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.
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