Now, let's start with the very left hand side as zero. Why zero at the left? Well, this page is written in english and we read english left to right. The is no other particularly good reason to have zero at the left rather than the right. So I have just made a choice.
I am not going to bother drawing in the circles that would result if we actually used a circle drawing device to make sure all the numbers are marked off equally. let's just say they are. I want to mark off numbers up to four. Why four? Well, \(\pi\) was between three and four, and \(e\) was between two and three, so if we go all the way to four, we can actually mark on all the numbers that we have talked about so far. The line will now look like this:
What we have marked out there are five "numbers". Or, four if you do not accept zero as a number. Or, actually, three if you do not accept one as a number. But that is really getting pretty semantic at this point. You can actually see what we are talking about. There is a dot marked two, which is twice as far from the zero as the dot marked one is. The dot marked four is four times as far away as the one, and twice as far again as the two. These are our whole, natural, numbers. These are the things that you count cows or apples in.
Our addition operation can also be visualised here. If we wanted to add two to one, we start at one and take two "hops" to the right ending up on three. It's the same thing with two plus two - you would end up at four.
When we looked at the different operations you could apply to these "natural" numbers we uncovered different kinds of numbers. For instance, we found fractions. How do these fit on the line? Easy, they just appear at some point between the natural numbers, like this:
In that picture the mark for a half is exactly half way between zero and one. What about irrational numbers? Can we show them? Yes, of course. Remember the square root of two? We cannot write it down as one whole number divided by another whole number. But it does have a size. If you use the random rooter spreadsheet that I made for the first blog post, you will find that the first five digits of the square root of two are \(1.41421\). We could get more and more precision, but that is fine for our number line:
There you go, between one and two, but not quite halfway. Now, what about our more esoteric numbers represented by the symbols \(e\) and \(\pi\)? We can just slot 'em on there:
We are probably, to be honest, getting a bit beyond what the Greeks could do with their number lines, but bear with me. There's \(e\) just before the three, and \(\pi\) just after the three. Those are the numbers that we are talking about - that's what they "look" like on the number line. This is all perfectly real. You can happily say that \(e\) is smaller than \(\pi\) because it is nearer the zero than \(\pi\).
OK, we have two other types of numbers that we found when looking at the operations you can perform, negative numbers and imaginary numbers. Can we make a mark on this number line to show either of these? Nope. Negative numbers are less than zero, but our line stops at zero. And imaginary numbers are ... something else entirely. Lets fix the negative numbers first. All we have to do is extend our line left beyond the zero:
And then we mark off our negative numbers:
Our operations to get negative numbers now make sense. To deduct three from two, we just start at two, and then take three "hops" left to get negative one. If we wanted to add two to one, we start at one and take two "hops" to the right ending up on three.
So now we come to imaginary numbers. We know already that they are neither positive or negative. How do we know that? Well remember back to when we tried to work out if the square root of negative one was positive or negative. We came to the conclusion that it was neither. This was because both a positive and a negative number square to a positive number. We need something new that squares to a negative. So the number \(i\) cannot be drawn on our line to either the left or right of the zero. So where do we draw it? We draw it ABOVE the zero:
In fact we draw a whole new \(i\) number line perpendicular to our \(1\) number line:
I had to move the zero digit to the left a bit, but it is still supposed to be the points where the lines cross over.
This is all a bit much to take in. It is tempting to ask ones self "how on earth there can be a whole new number line perpendicular to the original one?", or "It doesn't make any sense", or "What does it all mean?". Do not feel bad about asking these questions, but understand that these are exactly the sort of questions people would have asked about adding more number line to the left of the zero to accomodate the negative numbers. From the Egyptians to the century America won independence mathematicians pretended negative solutions to problems did not exist. They would have stared in disbelief at the existence of line to the left of the zero. So don't feel bad if that was also your first reaction to the up and down line.
Incidentally, just as there was no right 'end' to mark as zero, I could just as easily have drawn the line with the positive \(i\)'s underneath the real number line.
So what then does it mean to say that a number has real and imaginary parts? It basically means that instead of living in the one dimensional world of our original number line, actually numbers live in the two dimensional world of the plain delineated by the real number line and the imaginary number line. Say, what?
