Complex Operations and MORE TRIGONOMETRY

Right, we have complex numbers, we can draw them on our plane. We can convert between the two forms of notation that we are using, Polar and Rectangular. We also saw that multiplying by \(i\) had the effect of rotating a number by \(90^{\circ}\), or as we are now referring to angles, \(\tfrac{\pi}{2}\).

How, more generally, do we go about adding and multiplying and raising to an exponent complex numbers?

Adding is actually pretty easy. Lets put our number in rectangular notation. So it looks like \(a+bi\). Lets make a new number, \(c+di\). Now lets add them together:

\[(a+bi)+(c+di)\]

There is no multiplication there, apart from the \(b\) and \(d\) both being multiplied by \(i\). So we could just add all the terms up:

\[a+bi+c+di\]

And tidy:

\[a+c+bi+di\]

Now if we look closely, we can see that we have two numbers multiplied by \(i\). If you remember our red and blue baskets with apples and oranges in them, we can just pull out the \(i\)'s leaving the \(b\) and \(d\) in a bracket multiplied by \(i\):

\[a+c+i(b+d)\]

Hang on, we now have two real numbers added together, and then two other real numbers added together and multiplied by \(i\). That is very nearly back to our rectangular form of one real part, and one imaginary part represented by a real number multiplied by \(i\). We just stick another set of brackets in like this:

\[(a+c)+(b+d)i\]

And we are done, that looks just like a rectangular number. Adding numbers in rectangular form is as simple as just adding the real parts together, and then adding the imaginary parts together.

OK, what about multiplication?

\[(a+bi)\cdot (c+di)\]

Again, remember our baskets. We can write this out like this:

\[a\cdot c + a\cdot di + bi\cdot c + bi\cdot di\]

So we have each number in the first bracket multiplied by each number in the second bracket. It tidies up to look like:

\[a\cdot c + a\cdot di + bi\cdot c + b\cdot d\cdot i^2\]
\[a\cdot c + a\cdot di + bi\cdot c - bd\]
(Remember that when we multiply \(i\) by itself we get negative one, so the last term loses its \(i^2\) and is subtracted instead of added at the same time.) We can take away the dots as well:

\[ac + adi + bci - bd\]

And then rearrange to the real and imaginary parts:

\[ac - bd + adi + bci\]

Again we can take out the \(i\)'s, and bracket off the real and imaginary parts:

\[(ac - bd) + (ad + bc)i\]

Well, it IS a rectangular form complex number, but it looks horrid. You get no real feel for what is happening on the complex plane. What happens if we deal with multiplication in the polar form instead? Let's look at what happens to the absolute value of the number. We'll call \(r_{\scriptsize 1}\) the absolute value of the first number and \(r_{\scriptsize 2}\) the absolute value of the second number, and \(r_{\scriptsize 12}\) the absolute value of the pair of them multiplied together. And remember that the square of the absolute value of a complex number is the real part squared plus the imaginary part squared. So we get this:

\[r_{\scriptsize 1}^2=a^2+b^2\]
\[r_{\scriptsize 2}^2=c^2+d^2\]
\[r_{\scriptsize 12}^2=(ac-bd)^2+(ad+bc)^2\]

Let's take a very deep breath and try to expand the squares in the last line:

\[r_{\scriptsize 12}^2=acac-acbd-bdac+bdbd+adad+adbc+bcad+bcbc\]

Oh, boy. Lets try to tidy:

\[r_{\scriptsize 12}^2=a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2\]

Hmm. Not a great deal better. It's hard to spot but notice something about the four terms which are just \(abcd\). Two of them are subtracted, and two are added, so it could be written:

\[r_{\scriptsize 12}^2=a^2c^2-2\cdot abcd+2\cdot abcd+b^2d^2+a^2d^2+b^2c^2\]

But, of course if you add two somethings, and take two somethings away, then you end up with no somethings, so we can remove those parts altogether:

\[r_{\scriptsize 12}^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\]

Right. What do we do with that mess? The answer is red and blue baskets again. You spot that there are two parts multiplied by \(a^2\), namely \(a^2c^2\) and \(a^2d^2\), so we can take the \(a\)'s out and putting what remains in a bracket:

\[r_{\scriptsize 12}^2=a^2(c^2+d^2)+b^2d^2+b^2c^2\]

It should be easier to spot this time, but you can then do exactly the same with the \(b^2\) terms:

\[r_{\scriptsize 12}^2=a^2(c^2+d^2)+b^2(c^2+d^2)\]

And if we remember about the red and blue baskets, and we just want to know how many of everything we have, we can put the \(a^2\) and \(b^2\) in brackets as well:

\[r_{\scriptsize 12}^2=(a^2+b^2)(c^2+d^2)\]

Remember our definitions of the absolute values of the numbers we are multiplying?

\[r_{\scriptsize 1}^2=a^2+b^2\]
\[r_{\scriptsize 2}^2=c^2+d^2\]

So we can replace the contents of the first bracket with \(r_{\scriptsize 1}^2\) and the second bracket with \(r_{\scriptsize 2}^2\)

\[r_{\scriptsize 12}^2=(r_{\scriptsize 1}^2)(r_{\scriptsize 2}^2)\]

That can just be written as:

\[r_{\scriptsize 12}^2=r_{\scriptsize 1}^2\cdot r_{\scriptsize 2}^2\]
\[r_{\scriptsize 12}^2=r_{\scriptsize 1}\cdot r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot r_{\scriptsize 2}\]
\[r_{\scriptsize 12}^2=r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot r_{\scriptsize 1}\cdot r_{\scriptsize 2}\]
\[r_{\scriptsize 12}^2=(r_{\scriptsize 1}\cdot r_{\scriptsize 2})\cdot (r_{\scriptsize 1}\cdot r_{\scriptsize 2})\]
\[r_{\scriptsize 12}^2=(r_{\scriptsize 1}\cdot r_{\scriptsize 2})^2\]

And if we now take the square roots of both sides we get:

\[r_{\scriptsize 12}=r_{\scriptsize 1}\cdot r_{\scriptsize 2}\]

This tells us that when you multiply two complex numbers together, the absolute value of the result is the absolute values of the original two numbers multiplied together. So if the absolute values of your starting numbers were two and three, the absolute value of the result will be six.

