# A proof that the square root of 2 is irrational

Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. It is the most common proof for this fact and is by contradiction.

How do we know that square root of 2 is an irrational number? In other words, how do we know that √2 doesn't have a pattern in its decimal sequence? Maybe the pattern is very well hidden and is really long, billions of digits?

Here is where mathematical proof comes in. The proof that √2 is indeed irrational is usually found in college level math texts, but it isn't that difficult to follow. It does not rely on computers at all, but instead is a "proof by contradiction": if √2 WERE a rational number, we'd get a contradiction. I encourage all high school students to study this proof since it illustrates so well a typical proof in mathematics and is not hard to follow.

## A proof that the square root of 2 is irrational

Let's suppose √2 is a rational number. Then we can write it
√2 = *a/b* where *a*, *b* are whole numbers, *b* not zero.

We additionally assume that this *a/b* is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for *a/b* to be in simplest terms, both of *a* and *b* cannot be even. One or both must be odd. Otherwise, we could simplify *a/b* further.

From the equality √2
= *a/b* it follows that 2 = *a*^{2}/*b*^{2}, or *a*^{2} = 2 · *b*^{2}. So the square of *a* is an even number since it is two times something.

From this we know that *a* itself is __also__ an even number. Why? Because it can't be odd; if *a* itself was odd, then *a* · *a* would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!

Okay, if *a* itself is an even number, then *a* is 2 times some other whole number. In symbols, *a* = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute *a* = 2k into the original equation 2 = *a*^{2}/b^{2}, this is what we get:

2 | = | (2k)^{2}/b^{2} |

2 | = | 4k^{2}/b^{2} |

2*b^{2} | = | 4k^{2} |

b^{2} | = | 2k^{2} |

This means that *b*^{2} is even, from which follows again that *b* itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that *a/b* was simplified to lowest terms, and now it turns out that *a* and *b* both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.

### See also

Irrational Numbers; Rational Square Roots

How can you tell whether root 10 is a terminating or repeating decimal, or an irrational number? Are some square roots rational?

More proofs that square root of 2 is irrational

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