Irrational Square Roots

$√2$ is irrational

Let’s try to write $√2$ as a fraction, say $$√2=a/b$$ where $a$ and $b$ are integers. If we can do this at all, we can definitely do it with $$a/b$$ as a fraction in lowest terms. This means that $a$ and $b$ can’t both be even (since, if they were, we could divide both of them by 2 to get a smaller fraction with the same value).

On the other hand, if $$√2=a/b$$, then $$2=(a/b)^2={a^2}/{b^2}$$. Multiplying both sides by $b^2$, we get $2b^2=a^2$, so $a^2$ is even.

Since $a^2$ is even, then so is $a$. This means that $a=2c$ for some other integer $c$. So

But if $2b^2=4c^2$, then (by dividing by 2) $b^2=2c^2$, and so $b^2$ is even, and therefore $b$ is even. We already showed that $a$ was even, and that $a$ and $b$ can’t both be even. So it must be impossible to find any $a$ and $b$ with $$√2=a/b$$; that is, $√2$ is irrational.

The fundamental theorem of arithmetic

Any positive integer (whole number) can be written as a product of two other positive integers. For example, you can write $4=2(2)$, $5=1(5)$, or $12=3(4)$.

A prime number is an integer greater than 1 that can only be written as a product of two whole numbers in one way: as itself multiplied by 1. For example, 5 is a prime number: it isn’t a multiple of the smaller numbers 2, 3, or 4, so the only way to write it as a product is as $5=1(5)$. On the other hand, 12 is not a prime number, because you can write $12=3(4)$ as well as $12=1(12)$.

The few smallest prime numbers are 2, 3, 5, 7, 11, 13, and 17.

The prime numbers are useful to think about because of the fundamental theorem of arithmetic:

Any integer greater than 1 can be written as a product of prime numbers, ordered smallest to largest, in exactly one way.

For example, 60 can be written as $2(2)(3)(5)$, and in no other way as a product of prime numbers (ordered smallest to largest). 91 can be written as $7(13)$, and in no other way as an (ordered) product of prime numbers.

This unique way of writing a number is called its prime factorization.

$√n$ is irrational

In fact:

If $n$ is a positive integer and $√n$ is not an integer, then $√n$ is an irrational number.

We’ll use the fundamental theorem of arithmetic to show that $√12$ is irrational, but a very similar method would work for any such $n$.

Say that $√12$ can be written as a fraction, $$√12=a/b$$, for some integers $a$ and $b$ greater than 1. Just as in the first section, we can square this to get $$12={a^2}/{b^2}$$, and multiply both sides by $b^2$ to get the integer equation $12b^2=a^2$.

Now, we’ll look at the prime factorizations of the two sides of this equation. The first thing to notice is:

An integer greater than 1 is a square of a whole number if and only if in its prime factorization, each prime occurs an even number of times.

This is because, whenever we square a whole number $a$, $a^2$ can be written as a product of prime numbers by just using the prime numbers in $a$ twice. For example, $60=2(2)(3)(5)$, and so

On the other hand, say each prime occurs an even number of times in a number’s prime factorization. Then we can take that number’s square root by using each prime half as often. For instance, $324 = 4(81) = 2(2)(3)(3)(3)(3)$, so its square root is $2(3)(3) = 18$.

Because $12 = 2(2)(3)$ is not the square of a whole number, there is a prime that occurs an odd number of times (in this case, the prime 3). However, $a^2$ and $b^2$ are squares of whole numbers, so the prime 3 occurs an even number of times (which may be 0) in each of their prime factorizations. This means 3 occurs an odd number of times total in the prime factorization of $12b^2$.

So, if we could write $$√12=a/b$$, it would let us find two different prime factorizations of $a^2$ (which is also equal to $12b^2$): one where the number of 3s was even, and one where it was odd. Because the fundamental theorem of arithmetic tells us that no number can have two different prime factorizations, there can’t be any way to write $√12$ as a fraction. So $√12$ is irrational.