Where will the angle at the origin point to once the numbers have been multiplied? Well, our formula for working out the polar number from the rectangular number was a horrific mess. If we actually wanted to trail through that for the rectangular form \((ac - bd) + (ad + bc)i\) just to see what would happen, we would have to be nuts. Stark raving mad. Is there an easier way? Joyfully, yes. Remember our formula for working out the rectangular form of a polar number?

\[rcos\theta + risin\theta\]

That is a perfect halfway house between a rectangular form number and a polar form number. Why? Well it has a real part (distance along the real axis) and an imaginary part (distance along the imaginary axis), but it is actually written in terms of the size of the angle involved. So we can use this to see what happens to the angle when we multiply. The first thing to do is to spot that there are two entities multiplied by \(r\) and take the \(r\) out:

\[r(cos\theta + isin\theta)\]

Lets call that complex number one:

\[r_{\scriptsize 1}(cos\theta + isin\theta)\]

Lets make this next one complex number two:

\[r_{\scriptsize 2}(cos\alpha + isin\alpha)\]

If we multiply them together we get:

\[r_{\scriptsize 1}(cos\theta + isin\theta)\cdot r_{\scriptsize 2}(cos\alpha + isin\alpha)\]

Rearranging the \(r\)'s gives us:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \big(cos\theta + isin\theta\big)\cdot \big(cos\alpha + isin\alpha\big)\]

This should make sense to us, because look what we are doing with our absolute values: we are multiplying them together, just like we worked out above. If we multiply out all the stuff in the brackets, we will get this:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \big(
cos\theta cos\alpha+cos\theta isin\alpha+isin\theta cos\alpha+iisin\theta sin\alpha
\big)\]

Which simplifies (a bit), because the term at the end has \(i^2\) in it:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \big(
cos\theta cos\alpha+cos\theta isin\alpha+isin\theta cos\alpha-sin\theta sin\alpha
\big)\]

And then we can move the \(i\)'s out of the middle two terms, putting the remainder in brackets:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)-sin\theta sin\alpha
\Big)\]

I'll then do one further rearranging to get the real parts together:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

Now what in the name of sanity are we going to do with that mess in brackets? What we need to do is start drawing lines on the unit circle to see exactly what we are talking about. What IS the cosine of one angle multiplied by the cosine of another minus the sines of both multiplied together? To find out what this looks like, we are going to go back to the unit circle, and we are going to draw on the angle \(\theta\). I am marking the origin as point A, and the point where the radius hits the circle as B. We have seen all this before.

Forget for a moment that I am showing the angle down near the origin, or point A, rather than drawn onto the unit circle. This will make sense in due course. It doesn't really matter what size the angle is, because we are trying to come up with a general rule that will work for all angles. OK, next let's mark off the sine of the angle as a line going straight down from point B to the x axis:

We'll call the point where it meets the x axis C, and we will draw on a square to show that the angle at which it meets the x axis is a right angle:

And so we know what we are talking about, lets describe the line:

Good. Now what about the cosine of the angle \(\theta\)? That's just the x axis out to point C isn't it? So lets mark that on:

This is exactly the same picture that we have seen before, with a bit more writing on it for clarity's sake. For the purposes of this task I am going to zoom into the quarter circle between the positive x and y axes. That is the place where we are going to be playing, so we can just concentrate there:

OK, that's a bit clearer. Now what I am going to do is to take the point B and start to move it around the unit circle. But this time I am going to FIX the angle \(\theta\) so that does not change. That sounds very odd. What is that going to tell us? Well, wait and see. What is going to happen is that the whole triangle that we have just drawn, from A to B to C and back to A, is going to keep its shape but it is going to rotate about the point A. Lets see what we get:

So, you can see that we now have the triangle seemingly floating around. Point C is in empty space. The important fact is that I have ONLY rotated. The lengths of the sides have not changed. The angle at C is still a right angle. The length of A to B is still the radius of the unit circle, which is still one. What can I do now? Well I can show that I have rotated the first triangle by a new angle called \(\alpha\):

At this point, what I could do is draw a line all the way from the origin at A, through C and out to the unit circle to let us measure the sine and cosine of \(\alpha\). I am not going to do that though. What I am going to try and do is measure the sine and cosine of \(\alpha\) using the measurements we have already made for \(\theta\). You will see why shortly.

The first thing we need to do here is to draw on another line, to stand in for the sine of \(\alpha\). We want it to reach the x axis at right angles, just like a normal sine line. Because we are using the measurements of \(\theta\) to work out \(\alpha\) we need to use the whole of the line from A to C. So we will drop our "sine" of \(\alpha\) line down from C to the x axis, and we will mark the point where it hits as D, which is a right angle:

Right then, first BIG question. How do we work out the lengths of the line from A to D and from C to D? Just in case you are trying to figure this out I am talking about these distances:

If the line from A to C was actually the radius of a unit circle, then we could easily work out AD and CD. They would just be the cosine and sine of the angle \(\alpha\). So that's a start. Let's stick a circle on the diagram anyway, to see what that would look like:

So you can see that IF \(cos\theta\) was one, then the other two sides would just be the cosine and sine of \(\alpha\). What we are dealing with here is another problem of scaling. If you remember back to when we were first working out the rectangular form version of a complex number in polar form, what we did was to work out the sine and cosine on the unit circle and then we just multiplied them by the absolute value of the number. We did this because we had scaled the absolute value to one, so that we could work in the unit circle. Here we just need to do the same thing. Point C is not on the unit circle. It is like the complex number we are trying to find. The absolute value of that number is \(cos\theta\). So we just need to scale the ACTUAL sine and cosine of \(\alpha\) (whatever they are) by the factor \(cos\theta\), and that will give us the real and imaginary parts of the complex number C. Those real and imaginary parts are, of course, the lengths of the lines AD and CD that we are trying to find!

So we end up by saying that the length of AD is the cosine of \(\alpha\) MULTIPLIED by \(cos\theta\). Or \(cos\theta\cdot cos\alpha\). Hey! We've seen that before:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

It's the very first term of the mess in brackets that we are trying to sort out. Great, we are making progress! So what about the line CD? Well that is just going to be the sine of \(\alpha\) MULTIPLIED by \(cos\theta\). Or \(cos\theta\cdot sin\alpha\). And look, that's the third term in the mess in brackets. 50% done! Let's mark this up on our diagram:

I have taken the example unit circle away, because it was just making a mess, and we have finished with it. Right, we are halfway done now. We just need the other two terms of that mess so we can start to make sense of it. Looking at our diagram, we have now worked out the length of all our sides. What we need to make progress is ANOTHER triangle. No, I am not about to rotate the whole mess again, no more rotation. Instead I am going to look at the mess I am trying to make sense of and I am going to realise that to get the remaining two terms I need to start multiplying stuff by \(sin\theta\). I was able to multiply stuff by \(cos\theta\) because I made that the radius of a circle, and then scaled the other two sides by that factor. So lets make \(sin\theta\) the radius of a circle so we can play the same trick. How can we do that? Well if \(sin\theta\) is going to be the radius of a circle that makes either point B or point C the centre of the circle. Let's pick point C as the centre, and draw on the circle:

I am now going to twist the whole diagram through a quarter turn:

And I am going to draw on, in a dashed line because it is temporary, an x axis for this new circle. Now I can drop a line to this x axis from the point B, so as to meet the line at right angles at the new point E:

I will also draw in the line along the temporary x axis between C and E:

Now we can start working out the lengths of CE and BE. Hang on though, don't we need the size of the angle as well? Which angle? The angle marked \(\beta\):

So what size is this angle? The angle of a straight line is \(\pi\) radians. Is this right? Of course. If you take the unit circle, and you start at the normal place on the real axis, and you move along half the circle (\(\pi\) radians) you end up back on the real axis, on the negative side. You may as well have travelled in a straight line along the real axis and you would have ended up at the same place. Knowing what we know about angles on straight lines, what can we say about \(\beta\)?

