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# Is the square root of 8 a rational number?

Updated: 4/28/2022

Wiki User

6y ago

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 < âˆš8 < 3.

If you subtract all numbers by 2, you get:

0 < âˆš8 - 2 < 1.

If âˆš8 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 âˆš8. Therefore, âˆš8n must be an integer, and n must be the smallest multiple of âˆš8 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 âˆš8n by (âˆš8 - 2). This gives 8n - 2âˆš8n. Well, 8n is an integer, and, as we explained above, âˆš8n is also an integer; therefore, 8n - 2âˆš8n is an integer as well. We're going to rearrange this expression to (âˆš8n - 2n)âˆš8 and then set the term (âˆš8n - 2n) equal to p, for simplicity. This gives us the expression âˆš8p, which is equal to 8n - 2âˆš8n, and is an integer.

Remember, from above, that 0 < âˆš8 - 2 < 1.

If we multiply this inequality by n, we get 0 < âˆš8n - 2n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus âˆš8p < âˆš8n. We've already determined that both âˆš8p and âˆš8n are integers, but recall that we said n was the smallest multiple of âˆš8 to yield an integer value. Thus, âˆš8p < âˆš8n is a contradiction; therefore âˆš8 can't be rational and so must be irrational.

Q.E.D.

Wiki User

13y ago

Wiki User

6y ago

No.

Anonymous

Lvl 1
3y ago

yes