# A proof that the square root of 2 is irrational

How do we know that square root of 2 is an irrational number? In other words, how do we know that √2 wouldn't have a pattern in the decimal sequence? Maybe the pattern is very well hidden and is really long, billions of digits? Even if you check the decimal expansion till the first million digits, maybe the pattern is just longer than that?

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 it is a "proof by contradiction" — if √2
WERE a rational number, we'd get a contradiction. I encourage you to let your high school students study this proof since it is very illustrative of a typical proof in mathematics and is not very hard to follow.

## The proof that 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 the lowest terms, since that can obviously be done with any fraction. Notice that in order for *a/b* to be in its simplest terms, not both of *a* and *b* can 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 can 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 if you don't believe that!

Okay, if *a* itself is an even number, then *a* is 2 times some other whole number, or *a* = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming 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 b

^{2}is even, from which follows again that

*b*itself is an even number!!!

WHY is that a contradiction? Because we started the whole process saying that *a/b* is simplified to the lowest terms, and now it turns out that *a* and *b* would both be even. So
√2 cannot be rational.

### See also

Calculators and Irrational Numbers

When I square the square root of 11 on any calculator, I get the answer 11 (exactly). That seems to indicate that the square root of 11 is a rational number, but it's not. Can you explain this?

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

Provided by Cut-the-Knot.org.

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