Well, we can say that it is on a straight line, the line from D to E which passes through C. So the total angle on that straight line must be \(\pi\). We can see by looking that there are three angles that make up that total angle. They are the unmarked angle, the right angle (size \(\tfrac{\pi}{2}\)) and \(\beta\). So we can say:

\[\pi=\tfrac{\pi}{2}+\beta+unknown\]

Take away \(\tfrac{\pi}{2}\) from both sides and you get:

\[\tfrac{\pi}{2}=\beta+unknown\]

So what's the unknown angle? It is on the line, AND it is inside a right angled triangle. In case it is unclear it is a corner in this bold triangle:

If you add up all the angles in a right angled triangle in radian measure they total \(\pi\). Is this right? Yes, a right angled triangle is half of a rectangle. Each corner of the rectangle has an angle of \(\tfrac{\pi}{2}\), making a total of \(2\pi\). So one half of this shape will have angles that add up to one half of \(2\pi\) or \(\pi\). If they didn't two of them could not fit together to make a rectangle. (Actually this is true of any triangle drawn on a flat surface, but it is harder to explain why and involves Euclid and Parallel lines, which we do not need to concern ourselves with here).

The other two angles in that triangle are a right angle \(\tfrac{\pi}{2}\), and \(\alpha\), and all three angles must add up to \(\pi\). So we can also say that:

\[\pi=\tfrac{\pi}{2}+\alpha+unknown\]

Again, take away \(\tfrac{\pi}{2}\) from both sides:

\[\tfrac{\pi}{2}=\alpha+unknown\]

If we take \(\alpha\) away from both sides we get:

\[\tfrac{\pi}{2}-\alpha=unknown\]

So the unknown angle has size \(\tfrac{\pi}{2}-\alpha\). Let's put that back in our equation above, so:

\[\tfrac{\pi}{2}=\beta+unknown\]

Becomes:

\[\tfrac{\pi}{2}=\beta+\tfrac{\pi}{2}-\alpha\]

Now what can we do with that? First add \(\alpha\) to both sides:

\[\tfrac{\pi}{2}+\alpha=\beta+\tfrac{\pi}{2}\]

And finally, remove \(\tfrac{\pi}{2}\) from both sides:

\[\alpha=\beta\]

Hey presto, just like magic we have worked out that the angle \(\beta\) MUST be the same size as angle \(\alpha\). So lets mark that on our diagram:

Bingo. We can now deduce, using exactly the same logic we used for the lines AD and CD, the lengths of lines BE and CE. In particular, CE would be the cosine of \(\alpha\) IF you scaled it by the factor \(sin\theta\), so CE is \(sin\theta cos\alpha\). And in turn BE would be the sine of \(\alpha\) if you scaled it by the factor \(sin\theta\), so BE is \(sin\theta sin\alpha\). Lets draw those on:

Let's also get rid of the other circle, the temporary x axis, and rotate back to normal:

Now we have lines on our diagram that match all the parts of the mess in the brackets in this equation:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

Now we can try to answer the question what is the cosine of \(\theta\) multiplied by the cosine of \(\alpha\) minus the sine of \(\theta\) mutlipied by the sine of \(\alpha\). We know what \(cos\theta cos\alpha\) is, it is the line from A to D. We also know what \(sin\theta sin\alpha\) is, it is the line from B to E. So what do we get if we subtract the line BE from the line AD? First notice that the point E is directly above the point D. So if we start at D, and move back towards the origin at A until we are directly under B, the remaining distance to A will be AD-BE. The first thing we need to do is drop a line from B to the real axis so we can see exactly where that point will be:

I have moved the text down a bit showing the length of A to D. The angles are getting in the way a bit though, so I will shrink them down:

Can you see what has happened? The two angles we have been dealing with, \(\theta\) and \(\alpha\) can now be see as one big angle, which I can call \(\theta + \alpha\). What is the cosine of the angle \(\theta + \alpha\)? Well, first you would fix the point on the circle that corresponds to a journey around the circle of the size of the angle. That point is B isn't it? Remember that the line from A to B is the radius of the unit circle, so it is length one. That means that the cosine of the angle \(\theta + \alpha\) is the length from the origin along the real axis to the point directly under the point B. That's the point F we just drew isn't it? So \(cos(\theta + \alpha)\) is the length A to F. But how did we find the length A to F? We started with A to D, which was \(cos\theta cos\alpha\) and we subtracted the length from B to E, which was \(sin\theta sin\alpha\). So if AF=AD-BE, then:

\[cos(\theta + \alpha)=cos\theta cos\alpha-sin\theta sin\alpha\]

Fantastic! We can now go back to our horrible equation:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

and we can replace the first part of the mess with:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos(\theta + \alpha)+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

Well, that's better, and we are nearly finished. Let's tackle the second part. Back to the diagram, what is the sine of \(\theta + \alpha\)? Well given that the line from A to B is length one, the sine is just the height of the point B off the real axis, isn't it? Can you see a line that shows that height? Yes, it is the line from F to B. What OTHER line have we drawn that is the same height as F to B? Yes, D to E. So \(sin(\theta + \alpha)\) is the same as the length D to E. And look closely, what is D to E? It is \(cos\theta sin\alpha+sin\theta cos\alpha\). So:

\[sin(\theta + \alpha)=cos\theta sin\alpha+sin\theta cos\alpha\]

Fantastic times two! We can now finish clearing up our horrible equation by replacing the second mess in brackets with:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos(\theta + \alpha)+isin(\theta + \alpha)
\Big)\]

Just in case you think I fiddled the diagram with special angles, have a look at an animation of varying angles, and you'll set it holds for all of them:



What can we learn from this? Well, as we had already worked out, if you multiply two complex numbers in polar form, you multiply their absolute values together. That's the \(r_{\scriptsize 1}\cdot r_{\scriptsize 2}\) bit. You then just ADD the angles together - look you can see it happening in the brackets above. So the whole multiplication looks like this:

\[r_{\scriptsize 1}(cos\theta + isin\theta)\cdot r_{\scriptsize 2}(cos\alpha + isin\alpha)=
r_{\scriptsize 1}r_{\scriptsize 2}\Big(
cos(\theta + \alpha)+isin(\theta + \alpha)
\Big)\]

Formally then we can say that:

\[(a+bi)+(c+di)=(a+c)+(b+d)i\]

and

\[r_{\scriptsize 1}\angle \theta \cdot r_{\scriptsize 2}\angle \alpha = r_{\scriptsize 1}r_{\scriptsize 2}\angle (\theta + \alpha)\]

Hurrah! We've done multiplication and addition. Subtraction and division are essentially just the reverse. What about exponents?

