## Introduction: History of Proof of Irrationality

When this proof of irrationality of $\sqrt { 2 }$ was first discoverd, which was millenia ago, the person who came up with the proof was killed by the Pythagoreans. These people where convinced that the world is a rational place, and that everything makes sense. All the numbers that exist must also be rational of course, so they should all be neatly divisible by one another.

So, when someone stopped by to discliam that their entire worldview was wrong, this was of course enough reason to become outraged. Even though this sounds extreme, people have the tendency to become outraged when someone claims something that goes outside their worldview and picture of reality. People were outraged in the past when it was claimed that the earth was not flat. People are still outraged with religious disputes to this day. I think that this example from history illustrates very well this tendency we have to get mad about new things we might not understand yet.

Let us take a look at this infamous proof, and hopefully when I am done, you won’t get inclinations to stop by my house and hurt me in any way. If you do decide to stop by with nefarious intentions, I will try to persuade you with herbal tea and video games that I am not all bad.

### The Proof of Irrationality

Thus, without further ado let us jump into the proof. It goes like this. To begin, assume that $\sqrt { 2 }$ is rational. This would mean that it can be written as a fraction, in the form of $\frac { p }{ q }$, where p and q are integers and not further divisible. So the fraction is the lowest it can be. The reason we make this is assumption, is because this will lead to a contradiction, and that will imply that our starting assumptions were wrong. Thus, it will lead to us eventually being forced to conclude that $\sqrt { 2 }$ is not rational.

Now, since $\sqrt { 2 } = \frac { p }{ q }$, we can square both sides and get $2 = \frac { { p }^{ 2 } }{ { q }^{ 2 } }$.

By a rule of multiplication, we get that $2{ q }^{ 2 }=p^{ 2 }$.

Now, $2{ q }^{ 2 }$ is even, since it is something multiplied by 2, and everything multiplied by 2 is an even number. An odd number times 2 is even, and an even number times 2 is also even. Since $2{ q }^{ 2 }$ is even, $p^{ 2 }$ is even as well. That means that $p$ is even too, because an odd number times an odd number is odd (for example, $7\times 7=49$). Only an even number times an even number is even. So if ${ p }^{ 2 }$ is even, $p$ is even too.

If $p$ is even , we can write it as $2r$, that is to say, 2 times something. If we put this back in the original equation $2{ q }^{ 2 }=p^{ 2 }$, we get $2{ q }^{ 2 }=(2r)^{ 2 }$, which implies $2{ q }^{ 2 }=4r^{ 2 }$.

Divide both sides by 2 and we get ${ q }^{ 2 }=2r^{ 2 }$. Now, similarly as before, since ${ q }^{ 2 }$ is 2 times something, ${ q }^{ 2 }$ is even, and hence ${ q }$ is even as well.

And now comes the moment we have all been waiting for, we have arrived at the desired contradiction! By doing all this moving around, we have discovered that ${ p }$ and ${ q }$ are both even. We said at the outset (scroll back if you don’t believe me), that if $\sqrt { 2 }$ is rational it should be able to be written as a fraction $\frac { p }{ q }$, where p and q are integers and not further divisible. But it turns out that they are both even, so they should also be further divisible at least by the number 2! Therefore, if $\sqrt { 2 }$ is rational, the fraction should be further divisible and not further divisible at the same time.
This is of course impossible, since one excludes the other. We have no other choice than to conclude that out starting assumption was false, namely that $\sqrt { 2 }$ is rational. Thus, the opposite is true, that $\sqrt { 2 }$ is not rational. This is equivalent to saying that $\sqrt { 2 }$ is irrational. And there we go, it is proven once and for all, without a doubt.