Now we have learned a little bit about fractions and equations, we can try and satisfy ourselves that the square root of two cannot be written as a whole number divided by another whole number (a rational number). This is a small detour on the way to \(e\), but I did promise to deal with it. We should now have all the tools we need to work it out. The way we are going to do this is to simply assume that you can actually write the square root of two as one whole number divided by another, do some logic, and if something breaks then it means that our assumption was wrong. This is called proof by contradiction.
You can use proof by contradiction in your own life as well. When you wake up in the morning, and you can't remember if it is a work (or school) day or not, just assume it is the weekend, and go back to sleep. If nobody wakes you from your slumber, your assumption was correct. If your boss (or parent, or teacher), starts shouting at you to get up, your assumption was wrong, and you have to get up. This is an easy way of proving something by not doing very much, and waiting to see if it all goes tits up. It's the lazy approach to proof.
So, let's assume that the square root of two is actually a rational number. What would that look like? Well, it would be one number divided by another number. We don't know what those two numbers would be, so let's replace them with symbols in the meantime. Let's use 'a' for the top number and 'b' for the bottom number. So it would look like this:
\[\frac{a}{b}\]
Good start. Let's bring in the equals sign and the square root of two so we can see exactly what we are talking about:
\[\frac{a}{b}=\sqrt{2}\]
Excellent. Some fraction equals the square root of two. That's the assumption that we want to make. Nothing wrong so far. Let's now do a mathematical operation on this equation.
What we want to do is get the right hand side to just be two. This means that we need to multiply it by itself to get two. Obviously the square root of two multiplied by the square root of two is two - because that's where it came from. This looks like this:
\[\frac{a}{b}\neq 2\]
Notice that we have had to change the equals sign to a not equals sign. This is because we did something to the right hand side without doing the same to the left hand side. Now, we could just multiply the left hand side by \(\sqrt{2}\) as well, but that's not going to help us, because we want to get rid of the square root. So, instead lets multiply the left hand side by 'a divided by b' (because we have said that a divided by b is the same as \(\sqrt{2}\) and that's what we multiplied the right hand side by). If we do the same thing to both sides, we still get to use the equals sign. So:
\[\frac{a}{b}\cdot\frac{a}{b}=2\]
We also know what to do to multiply fractions together, so we get:
\[\frac{a\cdot a}{b\cdot b}=2\]
This is the same as:
\[\frac{a^2}{b^2}=2\]
For those last two steps, we were just rearranging the left hand side, not performing an operation on it, so we get to keep our equals sign. Now, we want to have only \(a^2\) on the left hand side. How do we achieve that? Well, we multiply the left hand side by \(b^2\). That looks like this:
\[b^2\cdot \frac{a^2}{b^2}\neq2\]
We have had to bring in the not equals sign again, so lets get the equals sign back:
\[b^2\cdot \frac{a^2}{b^2}=2\cdot b^2\]
Now we know that something times a fraction is just something times the top number of the fraction, leaving the bottom number untouched (two times one third equals two thirds). So the left hand side above is the same as:
\[\frac{b^2\cdot a^2}{b^2}=2\cdot b^2\]
It is important to note that, again, we are not actually doing anything to the left hand side just now. The number it equals stays the same. We are just rearranging the way we write the number. This means we do not lose our equals sign. Now, remember it does not matter in which order you multiply things, so \(b^2\cdot \frac{a^2}{b^2} =\frac{b^2\cdot a^2}{b^2} = \frac{a^2\cdot b^2}{b^2}=a^2\cdot\frac{b^2}{b^2}\). This all means that we can just take the \(a^2\) out on its own leaving:
\[\frac{b^2}{b^2}\cdot a^2=2\cdot b^2\]
Now what is any number divided by itself? One. So we get:
\[1\cdot a^2=2\cdot b^2\]
Which is the same as:
\[a^2=2\cdot b^2\]
Remember, it is worth mentioning again, that since multiplying both sides by \(b^2\) we have not actually changed the numbers at all - we have just rearranged the way they are written down. So in all the changes on the left hand side since we got our equals sign back again, we have not lost it. Now, what does that number above tell us? It tells us that \(a\) multiplied by itself is an even number. Why can we say that? Well, an even number is a number that can be divided by two. And look at the equation: \(a^2\) is two times some other number. So \(a^2\) divided by two IS that other number. So \(a^2\) CAN be divided by two, so \(a^2\) is even. So if \(a^2\) is even, can we draw any conclusions about \(a\)?
