# Proving The Square Root of 2 is Irrational

Like we said before, a sequence is just a list of numbers, and it can help us majorly prove some things about rational and irrational numbers. For starters, though, let's run through a quick proof showing *why* an irrational number is irrational.

Our good buddy is happy to step in. How can we prove that he's irrational? Besides pointing to his complicated conspiracy theories and fifty cats, we mean.

Let's start with an overview of the proof, and then we'll get into the guts of it. Remember to sterilize your hands first.

**Overview of the proof: **

This is a **proof by contradiction**. And yet, this is *not* a proof by contradiction. See what we did there?

We'll start out with the assumption that is rational. This means the statement (*) below has to be true. We'll then show that our assumption also implies the statement (*) has to be false. Since no statement can be both true and false, we must have made a bad assumption to start out with. This means can't be rational after all. Ready for us to prove ourselves wrong? Here we go...

**Guts of the proof:**

If were rational, we could write as a fraction . Since any fraction has an equivalent fraction in lowest terms, we can assume is in lowest terms (i.e. it can't be reduced). Someone must have already come by and simplified it for us. We'll have to send them a card.

(*) and there is no whole number that divides both *a* and *b*.

; therefore, by multiplying both sides by *b*, we get:

Let's show that *a* has to be even. Trust us on this one.

First we square both sides of the equation:

Okay, check this out: *b* is an integer, right? It would have to be, since we assumed at the beginning that is a rational number, which means *a* and *b* were both integers. So *b*^{2} is also an integer, 'cause we're not gonna get a fraction or decimal when we square an integer. So since , we know* a*^{2} is 2 times some integer *b*^{2}. That means *a*^{2} is an even number; 2 times *any *integer is an even number. This means *a* must also be even (think about it—whenever you square an even number, the result is always even). So *a *= 2 × *n* for some number *n*.

Now that that's settled, let's show that *b* has to be even. We do have a point, and we are getting to it. Bear with us.

Substitute 2 × *n* for *a* in the equation :

Again, square both sides:

Divide both sides by 2:

This means *b* must be even, for the same reasons that *a* had to be even. Come on, *b*, that's not cool—come up with your own reasons.

But if both *a* and *b* are even, then 2 would divide both *a* and *b*, which means the fraction isn't in lowest terms. Uh-oh. This is a contradiction: you're saying there's no whole number that divides both *a* and *b*, but 2 divides both *a* and *b*? Our fraction is reducible *and* not reducible? It's okay if this gives you a bit of a headache; it's hard to imagine things that are impossible! Like the idea of anyone enjoying figuring this stuff out for a living.

Since our initial assumption led to a contradiction, our initial assumption must have been false. This means can't be rational.

This is pretty much the opposite of a real-life scenario. Generally, when two people both have even tempers, they can be *quite* rational.