Raising a complex number to a real exponent is just the same as repeated multiplications, so that's fine. What about raising a real number to a complex exponent though? What does that look like? Answer a) it looks fucking horrible, is what it looks like. Answer b) it is the secret at the heart of finding out why \(e^{i\pi}+1=0\), because look, there is a complex number as an exponent of \(e\)! To get into that in any detail, we will have to work out a different way of writing out \(e\), which is going to be pretty traumatic. So as a diversion, lets look at something shiny instead.

Order of Operations, Part Two

OK, now we have learned about the opposites of the operations, does that make a difference to our rules about the order we do them in?

Sadly, and predictably, yes.

What is three minus two plus one? Is it three minus two (one) plus one (two)? Or is it two plus one (three) taken away from three (zero)?

Oh, bugger. But hang on, didn't I say that subtracting a number was the same as adding the negative of that number? So what if we just pretend, between us, that there is no such thing as subtraction. Lets concentrate on addition.

So what is three plus negative two plus one? Is it three plus negative two (one) plus one (two)? Or is it negative two plus one (negative one) plus three (two)? Hey - it's both! Yes, by getting rid of subtraction altogether the problem of the order goes away.

So what about multiplication? What is two multiplied by three divided by four? Is it two multiplied by three (six) divided by four (six fourths, or one and a half)? Or is it three divided by four (three quarters) multiplied by two (six quarters or one and a half)? Hey - it doesn't matter. Or does it? What about two divided by three times four. Is that two divided by three (two thirds) multiplied by four (eight thirds)? Or do you take three multiplied by four (twelve) and divide two by it (two twelfths, or one sixth)? Oh bugger, a sixth is not the same as eight thirds.

So the best idea is again to forget about division and instead stick with multiplication by the inverse. So for two divided by three multiplied by four, we can say two multiplied by a third (two thirds) multiplied by four (eight thirds), or a third multiplied by four (four thirds) multiplied by two (eight thirds)! Check.

Because we are now just adding (negatives) and multiplying (inverses) again we can stick to the same ordering conventions we decided upon before.

What about the opposite of exponents? Hmm, tricky. What is the square root of four plus one. Does that mean the square root of four (two) plus one (three)? Or does it mean the square root of five? Again, if you stick to square roots written as fractional exponents it clears this all up. You would say four to the power of a half plus one. And then ... bugger. Is that one plus a half (one and a half) to which you then raise four? Well, no because we have already decided that you do your exponents stuff first, and unless it is in brackets it is definitely four to the power of a half (two) plus one.

So, the opposites of addition, multiplication and exponents can cause problems with the ordering of operations, but if you just ignore the opposites and add, multiply or exponent (if that's a verb) by the new numbers we discovered then you are going to be OK with Brackets -> Exponents -> Multiplication -> Addition.

Now we are happily done with the different kinds of mathematical operations that we need to know about for this little adventure. It is time to get back on track with the subject at hand. We've looked at zero and one, so it is now time for one of those funny symbols.

The Opposite of Exponents

In the introduction we talked about Pythagoras theorem and the task of having to find the square root of a number. That involved finding which number, multiplied by itself, resulted in the number we started with. That task was a special case of the opposite of raising to the power. Raising a number to the power of another number, tells you how many times to multiply the first number by itself. The opposite operation is, when given the first number, and told how many times a second number has been multiplied by itself to get the first number, the task of finding the second number. This is known as finding the root of the first number. Multiplying a number by itself is squaring a number, so finding which number has been multiplied by itself is known as finding the square root. The same thing applies to cubing a number and finding the cubic root.

I explained how you go about doing that for the square root in the introduction. Finding any root (as opposed to the square root) is just the same except you need to do more divisions.

We use the \(\surd\) symbol to represent the operation of finding a root, and we add a small number to the left hand side of it to show what type of root we are trying to find. So, to represent the function of trying to find out which number you multiply by itself three times to get one hundred and twenty five, you would use the symbols \(\sqrt[3]{125}\). This is also known as finding the cube root of one hundred and twenty five. Technically to find which number you multiply by itself twice to get four you could use the symbols \(\sqrt[2]{4}\). However that is also trying to find the square root of four, and square roots are very very common. In fact they are so common that we do not use the small two, we just use the \(\surd\) symbol on its own to represent a square root.

As with addition and multiplication, and you may not be surprised, there is another way to perform this operation as a type of powering. Like multiplication and division, what you do is raise the number to the power of the INVERSE of the root you are trying to find. Sounds a bit complicated. Is actually simple. If you want to find the square root of four, you could also write \(4^{\frac{1}{2}}\). To find the cube root of one hundred and twenty five you would write \(125^{\frac{1}{3}}\).

What about raising a number to a non-unit fraction (something that isn't one half or a third or a blahth)? You work out which number was multiplied by itself the bottom number of the fraction times to get the original number to the power of the top number of the fraction. So if you raise something to the power of three halves (that is a three over a two) what you do is take your number and cube it (the 'three' in the exponent) and then work out the square root (the 'two' in the exponent) of the result. I have no idea why this works, and am not going to try and work it out now.

As with negative numbers and subtraction, and fractions with division, this operation also draws our attention to a new type of number. Another name for fractions is rational numbers. This is because one number divided by another is another way of expressing the ratio between the two numbers. So one divided by three is a third, which means that three is three times as big as one. That is a ratio. So a rational number is a fraction representing the ratio between two whole numbers. For instance two could be said to be a rational number because two is twice as big as one, you could represent this as the ratio \(\tfrac{2}{1}\). Can all numbers be represented this way?

The answer to that question is no. The number which, if multiplied by itself, makes two, cannot be written as a ratio between two whole numbers. We have to just leave it written as \(\sqrt{2}\). We call that number an irrational number because it cannot be represented by any fraction no matter how bit the numbers on top and bottom of the fraction actually are.

How do we know that no such number exists? We can work our way through a series of logic arguments the result of which answers the question completely. As long as the starting assumptions hold up, if your logic is good enough you can be said to have proved something mathematically. The proof of the irrationality of the square root of two is one of the oldest proofs that we know of. Nearly two and a half thousand years old. I will come back to this later on, and properly prove it.