Let's think about this. Any even number is any number that can be divided by two. That's the same as saying that any even number is some other number (the number you get when you divide your even number by two) MULTIPLIED by two. So lets call this number that you multiply by two to get an even number \(n\). Then any even number can be written as \(2\cdot n\) or \(2n\). If we then square that even number we get \(2\cdot n\cdot 2\cdot n\). Again remember that it doesn't matter what order we multiply things in. So that expression is the same as \(2\cdot 2\cdot n^2\). So, no matter what starting number \(n\) we choose, and in fact no matter whether the square of that number is odd or even, once we get the square of it we multiply it by two and then two again. Because we multiply by two to get the final number, what this tells us is that the final number has to be divisible by two. In turn this tells us that the square of any even number can be divided by two. And therefore that the square of any even number is an even number itself.
OK. But what about the square root of an even number (which is what we are considering). Just because every even number squares to an even number, it does not necessarily follow that the square root of an even number has to be even. That's like saying that because all sheep are fluffy white things, every fluffy white thing is a sheep. Which would be baaad news for clouds. Sorry.
Let's consider the possibility that an odd number could be the square root of an even number. Every number is capable of being multiplied by two, that's simple. So once you have multiplied every number by two, you end up with a list of all the even numbers. There is a gap between each even number which is exactly one number wide. That's because your original list which you multiplied had no gaps. If you multiplied all the original numbers by three, the gap would be two numbers wide. So between every even number there is a gap of exactly one number, and that number is an odd number, because it does not appear on the list of evens, so it cannot be divided by two. So for every even number if you add one, you move into that gap and land on an odd number. So, every odd number is an even number plus one. We have already agreed that every even number is \(2n\), so every odd number is \(2n+1\).
Good show. But what happens when we square an odd number? We get \((2n+1)(2n+1)\). You remember how you get rid of brackets don't you? We went through that whole tortuitous metaphor about the greengrocer with OCD. Once we have multiplied away the brackets we get:
\[2n\cdot 2n = 2\cdot 2\cdot n\cdot n = 4\cdot n^2\]
\[2n\cdot 1 = 2n\]
\[1\cdot 2n = 2n\]
\[1\cdot 1 = 1\]
And added all together those results look like this:
\[4n^2 + 2n + 2n +1\]
The \(n\)'s are the same thing, so we can just add them together:
\[4n^2 + 4n +1\]
Now we have added up everything, we can simplify it a bit by spotting that we have two things multiplied by four, so we can stick those things in a bracket and multiply the bracket by four:
\[4(n^2 + n)+1\]
Now, what can we tell from this? Well no matter what \(n^2\) actually is, we multiply it by four, which is two multiplied by two. Remember that as long as something is multiplied by two, the result can be divided by two, meaning that the result must be even. So no matter what goes on inside the brackets, when it gets multiplied by four it becomes even. What happens then? WE ADD ONE. We have already worked out that any even number plus one is an odd number. So no matter what \(n\) we choose to give us our original odd number, the square of that odd number will be odd because it is an even number plus one.
So we have now proved that any even number squared produces an even number, and any odd number squared produces an odd number. Now, you have to take one final thing on trust, because I have no proof for it. Other than zero and one, every other integer is either odd or even. There is no third type. And remember that because we are talking about the square root of two being one whole number divided by another whole number, all we care about are whole numbers.
So what does this tell us about \(a\)? Well we know that \(a^2\) is even (because it is some other number multiplied by two so it has to be divisible by two). If \(a^2\) is even can \(a\) be odd? No, because we just proved that any odd number squared is also an odd number. Can it be zero? No, because then the left hand side of the equation would be zero immediately, and the square root of two is not zero. Can it be one? No, because remember we are dealing with two whole numbers and 1 is not two times another whole number, one is two times a half. So if \(a\) is not odd, is not zero, is not one, it can only be even (because I have asked you to take as a given that there is no other option).
So if \(a\) is even, it is divisible by two, meaning that there is some other number that, if multiplied by two, gives \(a\). Lets call this other number \(c\). Important point, \(c\) is half of \(a\) so is smaller than \(a\). So we can now rewrite our equation by replacing a with two multiplied by \(c\). Remember the equation is:
\[a^2=2\cdot b^2\]
Replacing \(a\) for \(2\cdot c\) gives us:
\[(2\cdot c)^2=2\cdot b^2\]
Or:
\[2\cdot 2\cdot c\cdot c=2\cdot b^2\]
\[2\cdot 2\cdot c^2=2\cdot b^2\]
\[4\cdot c^2=2\cdot b^2\]
Notice that we have a two on the right hand side of the formula. We can divide that side by two to get rid of the two, as long as we divide the left hand side by two as well. If we do so, we get:
\[2\cdot c^2=b^2\]
Reverse the two sides and we get:
\[b^2=2\cdot c^2\]
What is the actual practical difference between that statement and the one we had before the substitution:
\[a^2=2\cdot b^2\]
The only difference is that we are dealing now with \(b\) and \(c\). We can still follow exactly the same logical steps with these two symbols as we did with \(a\) and \(b\). What will we end up with?
\[c^2=2\cdot d^2\]
with \(d\) equal to one half of \(b\). We could then go back to the beginning and end up with \(e\) and \(f\) with \(f\) half of \(d\) which is half of \(b\).
