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Q: Is the square root 6 rational or irrational?

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No. The square root of 6 is irrational.

irrational

4.6 is rational.

The square root of six is irrational. Do you mean a rational approximation?

Irrational: it amounts to roughly 2.449489743

It is irrational. * The square root of any positive integer, except of a perfect square, is irrational. * The product of an irrational number and a rational number (except zero) is irrational.

No because the square root of 36 is 6 which is a rational number

The square of 36 is 1,296 ... rational.The square root of 36 is 6 ... also rational.

The square root of 36 is 6 which is a rational number because it can be expressed as an improper fraction in the form of 6/1

No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.

Sometimes it is and sometimes it isn't. The square root of 2 and the square root of 3 are both irrational, as is their product, the square root of 6. The square root of 2 and the square root of 8 are both irrational, but their product, the square root of 16, is rational (in fact, it equals 4).

The square root of any positive square number is always rational as for example the square root of 36 is 6 which is a rational number.

The square roots of 36 are Â±6, therefore, sqrt(36) is a rational number.

Rational numbers:1, 2, 3, 4, 5, 6, 7, 8, 9, 10Irrational numbers:square root of (2)square root of (3)square root of (5)square root of (6)square root of (7)square root of (8)square root of (10)square root of (11)square root of (12)square root of (13)

It is known that the square root of an integer is either an integer or irrational. If we square root2 root3 we get 6. The square root of 6 is irrational. Therefore, root2 root3 is irrational.

It is an irrational number.

6 times the square root of 6

Yes. For example, the square root of 3 (an irrational number) times the square root of 2(an irrational number) gets you the square root of 6(an irrational number)

6 divided by sqrt(2) is an irrational number and so cannot be represented by a rational fraction.

No, they are not.

Yes. The square root of a fraction is the square root of the numerator over the square root of the denominator. First simplify the fraction (making mixed numbers into improper fractions). Now consider the numerator and denominator separately as whole numbers. Only perfect squares (the squares of whole numbers) have rational square roots. If either, or both, of the numerator and denominator is not a perfect square, the square root of the fraction will be irrational √(11/6) = (√11)/(√6). Neither 11 nor 6 is a perfect square, thus √(11/6) is irrational.

The square root of 6 is an irrational number; it can only be approximated as a decimal. √6 ≈ 2.449

For example, in the case of integers: the square root of a positive integer is either an integer (in the case of perfect squares, such as the square root of 1, 4, 9, 16, etc.), or an irrational number (such as the square root of 2, 3, 5, 6, etc.). Similar in the case of the square root of a rational number (fraction): if you don't specifically choose fractions with perfect squares in the numerator and denominator (e.g., 4/9), you will end up with an irrational square root.

No, it is not irrational because it is a square root of a negative number - which falls into the set of Complex numbers. Irrational numbers can not have an imaginary component.

the square root of 3, the square root of 5, the square root of 6, the square root of 7, the square root of 8 etc