no it is not
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.
No. The set of rational numbers is closed under addition (and multiplication).
Don't know about the "following" but any irrational added to its additive inverse is 0, which is rational. Therefore, the set of irrationals is not closed under addition.
None.
no it is not
Hennd
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.
No. The set of rational numbers is closed under addition (and multiplication).
The set of even numbers is closed under addition, the set of odd numbers is not.
No. For example, the square root of two plus (minus the square root of two) = 0, which is not an irrational number.
The numbers are not closed under addition because whole numbers, even integers, and natural numbers are closed.
Don't know about the "following" but any irrational added to its additive inverse is 0, which is rational. Therefore, the set of irrationals is not closed under addition.
None.
No. You can well multiply two irrational numbers and get a result that is not an irrational number.
Quite simply, they are closed under addition. No "when".
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.