To find a real number, all you need to have is one piece of information. Where it is positioned on the number line. However, to find an actual number with imaginary and real parts you need two pieces of information. First, where it lies on the real number line (the real part) and second where it lies on the imaginary number line (the imaginary part). If a number has a real part of two and an imaginary part of \(i\) it would be where the black dot is on this picture:
What you do is go along the real line to two, and then go up to \(i\). We write this number as \(2+i\). It has an imaginary part of one and a real part of two. All of the "natural" numbers exist on this plain, it's just that they are all in the form \(x+0i\). In other words their imaginary part is zero. But they still have one!
So how does all this fit into \(i\) being the square root of negative one? Simple. It turns out that multiplying by \(i\) is the same as rotating about the zero point by 90 degrees. Say, what? OK, OK. Lets look at what I mean. Take our number above \(2+i\) and lets multiply it by \(i\). How do we do that? Its just the same as any other time we multiply two equations in brackets. Remember that \(i\) also has a real component, which is just zero. So we just need to follow the standard procedure for multiplying brackets together (blue and red baskets with apples and oranges):
\[(0+i)\cdot(2+i)\]
We need to multiply the first two terms of each bracket together:
\[0\cdot 2=0\]
Then the first from the first bracket and the second from the second bracket:
\[0\cdot i=0\]
Then the second and the first:
\[i\cdot 2=2i\]
And finally the second and the second:
\[i\cdot i=-1\]
Remember that \(i\) is the square root of negative one, so of course if you multiply \(i\) by \(i\) you get negative one. Not let's add all that up:
\[0+0+(2i)+(-1)\]
We need to rearrange that a bit, because tradition has the real part coming first in our number:
\[-1+2i\]
What does that look like on our number plain?
Now we can multiply that number (\(-1+2i\)) by \(0+i\) again to see what happens:
\[(0+i)\cdot (-1+2i)\]
\[(0\cdot -1)+(0\cdot 2i)+(i\cdot -1)+(i\cdot 2i)\]
\[0+0+(-1\cdot i)+(2\cdot i\cdot i)\]
\[0+0-i+(2\cdot -1)\]
\[0+0-i-2\]
\[(-2-i)\]
See, we hve multiplied by \(i\) twice so we have come a total of 180 degrees from the start. We can get our last dot by running the multiplication again.
\[(0+i)\cdot (-2-i)\]
\[(0\cdot -2)+(0\cdot -i)+(i\cdot -2)+(i\cdot -i)\]
\[(0)+(0)+(-2i)+(i\cdot -i)\]
Hang on a minute. What is negative \(i\) multiplied by \(i\)? Well think of it like this. Negative \(i\) is the same as negative one times \(i\). So that sum in brackets that the end of the last line is the same as \(i\cdot -1\cdot i\). Which in turn is the same as \(-1\cdot i \cdot i\). We know what \(i\) multiplied by \(i\) is, so that gives us \(-1 \cdot -1\), which we already know to be the same as one. That gives us the next line:
\[(0)+(0)+(-2i)+(1)\]
\[(1-2i)\]
On our graph that is here:
We are nearly there, and we can check the seemingly inevitable destination by multipling once more. This time we can dispense with multiplying by zero because we know that gets us nowhere. So instead we have: \[\begin{multline}
i\cdot(1-2i)=(i\cdot 1)+(i\cdot -2i)=(i)+(-2\cdot i\cdot i)=\\
i+(-2\cdot -1)=i+2=2+i
\end{multline}\]
And that, of course, is exactly where we started.
Of course, we now know that if you start at \(i\) and multiply that number by \(i\) you rotate through 90 degrees counter clockwise, bringing you to ... drumroll ... negative one. That looks like this:
Right, brilliant, we can now see what it means to multiply something by \(i\), but is any of this actually real? What I mean is, is all this just an invention of mathematicians to fill a gap, or did imaginary numbers exists before mathematicians "found" them? You could ask this question of the whole of maths, and very earnest people with wild hair and ink spots on their shirt pockets often do. I am not going that far, but I am going to try and convince you that imaginary numbers do affect "normal" maths even when you are dealing purely with real numbers and not trying to find the square root of anything negative. To do that we are going to have to dive into some infinite sums and series.
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