One last word, if you remember when we first looked at exponents we say that if you had a number raised to and exponent raised to another exponent, you just multiplied the exponents together. If you weren't convinced before, you can consider this possibility: \(2^{2^{\tfrac{1}{2}}}\). That's two to the power of two to the power of a half. The answer is NOT \(2^{\sqrt{2}}\), but \(2^{2\cdot\tfrac{1}{2}}\) which is just \(2^1\) or \(2\). If you wanted it to be \(2^{\sqrt{2}}\), you would have to use brackets and write \(2^{(2^{\tfrac{1}{2}})}\)

Next we need to see if these opposite operations have an effect on the order in which you have to do them.

How do Fractional Operations work?

With division came fractions, in the form of an inverse to a number, and simply one number divided by another. It sounds really simple, but it is worth saying that one divided by three is a third. The mathematical notation version of that can look very similar:

\[1/3=\tfrac{1}{3}\]

Right, so how do we add together two fractions, say a half and a third? Well, to start with you need to be talking about the same type of things. Thirds and halves are not the same type of thing, apart from the fact they are both fractions. It is easy to add a third to a third, you get two thirds!

You can't directly add a half to a third though - what would you get? A hird? Or a Thralf? So we need to first of all make them the same thing. This may NOT be either halves or thirds, and it is going to change the numbers on top in each case.

A half is the same as two quarters, or six twelfths and so on. A third is also the same as two sixths or seven twenty-firsts, and so on. The game is to find the first type of thing (quarters, sixths, twenty firsts or whatever), that we can make both a half and a third fit into exactly.

So, first question, how many thirds in a half? One and a bit, because one third is less than a half, and two thirds is more than a half, so that is no good. What about quarters? How many quarters in a half - exactly two. Good. How many in a third? Again one and a bit for the same reason as halves. Fifths are also no good; there isn't an exact number of them in a half or a third.

So what about sixths? Well, there are two sixths in a third so good. And there are exactly three sixths in a half. Bingo. So our problem is now to add three sixths (a half) to two sixths (a third) and the answer is exactly five sixths.

So for addition, all you do is add the number of the type of thing you have together if it is all the same type of thing. If it isn't you need to make the type of things the same (which automatically changes the number of them you have) and then just add.

It is no coincidence that a sixth is half of a third by the way. The easiest way (although not necessarily the neatest) of converting two different types of things into the same type is just to multiply the types together. So two times three is six in our example.

For subtraction, you just do all that in reverse. Easy.

Multiplication seems a bit trickier, but is actually easier once we work out a shortcut. Lets say you have so many somethings or other and you want to multiply them by another amount of something else. Let's represent the 'so many' with the letter 'a' (just a placeholder, could be a picture of a pony or a guitar). Let's represent the somethings with 'b', the 'another amount' with 'c' and the 'something else's with 'd'. So what we get looks like this:

\[\frac{a}{b}\cdot\frac{c}{d}\]

So we are trying to multiply a number of b'ths by c number of d'ths. If that makes sense. This looks a bit daunting, so lets make it easier and say we are only going to find what one b'th of the second fraction is. Once we have worked out what one b'th is, we can just multiply that by a, and we will get the full answer we want. That would look like this:

\[\frac{1}{b}\cdot\frac{c}{d}\]

Now we are really just taking two numbers the 'c' and the 'b' and dividing 'c' by 'b'. Why are we doing that? Lets use some numbers to see what happens:

\[\frac{1}{2}\cdot\frac{4}{7}\]

So we are asking what is half of four sevenths? The obvious answer is two sevenths. Two sevenths plus another two sevenths is equal to four sevenths. So all we did was divide four by two and we got the answer. (Or you could say we multiplied four by a half or by the inverse of two ... it is all the same thing).

The problem is what happens if you cannot divide the 'c' number cleanly by the 'b' number. What about:

\[\frac{1}{2}\cdot\frac{5}{7}\]

That's going to leave two and a half on top of the right hand side fraction, which is a mess. How do we solve this? Just as with addition, we need to convert the top of the right hand number into something that the bottom of the left hand number can divide into. But, if we change either the top or bottom of a fraction, the other bit of it automatically changes as well. With addition the easiest way to get the numbers to fit together was to multiply them together. Same thing happens here. We need to multiply the 'c' number by the 'b' number. To maintain the ratio between the 'c' number and the 'd' number we also need to multiply the 'd' number by the 'b' number. So we get this in our example:

\[\frac{1}{2}\cdot\frac{5\cdot2}{7\cdot2}\]

\[\frac{1}{2}\cdot\frac{10}{14}\]

We can now happily divide ten by two to get five:

\[\frac{5}{14}\]

But hang on - the five is exactly the same as we started with. Why is that? Well remember that these are rational numbers. So if you double the number on the bottom without doubling the number on the top, the ratio is now half of what it used to be. If you triple the number on the bottom without changing the number on the top the ratio is now a third of the size it used to be. Remember I said that if you change either the top or bottom then the other part has to change automatically to remain the same? Well, here we do not WANT it to remain the same. So we can just multiply the bottom of the second fraction by the bottom of the first fraction and it will give us one b'th of the second fraction automatically.

Now we have one b'th of the second fraction, but we actually want 'a' number of b'th of the second fraction. Remember we simplified what we were going to do by ignoring the 'a'. Well that's easy to bring back in now. We know how big one b'th of the second fraction is, we just need 'a' number of them. So we just multiply the answer we got by 'a'. If you have one eighth and you multiply it by three you get three eighths. If you have two sevenths and you multiply it by two you get four sevenths. The eighth or seventh bit does not change, just the number of them. So all we do now is multiply the top number of the second fraction by 'a'. Remember the top number has not changed - we found out we could leave it as 'c'.

In summary then what we do is multiply the bottom of the second fraction by the bottom of the first, and then multiply the top of the second by the top of the first:

\[\frac{a}{b}\cdot\frac{c}{d}=a\cdot\frac{c}{b\cdot d}=\frac{a\cdot c}{b\cdot d}\]

What about division of fractions? This could be painful, really really painful. Think about this:

\[\frac{a}{b}/\frac{c}{d}=\frac{\frac{a}{b}}{\frac{c}{d}}\]

Eeeewwww. That looks messy. Remember our alternate rule about division? It is just multiplication by the inverse of the number you want to divide by? We know that the inverse of three is a third. Well the inverse of a third is three. So what is the inverse of two thirds? Do we (following what we did with fractions above) ask what is the inverse of one third (three) and then multiply that answer by two (getting six)? No, because we know what the inverse of six is (a sixth) and it isn't two thirds.

Remember that we said than an inverse was what happened when you asked a question like "how big is each piece if one is split up into so many equal pieces". That got us from three to the inverse of three being a third. We can also go in the other way. Starting with a third, we can get its inverse if we ask "what is one a third of?" The answer is obviously three, because there are three ones in three. How does this help with two thirds? The question we ask is "what number is one two thirds of?". The answer is one and a half, because there are three halves in one and a half, and two halves in one.

