How do we know for sure that the square root of two is an irrational number? Yes, it has a lot of digits in its decimal expansion, but since we will never see all of those digits, how do we know that they won't fall into a repeating pattern at some point? Maybe we just need to wait a little longer before we discover it has a repeating block of length 55 trillion or something like that.
In fact, we can prove that it's impossible to write the square root of two as a rational number. I don't need to see all of the digits in order to be sure of what they will do. I am certain that the square root of two cannot be rational. Let's start with some definitions:
A rational number can be written of the form a/b, with a and b both integers and b¹ 0.
The square root of two is the number that multiplies by itself to equal exactly two.
OK, that's all we need for our given information. This will be a proof by contradiction. I will start out by assuming that the square root of two is rational, and I will follow the consequences of that assumption. Eventually, the consequences will lead me to a conclusion that directly contradicts my opening assumption.
Assume that the square root of two is rational. Since rational numbers can be written in the form of integer/integer, I will write my square root of two like this:
Now here is a very important point of how I am writing my square root of two. I will deliberately choose the integers p and q such that this fraction is already reduced to lowest terms. All rational numbers can be reduced to lowest terms, so I definitely have the option to do this. Thus, p and q have no common factors. This will come back to haunt us later.
If p/q equals the square root of two, it must also be true that (p/q)2 = 2. After all, that is the definition of the square root of two. When you square it, you get two. Since squaring distributes over division, we can also say that p2/q2 = 2.
Okay, if p2/q2 = 2, I can multiply both sides by q2 to show that:
Now this is very interesting. It shows us that p2 must be an even number. An integer (q2) was multiplied by 2 to produce p2, so p2 is definitely a multiple of 2. If p2 is even, then p must also be even, because p2 consists of p times p. P certainly can't be odd if p2 is even.
I'm just continuing along the chain of consequences. We know now that p is an even number, in other words 2 times something. So I will rewrite p as 2m, where m is an integer. Now I will rewrite the p2 = 2q2 equation and substitute in 2m in place of p:
Then I divide both sides of the equation by 2 to produce this statement:
It all seems innocent enough, but I'm actually ready to move in for the kill. If 2 times m2 equals q2, then q2 must be an even number. And, of course, that compels q itself to be an even number. What's wrong with that, you say? Well, we already proved that p is an even number. If q is also an even number, then they have a common factor, namely 2. And we started off at the very beginning by saying that we would write the fraction p/q in lowest terms. So p and q did not have any common factors at the beginning of our argument. We followed all of the consequences of that assumption, and ended up showing that p and q do have a common factor. That is a direct contradiction. Since the assumption leads to a contradiction of itself, the assumption cannot possibly be true. Thus the square root of two (as we have defined it) cannot be a rational number (as we have defined that). If you accept both of those definitions, then you must accept the conclusion that the square root of two is irrational.
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