No. You can well multiply two irrational numbers and get a result that is not an irrational number.
No. sqrt(8) is irrational sqrt(2) is irrational but sqrt(8) /sqtr(2) = sqrt(4) = ±2 is not irrational.;
Yes. The empty set is closed under the two operations.
Yes.
No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.
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.
None.
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.
no it is not
It cannot be closed under the four basic operations (addition, subtraction, multiplication, division) because it is indeed possible to come up with two negative irrational numbers such that their sum/difference/product/quotient is a rational number, indicating that the set is not closed. You will have to think of a different operation.
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. You can well multiply two irrational numbers and get a result that is not an irrational number.
No, they are not. An irrational number subtracted from itself will give 0, which is rational.
No. sqrt(8) is irrational sqrt(2) is irrational but sqrt(8) /sqtr(2) = sqrt(4) = ±2 is not irrational.;
Hennd
Yes. The empty set is closed under the two operations.
+,-,X only