How long can we keep on doing this? How long can we keep running through this series of steps, getting new symbols each time? You may say, until we get to \(z\), but then we could move on to Greek letters, or hieroglyphs, or any other symbols you care to imagine, like pictures of clouds or puppies. Trust me we have plenty of symbols. The answer in maths in that once you have done something once, you can always do it again and again. I can add one to two to get three five hundred million times, and on the five hundred millionth and first time it is not going to suddenly be four. So we HAVE to be able to keep halving our variables each we run through, progressively getting smaller and smaller numbers.
But can we actually do that? Can we keep taking our original numbers and cut them in half forever and ever and never end up with a number less than one? Try it. Go on. Pick a number, any number. Cut it in half. Cut it in half again, and again, and do it forever. It will eventually get to a number that when you cut it in half it does not produce a whole number - the very last number, the finish line if you will, is the number one. If you hit that, you cut it in half and are left with a half.
(You may say, if you are being a smart arse, "A ha! I picked infinity, and I can keep cutting that in half forever." Tough. You can't do that. For a start it isn't really a number, but a concept. And secondly, you would need to have infinity on both the top AND bottom of the fraction. And what is any number divided by itself? One. And one multiplied by one is one, not two.)
So, you cannot keep reducing these starting numbers forever and ever, because you will eventually cut one into non-whole number sized pieces. But our original assumption, that the square root of two can be written as one whole number over another, says that we have to be able to just that. What is our conclusion? Our conclusion is that something has gone tits up, and broken. The office has phoned, and it's a work day, not the weekend. Our original assumption was false. Hence, the square root of two cannot be written as one whole number divided by another.
Monday, 30 May 2011
Monday, 23 May 2011
Functions & Algebra
OK, before we can progress look at \(e\) (and to the joys of proving the irrationality of the square root of two) we need to stop for a minute to think about functions. This is yet another fairly basic element of mathematics that was glossed over completely in my formal maths education.
So what is a function? In a basic sense, a function is a list of mathematical operations that you carry out on a variable. Hmm. I think what this scenario requires is an ill thought out analogy. Tradition dictates that it should involve some sort of hot drink.
So lets have our hot drink function. It would state:
1. Boil Kettle
2. Put boiled water in mug
3. Put drink flavouring in mug.
4. Stir mug
Those instructions are good enough to make coffee, tea, hot orange, or a cup-a-soup. The only bit of the instructions that would have to be changed is the "drink flavouring" bit. That could be instant coffee for coffee, a tea bag for tea, some orange squash for hot orange, or a sachet of cup-a-soup powder for the soup. All we have to do is change that bit, and follow the rest of the instructions to the letter, and we get a different drink at the end. The "drink flavouring" bit is the VARIABLE, meaning that it is the bit which can change, if we want to change the hot drink we end up with.
Once we have listed all these instructions once, it would be dull to write them out every time. So instead we use a shorthand system. Lets give the group of instructions a name: "hotbeverage". To show that the outcome of the hotbeverage instructions depends on the drink flavouring variable, we put that in brackets after the name: hotbeverage(drink flavouring). If you were reading the notation aloud, you would say "hotbeverage of drink flavouring". Which makes some sense.
That notation means that you can change what goes in the brackets (tea bag, soup powder etc) and you STILL FOLLOW exactly the same instructions, and you end up with something else. If you decided that you were going to make hot orange, you would write that as hotbeverage(orangesquash). You have replaced the placeholder with the ingredient. Someone reading that would know that you meant "follow the list of instructions that I have called hotbeverage replacing the variable 'drink flavouring' with orangesquash".
Now, lets bring some algebra into this. First question, what the fuck is algebra? Answer, a very useful system of mathematical thinking that Muslims came up with in the middle ages, based on earlier work of Hindus, all at a time when western society was trying to work out how the Romans had built the bloody aqueducts. You can tell the word has Arabic origins because of the 'al' - much like alchemy, and algorithm.
So what actually is it? Well, in the example above we had a variable - something that we could change to change the outcome of the function. Algebra is a way of working with these variables. Sometimes that are referred to as 'unknowns' which is coming at the thing from a different angle. Before we had the concept of algebra, mathematical thinking was really done using geometry. So if you saw an ancient mathematician scribbling away on papyrus, or a sand board, they would be drawing lines and circles and so on, NOT the kind of symbols and operations that we use these days. The difference that algebra brought to the table, so to speak, was the ability to think more abstractly.