The question may be a little bit more tricky, like "what is the inverse of five eighths". So we ask "what number is one five eighths of?". Now just take one, and split it into five equal pieces. Each one will be a fifth in size. One is now the same as five fifths. The question is now "what number is (five fifths) five eighths of?" The answer has to be eight fifths. Five fifths is five eighths of eight fifths. Sounds confusing? Work backwards. Eight fifths are eight things. Five of them will be five eighths of the whole. But each one of the things is one fifth in size. Five fifths is also one. So the inverse of five eighths is eight fifths.

Hey there is a symmetry there! What about two thirds and one and a half? Well remember that one and a half is three halves. So the inverse of two thirds is three halves. That's symmetrical as well. The rule we have found is that the inverse of a fraction is just the fraction turned upside down. That fits with our earlier, simpler, inverse of whole numbers, because three is equal to three ones, flip the one and the three, and you get one third.

So, our horrendous expression:

\[\frac{a}{b}/\frac{c}{d}=\frac{\frac{a}{b}}{\frac{c}{d}}\]

can actually be written as:

\[\frac{a}{b}/\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}\]

because dividing by c number of d'ths is the same as multiplying by the inverse of c number of d'ths, which as we have just seen is d number of c'ths. Simple.

You can also do things with fractions as exponents, but I will leave that until next time.

The Opposite of Multiplication

Following in from addition and subtraction, the opposite of multiplication is division. This is the process of taking a group of things and asking, if I have to separate these things into a certain number of equally sized groups, how many will be in each group. So if you have six apples and three hungry people, you would want to know how many apples each hungry person could have if they all had to get the same amount. This is the process of dividing six by thee. Each hungry person will get two apples.

Like subtraction and addition, we can treat the act of division as a special type of multiplication. But my feeling is that this gets hopelessly circular very quickly. The logic goes that if you divide a number by another number that is the same as multiplying the first number by the inverse of the second number.

Well, what the hell is an inverse? Good question. An inverse is the size of the individual pieces you get if you split the unit number (for us the number one remember) up into the same number of equal pieces as the original number. If you consider the apples and hungry people example, we are dividing six by three. The inverse of three is a third, because if you split the number one up into three equal pieces, each piece is a third in size. So instead of asking how many ones are in three, you ask the INVERSE of the question, which is how big is each piece if one is split up into three equal pieces.

What you then do, according to the multiplication theory of division, is multiply six by one third. Remember that multiplication is replicating one collection of things so that you end up with a second number of those collections. I did not say that you had to use WHOLE numbers.

So what is a whole number, and more importantly, what is NOT a whole number? As with negative numbers and addition, multiplying by an inverse forces us to consider another new type of number. A whole number is a number that is made up of ones and ones alone. It could have one one in it, or two, or three or fifteen billion, or negative twenty two, but it is only made up of ones. Another name for this type of number is an integer. If it is a positive integer (more than zero) then it is a natural number (but still an integer).

A number which is NOT a whole number is made up of PARTS OF ones. It may be made of some whole ones as well, but it will have parts of one in there too. So, the quantity two and a half is made up of two ones, and a half of one. The time a quarter past three is three hours (past midnight or midday) and a quarter part of an hour. The question is are these quantities or times really numbers? We quickly find ourselves back at the same kind of philosophical problem as we met at the negative number stage. Is it meaningless to take about half a kilogram of flour? No. Is it meaningless to talk about half an egg? Maybe - after all you would never go into a shop and try to buy three and a half eggs.

Well the answer, as you may have expected, is that we are quite comfortable now with the idea of these parts of one being numbers in their own right. They are just not whole numbers. They are also much less abstract than negative numbers. If you have half a cow, you can at least see and smell it, albeit the experience may not be a pleasant one.

So this part, or fraction, of one is a number is its own right. You can then multiply another number by that number. What does it mean to multiply a number by a fraction? Well remember that multiplying a group of things by a number means you end up with that number of groups of things. So all that happens if you multiply by a fraction is that you end up with a fraction of the original group. So if you have a group of twenty things, and you multiply that by a half, you end up with ten things. It still does not matter what order you do this in, so if you gather up twenty halves, and put them all together, each one finds a friend and you end up with ten whole things. In that example, of course, one half is the INVERSE of two, because there are two equal halves in one, rather than two ones in two.

So why is all this circular? Because we represent fractions as one number divided by another. So we show a half as one divided by two. So we are back to division already. Brilliant. So to be able to multiply by an inverse instead of dividing the first thing you have to do is divide to get the value of the inverse. Seemingly pointless. However it will be surprisingly useful when we come to try to multiply fractions together. Lets have a look at that kind of thing before we look at our last opposite operation.

The Opposite of Addition

Last time we looked at three mathematical operations, addition (combining two groups of things into one big thing and counting how many things you have in the big thing), multiplication (a special type of addition where you add together a certain number of equal groups of things) and raising to the power (a special type of multiplication where you multiply a number by itself a certain number of times).

Each of these operations has its opposite, and I want to say a few words about those. These do not arise in the formula that we are looking at, but they do crop up in some of the explanations of the individual components of the formula.

Firstly the opposite of addition is subtraction. You start with some number of things, remove a certain amount of them, and count how many are left. You have subtracted the certain amount from the number of things.

What you can also do is just consider this to be a special type of addition where you add a negative number to your starting number. At this point we encounter our first real philosophical problem. What is a negative number? We have been imagining lots of "things" in our discussions of addition, subtraction, and raising to the power. These things could have been anything - carrots, tractors, stamps, people, bacteria - it doesn't really matter. But you could picture the thing in your minds eye - you could touch it, look at it - it was real. We call these numbers of things, NATURAL numbers. But what can it possibly mean to talk about a negative tractor? What colour would it be?

Well, this is a thorny problem that ancient mathematicians dealt with by ignoring it and hoping it would go away. For instance the ancient Egyptian mathematicians said there was no sensible answer to the question "what number, if you multiply it by four and then add twenty, equals zero". Today we would say that the answer is negative five, because four lots of negative five make negative twenty, and twenty added to negative twenty makes zero. In the middle ages, Hindu and Muslim mathematicians worked out how to use negative numbers, primarily as a reflection of a debt. Meanwhile the western world was trying to work out how to stop the barbarians from clubbing them over the head and being eaten. As recently as the Eighteenth century (when America got its independence) mathematicians would ignore negative answers to problems because they were meaningless.

Nowadays we are relatively comfortable with the idea of negative numbers. If you gift me your only tractor for my birthday, then you have zero tractors. If, however, I borrow your only tractor you have a negative tractor because I am due to return your tractor to you.

So if you have five of something, and you add negative three somethings to it you end up with two of the somethings left.

