Is the square root of 30 an irrational number?

Answer 1

Yes, here's the proof.

Let's start out with the basic inequality 25 < 30 < 36.

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

5 < √30 < 6.

If you subtract all numbers by 5, you get:

0 < √30 - 5 < 1.

If √30 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 √30. Therefore, √30n must be an integer, and n must be the smallest multiple of √30 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 √30n by (√30 - 5). This gives 30n - 5√30n. Well, 30n is an integer, and, as we explained above, √30n is also an integer, so 5√30n is an integer too; therefore, 30n - 5√30n is an integer as well. We're going to rearrange this expression to (√30n - 5n)√30 and then set the term (√30n - 5n) equal to p, for simplicity. This gives us the expression √30p, which is equal to 30n - 5√30n, and is an integer.

Remember, from above, that 0 < √30 - 5 < 1.

If we multiply this inequality by n, we get 0 < √30n - 5n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √30p < √30n. We've already determined that both √30p and √30n are integers, but recall that we said n was the smallest multiple of √30 to yield an integer value. Thus, √30p < √30n is a contradiction; therefore √30 can't be rational and so must be irrational.

Q.E.D.