^3*i=-i*i=-(-1)=1$$ $$i^5=i^4*i=1*i=i$$

And then repeating in cyclical fashion.

We treat $i$ as its own variable almost, just like we cannot mix $3x+3$.

Numbers of the form $ai$ are said imaginary. The set of imaginary numbers is declared as such:

$$\mathbb{I} = \{ a*i | a \in \mathbb{R}\}$$

On its own, not necessarily useful (somehow, zero is also imaginary, as $0*i=0$).

More important, is the set of complex numbers.

$$\mathbb{C} = \{ x + y | x \in \mathbb{R}, y \in \mathbb{I}\}$$

These types of numbers are of the form $a+b*i$. They have both a real part $a$, and an imaginary part $b*i$.

The reason other forms are not as interesting, is due to $x*y$ just ending up in imaginary part, so only addition of two different "units" is interesting. When learning a bit more about graphing and geometry, we'll see that there's two main ways to represent complex numbers, and this current form of keeping track $a$ and $b$ is called "cartesian".

As just an extra, here's what happens if we add two imaginary numbers together:

$$(a+b*i) + (c+d*i)$$ $$a + b*i + c + d*i$$ $$a + c + b*i + d*i$$ $$(a + c) + i*(b+ d)$$ $$(a+c) + (b+d)i$$

Therefore, we add up together the real and imaginary parts together, separately.

For multiplication:

$$(a+b*i)*(c+d*i)$$

Doing a multiplication of an addition, using our usual distributive rules, means the same as splitting each term and multiplying it for the whole second one.

$$(x+y)*z = x*z+y*z$$

Therefore, we take the same approach here (I am going to remove the multiplication symbols, so $abc=a*b*c$, as otherwise it gets really difficult to read):

$$(a+bi)*(c+di)$$ $$a(c+di)+bi(c+di)$$ $$ac + adi + bic+bidi$$ $$ac + adi + bci + bdi^2$$

We know $i^2=-1$ therefore we substitute it:

$$ac + adi + bci - bd$$

We now group up the terms with $i$ and the ones without:

$$(ac-bd) + (ad+bc)i$$

And therefore that's the end result of a multiplication. It helps considering i as something we cannot reduce, except for the cases in the series above, where $i^n$ can be substituted.