Is the square root of 13 an irrational number?

Answer 1

Yes, here's the proof.

Let's start out with the basic inequality 9 < 13 < 16.

Now, we'll take the square root of this inequality:

3 < √13 < 4.

If you subtract all numbers by 3, you get:

0 < √13 - 3 < 1.

If √13 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √13. Therefore, √13n must be an integer, and n must be the smallest multiple of √13 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply √13n by (√13 - 3). This gives 13n - 3√13n. Well, 13n is an integer, and, as we explained above, √13n is also an integer, so 3√13n is an integer too; therefore, 13n - 3√13n is an integer as well. We're going to rearrange this expression to (√13n - 3n)√13 and then set the term (√13n - 3n) equal to p, for simplicity. This gives us the expression √13p, which is equal to 13n - 3√13n, and is an integer.

Remember, from above, that 0 < √13 - 3 < 1.

If we multiply this inequality by n, we get 0 < √13n - 3n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √13p < √13n. We've already determined that both √13p and √13n are integers, but recall that we said n was the smallest multiple of √13 to yield an integer value. Thus, √13p < √13n is a contradiction; therefore √13 can't be rational and so must be irrational.

Q.E.D.