Is the square root of 8 a rational number?

Answer 1

No, it is irrational. Here is the proof:

Let's start out with the basic inequality 4 < 8 < 9.

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

2 < √8 < 3.

If you subtract all numbers by 2, you get:

0 < √8 - 2 < 1.

If √8 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 √8. Therefore, √8n must be an integer, and n must be the smallest multiple of √8 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 √8n by (√8 - 2). This gives 8n - 2√8n. Well, 8n is an integer, and, as we explained above, √8n is also an integer; therefore, 8n - 2√8n is an integer as well. We're going to rearrange this expression to (√8n - 2n)√8 and then set the term (√8n - 2n) equal to p, for simplicity. This gives us the expression √8p, which is equal to 8n - 2√8n, and is an integer.

Remember, from above, that 0 < √8 - 2 < 1.

If we multiply this inequality by n, we get 0 < √8n - 2n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √8p < √8n. We've already determined that both √8p and √8n are integers, but recall that we said n was the smallest multiple of √8 to yield an integer value. Thus, √8p < √8n is a contradiction; therefore √8 can't be rational and so must be irrational.

Q.E.D.

Answer 2

No.

Answer 3

yes