There is an important difference between ACTUAL subtraction and adding a negative number. Remember that with addition, it does not matter which way round you do it. So, if you have four and you add negative three to that you get one. If you have negative three and add four to it you get one again. Just as we would have expected. BUT if you have four and SUBTRACT three you get one, WHILE if you have three and SUBTRACT four you get negative one. So watch with this - if subtracting you need to get the numbers in the right order, but if you just add using negative numbers it does not matter.

We use the symbol \(-\) to show the action of subtracting, and we put the same symbol to the left of a number to show that it is negative: \(-5\).

Order of Operations, Part One

OK, before we go any further we need to think about the order that the operations we looked at are done in. What do we mean? Well one plus two can only be done two ways. You can add one to two, or you can add two to one. Either way it makes no difference.

Think about it, if you have a table with one apple towards the left hand side of the table and two to the right hand side, then you have three apples on the table. If you do not touch the apples, but instead walk around to the other side of the table, you still have three apples on the table, despite the fact that now you are looking at two on the left and one on the right. Changing your perspective doesn't change the number of apples on the table.

OK, so what about three multiplied by two plus one. Lets look at that. It sounds simple. Surely the answer is seven, because three multiplied by two is six, which if you add one is seven. But hang on. What if you metaphorically walk around to the other side of the table. Then you are looking at one plus two multiplied by three. That is surely nine, because one plus two is three, and three multiplied by three is nine.

Hmm. We have already seen it doesn't matter what order you multiply things in, or what order you add things in, but apparently it DOES matter what order you do both in. We had better agree on what we mean when we say "three multiplied by two plus one". Do we mean three groups of two objects, the total of which we add one to? Or do we mean two added to one three times?

Well, by convention we mean the first option. We do multiplications first. So you first look at which numbers need multiplied together, do that, and then do any additions. I say, by convention but it turns out if you accept a certain set of axioms (stuff you just accept to be true to get on with the argument, like Hitler was a bad guy, or Jar Jar Binks is an idiot), then you can actually prove that it has to be multiplication first. Once can read about this here, but when one understands that this was only proven 150 years ago, and for the two millennia before that mathematicians just had to get by with "its convention", then one can be quite comfortable not bothering with this proof.

What about the third operation - exponents? Now we are talking about what three to the power of two plus one amounts to. Is that three to the power of two (nine) plus one (ten)? Or is it one plus two (three), to which power you then raise three (twenty seven)? Conventionally we do the exponential bit first. So the answer is ten not twenty seven. Actually it is a bit easier with exponents than addition or multiplication, because the symbols give you the clues. With the superscript notation you would write \(3^2+1\) for the first example and \(3^{2+1}\) for the second. So the actual notation is different which was not the case with addition and multiplication. A final issue which will be important is what and exponent to an exponent is. Huh? Well, what about \(2^{2^{2}}\)? Actually thats quite easy if you just think it through. THe first \(2^2\) is just \(2\cdot 2\). And then the second, tiny, power of two multplies that by itself as well. So you get \((2\cdot 2)\cdot(2\cdot 2)\). And that is just \(2\cdot 2\cdot 2\cdot 2\). Which is just \(2^4\). THat's a general rule that works quite well. All it says is that if you have a number to an exponent, then you end up with the exponent number of the numbers multiplied together. If you then duplicate that group by however many times as the second, smaller, exponent, then you are really just adding more and more of the number being multiplied. This means that you can just multiply the two exponents together to get one big exponent.

If you really want to do something out of order then you put brackets round it. So if we said three added to one multiplied by two, and we actually meant take three things, add one to them to make four and then double it, then you would say (three plus one) multiplied by two. That would look like \((3+1)\cdot 2\). Anything in brackets we do first. Even before exponents. So the system goes brackets -> exponents -> multiplication -> addition.

One more thing about brackets. Some of the stuff we are going to look at has stuff in a bracket multiplied by stuff in ANOTHER bracket. We may not want that, and we need a way to get rid of the brackets. How do we do that? Back to the fruit. Let's say that you have baskets of apples and oranges. You have the same number of apples and the same number of oranges in each basket. We are saying that you are a greengrocer with OCD, but just run with it. Let's also say that you have red baskets and blue baskets. It doesn't matter what colour the baskets are, they still have the same number of apples and the same number of oranges in them.

So how do we work out how many pieces of fruit you have in total? Firstly, you could just tip all the fruit out into a big pile and count them. Sensible enough approach, because it is going to get you the answer eventually. Lets call that the brute force approach. It may not be wise though because, having OCD, you are going to have to put all the fruit back into the baskets afterwards. A more efficient approach would be to count up how many apples and oranges you have in each basket. You then count up how many red baskets, and how many blue baskets you have. You can then ask, how many apples in blue baskets do I have, how many oranges in blue baskets do I have, how many apples in red baskets do I have and how many oranges in red baskets do I have? You would write that down like this:

\[\begin{multline}
blue baskets\cdot apples + blue baskets\cdot oranges \\
+ red baskets\cdot apples + red baskets\cdot oranges
\end{multline}
\]

That is four multiplications and three additions, which is quite a lot. If you think about it, we are just interested in the total numbers of bits of fruit, we are not bothered about the differences between apples and oranges. So we could just say, how many bits of fruit in the blue baskets and how many in the red baskets? The total number of bits of fruit in each basket is just the number of apples plus the number of oranges:

\[(apples + oranges)\]

So we could replace our previous question with this:

\[blue baskets\cdot (apples + oranges) + red baskets\cdot (apples + oranges)\]

That is still going to give us the same answer, which is the total number of bits of fruit. In fact, going further we could say, we don't need to know about how many bits of fruit are in red baskets and how many are in blue baskets - we just want the total number of bits of fruit in ALL the baskets. To do that we would need to work out the total number of baskets we have, which is:

\[(blue baskets + red baskets)\]

You would then just multiply that with the statement of the total number of bits of fruit in each basket:

\[(blue baskets + red baskets)\cdot (apples + oranges)\]

All you have to do now is two additions and one multiplication, which is much easier. The baskets you have and the total number of bits of fruit in them have not changed at all in this process. We have actually gone through this process in reverse. What we wanted to find out was how to multiply two brackets together. What we have done though is end up with two brackets. The good news is that to get the answer we wanted, we just run that process in reverse. So if you ever have two items in one bracket, and two in another bracket, and the brackets have to be multiplied, you can see that you just need to take the first item in the first bracket and multiply it by each of the items in the second bracket and add the answers. Then you take the second item in the first bracket and multiply it by each of the items in the second backet, and add the answers. Then you add both sets of answers, and hey presto, the brackets have gone.

What are Mathematical Operations?

For our restricted purposes here, a mathematical operation is some way of combining two numbers to get a third number. The third number you get will depend on the type of operation you use. For consideration of our formula, we need to look at addition, multiplication and raising to the power.

