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Irrational Numbers are not closed under any of the fundamental operations. You can always find cases where you add two irrational numbers (for example), and get a rational result. On the other hand, the set of real numbers (which includes both rational and irrational numbers) is closed under addition, subtraction, and multiplication - and if you exclude the zero, under division.

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Q: What are irrational numbers closed under?

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no it is not

None.

No, they are not. An irrational number subtracted from itself will give 0, which is rational.

No. The set of rational numbers is closed under addition (and multiplication).

No. For example, the sum of pi and -pi is zero, which is rational - while each of the addends is irrational.

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no it is not

None.

No. You can well multiply two irrational numbers and get a result that is not an irrational number.

No; here's a counterexample to show that the set of irrational numbers is NOT closed under subtraction: pi - pi = 0. pi is an irrational number. If you subtract it from itself, you get zero, which is a rational number. Closure would require that the difference(answer) be an irrational number as well, which it isn't. Therefore the set of irrational numbers is NOT closed under subtraction.

Hennd

No. Sqrt (2) is irrational. Square it, or raise it to any even power, and it becomes rational. The set is not closed under exponentiation.

No, they are not. An irrational number subtracted from itself will give 0, which is rational.

No. 4 root 2 and 2 root 2 are both irrational. Divide the first by the second you get 2. Which is not a member of the set of irrational numbers.

No. The set of rational numbers is closed under addition (and multiplication).

No. For example, the sum of pi and -pi is zero, which is rational - while each of the addends is irrational.

No. To say a set is closed under multiplication means that if you multiply any two numbers in the set, the result is always a member of the set. If, say, the 2 numbers are radical 2 and radical 2 we have (1.4142...)(1.4142...) which by definition equals 2. The result is not an irrational number, so the set is not closed.

No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.

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