The Muslims made the leap from lines and circles to abstract ideas, but they could only express those ideas in long wordy sentences. Such as "if you take the third part of the first party raised to the second power and then find the root of the difference between that and the fifth part of the....". You get the drift. Much later than the Muslims, Western Europeans started using letters of the alphabet in place of the unknowns. They also started using the symbols we have already looked at to show what operations you were doing to numbers or these unknowns. So finally instead of drawings, or long wordy sentences, we finally had the numbers, letters and symbols that we now recognise as 'algebra'.
Tradition has it that we use letters from the end of the alphabet to represent these variables or unknowns, starting with \(x\). We have already used a letter from the Greek alphabet to represent a number - \(\pi\). Is \(x\) the same? No. \(\pi\) is always the same number a little bit more than 3, roughly \(\tfrac{22}{7}\). It does not vary. We just use the letter symbol, because we would never finish writing out the number otherwise. \(\pi\) does not vary - it remains constant, and so we call it a ... constant. It is like the symbol for two, '2', or five, '5'. Those symbols always means two or five. In the same way \(\pi\) always means just a little bit more than three.
So we normally use \(x\) to represent the first variable, and then \(y\) and \(z\) if there are more variables. In other subjects where algebra is bring used, you find different letters, or even other Greek letters. Physics is bloody littered with different letters ('s' usually stands for a variable which is the speed of something, 'd' for distance, 'v' for velocity (not the same as speed but never mind that now) and so on). So, a typical algebraic equation would look like this:
\[x+1=3\]
We are now supposed to follow some formal algebraic rules to get a statement that starts with:
\[x=\]
The bit on the right side of the = tells us what \(x\) actually is in this example. In this case, the rules we follow are to deduct 1 from each side of the equation:
\[x=2\]
We now know what \(x\) is. We can plug 2 into the place of \(x\) in the first equation:
\[2+1=3\]
Yep, that is correct. Richard Feynman explained algebra best when he said that it is just a puzzle game, where the goal is to find out what \(x\) is! You can look back at the original statement \(x+1=3\) and rephrase it as the question, "what number, if you add one to it, makes three?" Then it is just a puzzle which is easily solved.
That was an example of an algebraic equation. That English language question I translated it into was how algebra was done before the symbols were invented. Could ancient mathematicians have solved this using their lines and circles? Yes. What you do is draw a long line. And then take a compass set to a specific, and completely arbitrary, width. Stick the point anywhere along the long line. Draw a circle. The circle will cross the line at two points, the same distance from the pointy end of the compass. You now have three points, the original bit where you stuck the compass in, and two where the circle crosses the line. Can we answer the question what plus one is three? No, because we only have two identical line segments.
So now stick the compass pointy bit on either of the new points and (without changing the width of the compass) draw another circle. One crossing point will be the very first pointy bit, and the second will be a new point on the line. We now have four points and three equal segments of line. We now have three identical line segments. In this step we added one identical line segments, so the answer to the question must be "how many line segments did we have before we added this third one", and the answer as we saw a second ago was "two".
It should be obvious, but it bears repeating. Whenever you hear of an ancient mathematical proof or theorem, and you think "that's a doddle, I could have done that, and I am an idiot", remember they didn't have symbolic algebra. They had tools to draw circles, and tools to draw straight lines. That was it. And they still managed to come up with good stuff. Geniuses.
Anyway, we started to talk about functions. We have seen a function written out in normal language, so what does a function look like in symbolic algebra? Well, we have already seen one of those as well. A function is basically just the side of the equation that \(x\) is on. So in the above example, the function is:
\[x+1\]
The instructions we follow are really just to add 1 to our variable. We need to give our function a name, and traditionally we give it a name one letter long. We start using the letter \(f\), for the first function we use. If we need to use more than one to solve a problem, we give the next one the name \(g\). And so on. It would be terribly complex if we chose letters nearer \(xyz\) because you would start to get confused between the names of functions and the names of variables. That would be bad.
There is nothing special about the letters chosen, by the way. A variable called \(x\) is not one greater or less than a variable called \(y\). We could just as well be using pictures of clouds, buses, or lawnmowers. It is nothing more than a marker. Same thing applies to the names chosen for functions. I said earlier that we put the variable in brackets after the name of the function. So we end up with:
\[f(x)\]
That means the function is called \(f\) and it involves a variable called \(x\). We know what the function looks like, so we can describe the whole thing as:
\[f(x)=x+1\]
We can then replace the \(x\) with two like this:
\[f(2)=2+1\]
\[f(2)=3\]
We can now talk about \(f(x)\) rather than repeating all the steps every time. This is not too onerous with \(x+1\)) but we will see some more insane functions in due course. Much like the beverage example above you would say that this is "eff of ecks".
