What is the proof that the square root of 3 is an irrational number?

Answer 1

Let's start out with the basic inequality 1 < 3 < 4.

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

1 < √3 < 2.

If you subtract all numbers by 1, you get:

0 < √3 - 1 < 1.

If √3 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 √3. Therefore, √3n must be an integer, and n must be the smallest multiple of √3 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 √3n by (√3 - 1). This gives 3n - √3n. Well, 3n is an integer, and, as we explained above, √3n is also an integer; therefore, 3n - √3n is an integer as well. We're going to rearrange this expression to (√3n - n)√3 and then set the term (√3n - n) equal to p, for simplicity. This gives us the expression √3p, which is equal to 3n - √3n, and is an integer.

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

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

Q.E.D.