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.
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