The important difference here is that an equation has an equals sign in it, whereas a function does not. OK, pedant, yes the example just about has an equals sign, but that is just telling you what the function is. The actual function (the bit on the right of the equals sign) only has a variable, the sign for addition, and a number. It has no equals sign. An equation, on the other hand, tells you that two different looking thinks are equivalent to each other, whereas a function is list of things that you do to \(x\), say, to get a result.
So what is a function? In a basic sense, a function is a list of mathematical operations that you carry out on a variable. Hmm. I think what this scenario requires is an ill thought out analogy. Tradition dictates that it should involve some sort of hot drink.
So lets have our hot drink function. It would state:
1. Boil Kettle
2. Put boiled water in mug
3. Put drink flavouring in mug.
4. Stir mug
Those instructions are good enough to make coffee, tea, hot orange, or a cup-a-soup. The only bit of the instructions that would have to be changed is the "drink flavouring" bit. That could be instant coffee for coffee, a tea bag for tea, some orange squash for hot orange, or a sachet of cup-a-soup powder for the soup. All we have to do is change that bit, and follow the rest of the instructions to the letter, and we get a different drink at the end. The "drink flavouring" bit is the VARIABLE, meaning that it is the bit which can change, if we want to change the hot drink we end up with.
Once we have listed all these instructions once, it would be dull to write them out every time. So instead we use a shorthand system. Lets give the group of instructions a name: "hotbeverage". To show that the outcome of the hotbeverage instructions depends on the drink flavouring variable, we put that in brackets after the name: hotbeverage(drink flavouring). If you were reading the notation aloud, you would say "hotbeverage of drink flavouring". Which makes some sense.
That notation means that you can change what goes in the brackets (tea bag, soup powder etc) and you STILL FOLLOW exactly the same instructions, and you end up with something else. If you decided that you were going to make hot orange, you would write that as hotbeverage(orangesquash). You have replaced the placeholder with the ingredient. Someone reading that would know that you meant "follow the list of instructions that I have called hotbeverage replacing the variable 'drink flavouring' with orangesquash".
Now, lets bring some algebra into this. First question, what the fuck is algebra? Answer, a very useful system of mathematical thinking that Muslims came up with in the middle ages, based on earlier work of Hindus, all at a time when western society was trying to work out how the Romans had built the bloody aqueducts. You can tell the word has Arabic origins because of the 'al' - much like alchemy, and algorithm.
So what actually is it? Well, in the example above we had a variable - something that we could change to change the outcome of the function. Algebra is a way of working with these variables. Sometimes that are referred to as 'unknowns' which is coming at the thing from a different angle. Before we had the concept of algebra, mathematical thinking was really done using geometry. So if you saw an ancient mathematician scribbling away on papyrus, or a sand board, they would be drawing lines and circles and so on, NOT the kind of symbols and operations that we use these days. The difference that algebra brought to the table, so to speak, was the ability to think more abstractly.
The Muslims made the leap from lines and circles to abstract ideas, but they could only express those ideas in long wordy sentences. Such as "if you take the third part of the first party raised to the second power and then find the root of the difference between that and the fifth part of the....". You get the drift. Much later than the Muslims, Western Europeans started using letters of the alphabet in place of the unknowns. They also started using the symbols we have already looked at to show what operations you were doing to numbers or these unknowns. So finally instead of drawings, or long wordy sentences, we finally had the numbers, letters and symbols that we now recognise as 'algebra'.
Tradition has it that we use letters from the end of the alphabet to represent these variables or unknowns, starting with \(x\). We have already used a letter from the Greek alphabet to represent a number - \(\pi\). Is \(x\) the same? No. \(\pi\) is always the same number a little bit more than 3, roughly \(\tfrac{22}{7}\). It does not vary. We just use the letter symbol, because we would never finish writing out the number otherwise. \(\pi\) does not vary - it remains constant, and so we call it a ... constant. It is like the symbol for two, '2', or five, '5'. Those symbols always means two or five. In the same way \(\pi\) always means just a little bit more than three.
So we normally use \(x\) to represent the first variable, and then \(y\) and \(z\) if there are more variables. In other subjects where algebra is bring used, you find different letters, or even other Greek letters. Physics is bloody littered with different letters ('s' usually stands for a variable which is the speed of something, 'd' for distance, 'v' for velocity (not the same as speed but never mind that now) and so on). So, a typical algebraic equation would look like this:
\[x+1=3\]
We are now supposed to follow some formal algebraic rules to get a statement that starts with:
\[x=\]
The bit on the right side of the = tells us what \(x\) actually is in this example. In this case, the rules we follow are to deduct 1 from each side of the equation:
\[x=2\]
We now know what \(x\) is. We can plug 2 into the place of \(x\) in the first equation:
\[2+1=3\]
Yep, that is correct. Richard Feynman explained algebra best when he said that it is just a puzzle game, where the goal is to find out what \(x\) is! You can look back at the original statement \(x+1=3\) and rephrase it as the question, "what number, if you add one to it, makes three?" Then it is just a puzzle which is easily solved.
