Is the square root of 50 irrational or rational?

Answer 1

It is irrational, here's the proof.

Let's start out with the basic inequality 49 < 50 < 64.

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

7< √50 < 8.

If you subtract all numbers by 7, you get:

0 < √50 - 7 < 1.

If √50 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 √50. Therefore, √50n must be an integer, and n must be the smallest multiple of √50 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 √50n by (√50 - 7). This gives 50n - 7√50n. Well, 50n is an integer, and, as we explained above, √50n is also an integer, so 7√50n is an integer too; therefore, 50n - 7√50n is an integer as well. We're going to rearrange this expression to (√50n - 7n)√50 and then set the term (√50n - 7n) equal to p, for simplicity. This gives us the expression √50p, which is equal to 50n - 7√50n, and is an integer.

Remember, from above, that 0 < √50 - 7 < 1.

If we multiply this inequality by n, we get 0 < √50n - 7n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √50p < √50n. We've already determined that both √50p and √50n are integers, but recall that we said n was the smallest multiple of √50 to yield an integer value. Thus, √50p < √50n is a contradiction; therefore √50 can't be rational and so must be irrational.

Q.E.D.

Answer 2

√50 is irrational.

Answer 3

Irrational.