Is the square root of 1000 rational or irrational?

Answer 1

It's irrational, here's the proof.

Let's start out with the basic inequality 961 < 1000 < 1024.

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

31 < √1000 < 32.

If you subtract all numbers by 31, you get:

0 < √1000 - 31 < 1.

If √1000 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 √1000. Therefore, √1000n must be an integer, and n must be the smallest multiple of √1000 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 √1000n by (√1000 - 31). This gives 1000n - 31√1000n. Well, 1000n is an integer, and, as we explained above, √1000n is also an integer, so 31√1000n is an integer too; therefore, 1000n - 31√1000n is an integer as well. We're going to rearrange this expression to (√1000n - 31n)√1000 and then set the term (√1000n - 31n) equal to p, for simplicity. This gives us the expression √1000p, which is equal to 1000n - 31√1000n, and is an integer.

Remember, from above, that 0 < √1000 - 31 < 1.

If we multiply this inequality by n, we get 0 < √1000n - 31n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √1000p < √1000n. We've already determined that both √1000p and √1000n are integers, but recall that we said n was the smallest multiple of √1000 to yield an integer value. Thus, √1000p < √1000n is a contradiction; therefore √1000 can't be rational, and so must be irrational.

Q.E.D.