That was an example of an algebraic equation. That English language question I translated it into was how algebra was done before the symbols were invented. Could ancient mathematicians have solved this using their lines and circles? Yes. What you do is draw a long line. And then take a compass set to a specific, and completely arbitrary, width. Stick the point anywhere along the long line. Draw a circle. The circle will cross the line at two points, the same distance from the pointy end of the compass. You now have three points, the original bit where you stuck the compass in, and two where the circle crosses the line. Can we answer the question what plus one is three? No, because we only have two identical line segments.
So now stick the compass pointy bit on either of the new points and (without changing the width of the compass) draw another circle. One crossing point will be the very first pointy bit, and the second will be a new point on the line. We now have four points and three equal segments of line. We now have three identical line segments. In this step we added one identical line segments, so the answer to the question must be "how many line segments did we have before we added this third one", and the answer as we saw a second ago was "two".
It should be obvious, but it bears repeating. Whenever you hear of an ancient mathematical proof or theorem, and you think "that's a doddle, I could have done that, and I am an idiot", remember they didn't have symbolic algebra. They had tools to draw circles, and tools to draw straight lines. That was it. And they still managed to come up with good stuff. Geniuses.
Anyway, we started to talk about functions. We have seen a function written out in normal language, so what does a function look like in symbolic algebra? Well, we have already seen one of those as well. A function is basically just the side of the equation that \(x\) is on. So in the above example, the function is:
\[x+1\]
The instructions we follow are really just to add 1 to our variable. We need to give our function a name, and traditionally we give it a name one letter long. We start using the letter \(f\), for the first function we use. If we need to use more than one to solve a problem, we give the next one the name \(g\). And so on. It would be terribly complex if we chose letters nearer \(xyz\) because you would start to get confused between the names of functions and the names of variables. That would be bad.
There is nothing special about the letters chosen, by the way. A variable called \(x\) is not one greater or less than a variable called \(y\). We could just as well be using pictures of clouds, buses, or lawnmowers. It is nothing more than a marker. Same thing applies to the names chosen for functions. I said earlier that we put the variable in brackets after the name of the function. So we end up with:
\[f(x)\]
That means the function is called \(f\) and it involves a variable called \(x\). We know what the function looks like, so we can describe the whole thing as:
\[f(x)=x+1\]
We can then replace the \(x\) with two like this:
\[f(2)=2+1\]
\[f(2)=3\]
We can now talk about \(f(x)\) rather than repeating all the steps every time. This is not too onerous with \(x+1\)) but we will see some more insane functions in due course. Much like the beverage example above you would say that this is "eff of ecks".
The important difference here is that an equation has an equals sign in it, whereas a function does not. OK, pedant, yes the example just about has an equals sign, but that is just telling you what the function is. The actual function (the bit on the right of the equals sign) only has a variable, the sign for addition, and a number. It has no equals sign. An equation, on the other hand, tells you that two different looking thinks are equivalent to each other, whereas a function is list of things that you do to \(x\), say, to get a result.
Monday, 16 May 2011
It's All Greek to Me
Yes \(\pi\) is a letter. In actual fact it is a letter of the greek alphabet. Quite a lot of mathematical place markers are. \(\theta\) (pronounced theta) is commonly used to represent and unknown or varying angle in problems. \(\pi\) is a bit different because it does not represent an unknown or fluctuating quantity, but instead it represents a single, solitary number. The number it represents is a ratio, an unchanging ratio - at least in our universe.
Let us consider a circle. A useful definition of a circle is the set of all points which lie the same distance from a single point. You can easily make a circle by using a pointy sharp thing with a fixed length of string attached to it, and with some sort of device that leaves a mark attached to the end of the string. You also need a flat surface. You can draw circles on non-flat surfaces, like footballs or saddles, but that way lies a special kind of madness called non euclidean geometry, which I am not touching with a bargepole. You stick the pointy sharp bit into the surface you wish to draw a circle on, pull the string tight, and place the marking device on the surface. What you have marked is a point which is the string's length away from the pointy bit. Now if you lift the marking device up and place it down (with the string still held taught) ANYWHERE else, you will mark another point the string's length away from the pointy bit. If you keep doing this randomly, you will eventually see the outline of a circle start to form - defined by all the individual points.