Addition is pretty much the easiest thing you can do if you are given two numbers. If you are given three apples (for some reason explanations like this always involve fruit) and then four more apples, then how many apples do you have in total? The process you go through to answer that question is called addition. (The answer is seven apples).

So basically, if you have two or more groups of things, then to work out how many things there are in all the groups combined you add them together. Simples.

We represent the process of adding two numbers by putting the symbol [+] between them.

What is Multiplication?

Multiplication is a special type of addition. If you have two numbers, then if you multiply them together it means that you take the first number and duplicate it so that you end up with the second number of individual instances of the first number. You then add up the results. So 3 multiplied by 4 is 3+3+3+3 - notice there are 4 threes in that list. It doesn't matter which way round you do the operation. So 4+4+4 is the same as 3+3+3+3. That is pretty much all there is to multiplication for our purposes.

The representation of multiplying two numbers is a bit trickier than adding, where the symbol is [+]. We are first taught in school to use [x] to show we are multiplying. So three multiplied by four is 3x4. That's the symbol you will find on your calculator to multiply. This is fine, until we get to algebra. That is a subject for another day, but suffice to say a big part of it is using variables to represent unknown numbers. Now these variables could be represented by a pictogram of a gerbil, or a house, or a dragon or anything you like really. Guess what one of the most common symbols chosen to represent these variables is? Yes, [x]. So if you were multiplying [x] by three you would find yourself writing 3xx. Not ideal. In fact far from ideal. So, there are two other ways to symbolise multiplication. The first is to put a dot between the two numbers (or variables) like this \(3\cdot x\). The second is just to forget about a symbol altogether giving 3x. Obviously the latter option cannot be used for two numbers on their own because we would not be able to tell the difference between the number 34 and three times four.

What is raising to the power?
This is a special type of multiplication, almost in the same way as multiplication is a special type of addition. Again you are given two numbers. This time you gather up a second number of first numbers and then multiply them together rather than adding. So two to the power of two is 2x2 or 2+2. Two to the power of 3 is 2x2x2, or (2+2)+(2+2). The brackets there do not do anything mathematically, they are just there to show the two sets of (2+2) that are being added together. The change from power of two in the last example to power of three in this example just means multiple everything by two again. So given that two to the power of two is (2+2), two to the power of three just means double that, or (2+2)+(2+2). So you have two groups of two groups of twos.

In the same way three to the power of four is 3x3x3x3, or ((3+3+3)+(3+3+3)+(3+3+3))+((3+3+3)+(3+3+3)+(3+3+3))+((3+3+3)+(3+3+3)+(3+3+3)). I have used double brackets to show that there are three groups of three groups of threes. If that looks a bit complicated, consider this:

three to the power of one = 3 (you only have one number so you cannot do anything with it, it just sits there)
three to the power of two = three copies of the last answer = 3+3+3
three to the power of three = three copies of the last answer = (3+3+3)+(3+3+3)+(3+3+3)
three to the power of four = three copies of the last answer = ((3+3+3)+(3+3+3)+(3+3+3))+((3+3+3)+(3+3+3)+(3+3+3))+((3+3+3)+(3+3+3)+(3+3+3))

An important point to note here is that we are just adding up to get the answer because 'powering' a number is a type of multiplication, which in turn is just a type of addition.

One difference between powering and multiplication is that it DOES matter which number does what. So while 2x3 (3+3) = 3x2 (2+2+2), two to the power of three, 2x2x2=(2+2+2+2)=8, DOES NOT equal three to the power two, 3x3=(3+3+3)=9.

Using a computer keyboard the symbol for powering numbers is [^], so 2^3 is two to the power of three. Normally though we do not use that symbol, or indeed any symbol. Instead we just write the number to be powered first, then the value of the power second. To distinguish this from multiplying (where we also usually do not use a symbol) we write the second number in superscript (which means a bit smaller and higher up). So two to the power of three is 2^3 or \(2^3\). We also call the small number, the power to which we are raising, the exponent. From this we get phrases like "exponential increase".

If you raise the number to the power of two, that is known as 'squaring' the number. This is because to find the area of a square you multiply the length of the sides by itself. If you raise a number to the power of three that is known as "cubing" the number because, similarly, you would need to multiply together three lengths of the side of the cube to get the volume of it. We really stop using special numbers after squares and cubes because our brains do not work in the fourth dimension.

We need to say a quick word about something that is not immediately obvious. One plus zero, is one. If you have something and don't add anything you have the thing you started with. Two multiplied by zero is zero. If you want zero individual instances of two, then when you come to add them up you will get nothing. So what is three to the power of zero? Is it three, or is it zero? Well, it isn't three, because as we have seen that is three to the power of one. Can it be zero? Hmmm.

Lets look at three to the power of three. That is three multiplied together three times or three multiplied by three multiplied by three. What is three to the power of one? Three. And what is three to the power of two? Three multiplied by three. So what is (three to the power of one) MULTIPLIED by (three to the power of two)? Well that is (three) MULTIPLIED by (three multiplied by three). Or just three multiplied by three multiplied by three. So we can say that:

\[3^3=3^2\cdot3^1=3^{2+1}\]

This is going to work for whatever numbers we choose, because raising to the power is just telling us how many of the chosen numbers we are multiplying together. We still end up just doing a multiplication. We have seen that it doesn't matter which order we do our multiplication in - and certainly not if it is just the same number being multiplied together again and again. So if we have a much bigger number like five to the power of six, that is just six fives in a row getting multiplied together. You could also look at it as a group of four fives multiplied together and then multiplied by a group of two fives multiplied together. It really doesn't matter where you draw the lines.

\[5\cdot5\cdot5\cdot5\cdot5\cdot5=(5\cdot5\cdot5)(5\cdot5\cdot5)=(5\cdot5)(5\cdot5\cdot5\cdot5)\]

Good. Makes sense. So what is three to the power of zero? Well if we want our rule to hold, that you just add the exponents, then the answer has to be something that satisfies this formula:

\[3^3=3^{3+0}=3^3\cdot3^0\]

If we make three to the power zero equal to zero then that doesn't work. The bit at the end will break because you are multiplying something BY zero, which makes the whole thing zero. So what we want three to the power of zero to be is a number that DOES NOT change other numbers when it multiplies them. What number is that? One. One times anything is just the anything you started with. So:

\[3^3=3^{3+0}=3^3\cdot3^0=3^3\cdot1=3^3\]

We can safely conclude then that anything to the power of zero is the number one. Well, OK, maybe it wasn't that QUICK a word.

So, in our formula, we have addition when we add one:

\[+1\]

We have multiplication where we multiply \(\pi\) by \(i\) (we will come on to what these symbols mean later):

\[\pi i\]

And we then raise \(e\) to the power of \(\pi i\):

\[e^{\pi i}\]