Is square root 48 a rational or irrational number?

Answer 1

Yes, here's the proof.

Let's start out with the basic inequality 36 < 48 < 49.

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

6 < √48 < 7.

If you subtract all numbers by 6, you get:

0 < √48 - 6 < 1.

If √48 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 √48. Therefore, √48n must be an integer, and n must be the smallest multiple of √48 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 √48n by (√48 - 6). This gives 48n - 6√48n. Well, 48n is an integer, and, as we explained above, √48n is also an integer, so 6√48n is an integer too; therefore, 48n - 6√48n is an integer as well. We're going to rearrange this expression to (√48n - 6n)√48 and then set the term (√48n - 6n) equal to p, for simplicity. This gives us the expression √48p, which is equal to 48n - 6√48n, and is an integer.

Remember, from above, that 0 < √48 - 6 < 1.

If we multiply this inequality by n, we get 0 < √48n - 6n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √48p < √48n. We've already determined that both √48p and √48n are integers, but recall that we said n was the smallest multiple of √48 to yield an integer value. Thus, √48p < √48n is a contradiction; therefore √48 can't be rational and so must be irrational.

Q.E.D.