Of course, that is not the sensible way to draw a circle. Instead of lifting the marking device up every time, you just leave it touching the surface and, again keeping the string tight, you move it in either direction. I say either direction because you will find that you are constrained to only move two ways, clockwise or counter clockwise. Once you have moved back to the point you started from you have drawn a circle. Well done. Now, we can say five things about the circle you have marked. We can of course say what colour it is, but that is irrelevant for geometrical purposes. Secondly we can say how thick the line is that marks out the circle. This will depend on your choice of marking device. A felt tip pen or highlighter will leave a thicker line than a biro, and a crayon or a piece of chalk will leave a thicker line that all of the foregoing.
In pure geometry though, lines do not have thickness, and points do not have an area. A line, including the outline of the circle we drew, only extends in one direction. This makes sense, because of our definition., The circle is the set of all points EXACTLY the same distance from another point, not sort of the same distance depending on how big your marking device is. So the thickness of the line marking the circle, as with its colour, is irrelevant to us.
Thirdly, we can say how big the area is inside the circle. This is going to be measured on a two dimensional surface, so the answer will be some number of whatever units you like to the power of two. By that I mean the units are squared, not the number of them. So if your chosen measure of length is the flangit, then your chosen measurement of area will be square flangits. A square flangit is just a square whose length is exactly a flangit. Squaring a number is the same as raising it to the power of two. The area in the circle is therefor some number of square flangits.
Fourthly, we could talk about the length of the line we have drawn on the surface. If we took it in our minds eye and straightened it out, and measured it, how long would it be? Fifthly and finally we can talk about the length of the bit of string between the pointy sharp bit and the marking device.
As it turns out, the area in the circle and the length of the line you draw are related ONLY to the length of that bit of string. If that seems remarkable, remember that it doesn't matter where on your surface you poke your pointy sharp thing your circle will still look the same. The only determining factor of the size of the circle you create is the length of string you allow between the pointy sharp bit and the marking device.
So what does all this have to do with \(\pi\)? Well, if you take your length of string, and multiply it by \(2\pi\) you get the length of the line you draw. So \(\pi\) is the RATIO between the length of the string and the length of the line around the outside of the circle you draw with that string. It matters not a jot how long your bit of string is, the circle that results will ALWAYS have a line that is \(2\pi\) times the length of that string long around its outside.
So why do we say \(2\pi\) and not just \(\pi\) for this ratio? Good question. It turns out that it is a lot more useful to work with \(2\pi\) because \(\pi\) itself turns up in far more places on its own. If we went with \(\pi\) then we would keep having to half it. For example, the area of the circle that we can describe? It is the length of the string times itself (this makes the units squares remember) multiplied by \(\pi\). If we defined \(\pi\) as the direct ratio between the length of the string and the line, that number used to find the area would have to be half of \(\pi\). Which looks messy.
The line which forms the circle is called the circumference of the circle. Another word for something that runs around the outside of something else is the perimeter. The word for perimeter in ancient greek started with a letter in the ancient greek alphabet. Can you guess which letter? The length of the bit of string is called the radius of the circle. The point we made with the pointy sharp thing is the centre of the circle.
There is a line which is double the length of the radius, and is described as the straight line from one point on the circle, through the centre, to another point on the circle (which will inevitably be exactly opposite the starting point). This line is the diameter of the circle. Because this line is the radius multiplied by two, and because (as we have seen) it does not matter which order you multiply things in, you can get rid of the two from the \(2\pi\) when you talk about the ratio between the diameter and the circumference. The circumference is therefor just \(\pi\) times the diameter. This is because the radius times two times \(\pi\) is the same as two times the radius times \(\pi\), and two times the radius is the diameter.
What does this all look like? Glad you asked. Have a gander at this snazzy diagram:
I don't like the diameter though, because it has nothing to do with the construction of the circle. The diameter comes about once a circle has been made, while the radius is used in making the circle in the first place. So I prefer to think of \(\pi\) as it relates to the radius not the diameter.
The number itself is a bit more than three. It is not a whole number you can count on your fingers, and neither is it a fraction (although \(\tfrac{22}{7}\) comes pretty close. If you were to write out \(\pi\) as a decimal number as three point something something something, you would be writing "somethings" for ever, because there are an infinite amount of them. So instead of worrying about all those "somethings" we just write \(\pi\). This is a fantastic animation of the circumference "unrolling" to show it is \(pi\) times the diameter:
That animation was created (and GPL licensed) by wikipedia user John Reid.
Let us consider a circle. A useful definition of a circle is the set of all points which lie the same distance from a single point. You can easily make a circle by using a pointy sharp thing with a fixed length of string attached to it, and with some sort of device that leaves a mark attached to the end of the string. You also need a flat surface. You can draw circles on non-flat surfaces, like footballs or saddles, but that way lies a special kind of madness called non euclidean geometry, which I am not touching with a bargepole. You stick the pointy sharp bit into the surface you wish to draw a circle on, pull the string tight, and place the marking device on the surface. What you have marked is a point which is the string's length away from the pointy bit. Now if you lift the marking device up and place it down (with the string still held taught) ANYWHERE else, you will mark another point the string's length away from the pointy bit. If you keep doing this randomly, you will eventually see the outline of a circle start to form - defined by all the individual points.
