0

# Is the square root of 1000 rational or irrational?

Updated: 10/25/2022

Wiki User

13y ago

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 < âˆš1000 < 32.

If you subtract all numbers by 31, you get:

0 < âˆš1000 - 31 < 1.

If âˆš1000 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 âˆš1000. Therefore, âˆš1000n must be an integer, and n must be the smallest multiple of âˆš1000 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 âˆš1000n by (âˆš1000 - 31). This gives 1000n - 31âˆš1000n. Well, 1000n is an integer, and, as we explained above, âˆš1000n is also an integer, so 31âˆš1000n is an integer too; therefore, 1000n - 31âˆš1000n is an integer as well. We're going to rearrange this expression to (âˆš1000n - 31n)âˆš1000 and then set the term (âˆš1000n - 31n) equal to p, for simplicity. This gives us the expression âˆš1000p, which is equal to 1000n - 31âˆš1000n, and is an integer.

Remember, from above, that 0 < âˆš1000 - 31 < 1.

If we multiply this inequality by n, we get 0 < âˆš1000n - 31n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus âˆš1000p < âˆš1000n. We've already determined that both âˆš1000p and âˆš1000n are integers, but recall that we said n was the smallest multiple of âˆš1000 to yield an integer value. Thus, âˆš1000p < âˆš1000n is a contradiction; therefore âˆš1000 can't be rational, and so must be irrational.

Q.E.D.

Wiki User

13y ago