# Proving square root of 2 is irrational with well ordering principle?

1. Nov 5, 2012

### symaticc

1. The problem statement, all variables and given/known data
I know how to prove that square root of 2 is irrational using the well ordering principle but what i'm wondering is, how can we use the well ordering principle to prove this when the square root of two isn't even a subset of the natural numbers? Doesn't the well ordering principle only hold for natural numbers and not rational numbers? If we prove that the well ordering principle does not hold for the square root of 2, aren't we just proving that the square root of 2 is not a natural number?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 5, 2012

### micromass

Staff Emeritus
Can you include the proof here?? Where exactly did you use well-ordering of rational numbers??

3. Nov 5, 2012

### symaticc

Okay so what I did was first, assume that square root of 2 is rational. let √2 = a/b
2 = a^2/b^2
2b^2 = a^2
a^2 is even therefore a is even so it can be written as 2n...
2b^2 = (2n)^2
2b^b = 4n^2
b^2 = 2n^2
2 = b^2 / n^2
√2 = b/n
Since we started with √2 = a/b and then got √2 = b/n where b/n < a/b, if we repeat the step above and keep repeating it, we will always get smaller representations of √2 which means that there is no smallest element and there for √2 is irrational by the well ordering principle which states that every non-empty subset of N has a least element.

Im wondering, if the well ordering principle is only for subsets of N and it doesn't hold for √2, doesn't that just mean that √2 is not a natural number? How does that prove that √2 is irrational?

4. Nov 5, 2012

### Petek

By definition, a set S is well ordered if every non-empty subset of S contains a least element. The set of positive rational numbers is not well-ordered because, for example,

$$\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \}$$

has no least element.

I take it that you've been asked to prove that $\sqrt{2}$ is irrational by using the well ordering property of the positive integers. Here's a hint to get you started. Assume that $\sqrt{2} = \frac{a}{b}$ where a and b are positive integers. Thus, $b\sqrt{2} = a$ is a positive integer. This implies that the set S, defined by

$$S = \{n \in \mathbb{N} : n\sqrt{2} \in \mathbb{N}\}$$

is non-empty. What does the well ordering principle tell you about S?