Of course, that is not the sensible way to draw a circle. Instead of lifting the marking device up every time, you just leave it touching the surface and, again keeping the string tight, you move it in either direction. I say either direction because you will find that you are constrained to only move two ways, clockwise or counter clockwise. Once you have moved back to the point you started from you have drawn a circle. Well done. Now, we can say five things about the circle you have marked. We can of course say what colour it is, but that is irrelevant for geometrical purposes. Secondly we can say how thick the line is that marks out the circle. This will depend on your choice of marking device. A felt tip pen or highlighter will leave a thicker line than a biro, and a crayon or a piece of chalk will leave a thicker line that all of the foregoing.
In pure geometry though, lines do not have thickness, and points do not have an area. A line, including the outline of the circle we drew, only extends in one direction. This makes sense, because of our definition., The circle is the set of all points EXACTLY the same distance from another point, not sort of the same distance depending on how big your marking device is. So the thickness of the line marking the circle, as with its colour, is irrelevant to us.
Thirdly, we can say how big the area is inside the circle. This is going to be measured on a two dimensional surface, so the answer will be some number of whatever units you like to the power of two. By that I mean the units are squared, not the number of them. So if your chosen measure of length is the flangit, then your chosen measurement of area will be square flangits. A square flangit is just a square whose length is exactly a flangit. Squaring a number is the same as raising it to the power of two. The area in the circle is therefor some number of square flangits.
Fourthly, we could talk about the length of the line we have drawn on the surface. If we took it in our minds eye and straightened it out, and measured it, how long would it be? Fifthly and finally we can talk about the length of the bit of string between the pointy sharp bit and the marking device.
As it turns out, the area in the circle and the length of the line you draw are related ONLY to the length of that bit of string. If that seems remarkable, remember that it doesn't matter where on your surface you poke your pointy sharp thing your circle will still look the same. The only determining factor of the size of the circle you create is the length of string you allow between the pointy sharp bit and the marking device.
So what does all this have to do with \(\pi\)? Well, if you take your length of string, and multiply it by \(2\pi\) you get the length of the line you draw. So \(\pi\) is the RATIO between the length of the string and the length of the line around the outside of the circle you draw with that string. It matters not a jot how long your bit of string is, the circle that results will ALWAYS have a line that is \(2\pi\) times the length of that string long around its outside.
So why do we say \(2\pi\) and not just \(\pi\) for this ratio? Good question. It turns out that it is a lot more useful to work with \(2\pi\) because \(\pi\) itself turns up in far more places on its own. If we went with \(\pi\) then we would keep having to half it. For example, the area of the circle that we can describe? It is the length of the string times itself (this makes the units squares remember) multiplied by \(\pi\). If we defined \(\pi\) as the direct ratio between the length of the string and the line, that number used to find the area would have to be half of \(\pi\). Which looks messy.
The line which forms the circle is called the circumference of the circle. Another word for something that runs around the outside of something else is the perimeter. The word for perimeter in ancient greek started with a letter in the ancient greek alphabet. Can you guess which letter? The length of the bit of string is called the radius of the circle. The point we made with the pointy sharp thing is the centre of the circle.
There is a line which is double the length of the radius, and is described as the straight line from one point on the circle, through the centre, to another point on the circle (which will inevitably be exactly opposite the starting point). This line is the diameter of the circle. Because this line is the radius multiplied by two, and because (as we have seen) it does not matter which order you multiply things in, you can get rid of the two from the \(2\pi\) when you talk about the ratio between the diameter and the circumference. The circumference is therefor just \(\pi\) times the diameter. This is because the radius times two times \(\pi\) is the same as two times the radius times \(\pi\), and two times the radius is the diameter.
What does this all look like? Glad you asked. Have a gander at this snazzy diagram:
I don't like the diameter though, because it has nothing to do with the construction of the circle. The diameter comes about once a circle has been made, while the radius is used in making the circle in the first place. So I prefer to think of \(\pi\) as it relates to the radius not the diameter.
The number itself is a bit more than three. It is not a whole number you can count on your fingers, and neither is it a fraction (although \(\tfrac{22}{7}\) comes pretty close. If you were to write out \(\pi\) as a decimal number as three point something something something, you would be writing "somethings" for ever, because there are an infinite amount of them. So instead of worrying about all those "somethings" we just write \(\pi\). This is a fantastic animation of the circumference "unrolling" to show it is \(pi\) times the diameter:
That animation was created (and GPL licensed) by wikipedia user John Reid.
Monday, 9 May 2011
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.
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.
Monday, 2 May 2011
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.
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.
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