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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.

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Q: Is the square root of 8 a rational number?

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No, in fact unless an integer has an integral square root, the square root is always irrational. Note that some rational fractions can have rational square roots, though. Example: sqrt(4/9) = sqrt(4)/sqrt(9) = 2/3, which is rational. Or sqrt(9/16) = 3/4

The square of a negative number is always positive. The square of -8 = 64, a rational number

No. Square roots of whole numbers that don't result in whole numbers (e.g. the square root of 64 is 8), are considered to be "irrational numbers".

Square root of 59 is 7.681145748 Nearest whole number would be 8

Sure. Let's take an example with numbers instead of variables. 8/9 is a rational number - the ratio of two integers. If you square it, you get (8/9)2 = 82/92 = 64/81; that is, if you square each integer, you get an integer again. Therefore, the resulting number is rational again - a number that can be expressed as the ratio of two integers. Replace with any other integers; you will still have a ratio of integers if you square it.

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The square root of 64 is 8 which is a rational number

It is a rational number because the square root of 64 is 8 which is a rational number

No, they are not.

The square root of 64 is 8, which is a rational number.

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Rational. It is a perfect square with answer exactly 8The square root of 64 is the integer 8 (8 x 8 = 64) so it is rational.

no! the square root of 8 is not retional but the cube root is retional