Is 87 an Irrational or Rational Number...Explain?

Answer 1

Yes, here's the proof.

Let's start out with the basic inequality 81 < 87 < 100.

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

9 < √87 < 10.

If you subtract all numbers by 9, you get:

0 < √87 - 9 < 1.

If √87 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 √87. Therefore, √87n must be an integer, and n must be the smallest multiple of √87 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 √87n by (√87 - 9). This gives 87n - 9√87n. Well, 87n is an integer, and, as we explained above, √87n is also an integer, so 9√87n is an integer too; therefore, 87n - 9√87n is an integer as well. We're going to rearrange this expression to (√87n - 9n)√87 and then set the term (√87n - 9n) equal to p, for simplicity. This gives us the expression √87p, which is equal to 87n - 9√87n, and is an integer.

Remember, from above, that 0 < √87 - 9 < 1.

If we multiply this inequality by n, we get 0 < √87n - 9n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √87p < √87n. We've already determined that both √87p and √87n are integers, but recall that we said n was the smallest multiple of √87 to yield an integer value. Thus, √87p < √87n is a contradiction; therefore √87 can't be rational and so must be irrational.

Q.E.D.

The question asks if 87 is rational, not √87. 87 is rational because it can be expressed as the ratio of two integers i.e. 87 = 87/1.