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no it is not

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Q: Is the set of irrational numbers closed under addition?
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What answer choice shows that the set of irrational numbers is not closed under addition?

Hennd


What are irrational numbers closed under?

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.


Can you add two rational numbers and get an irrational number?

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


Are irrational numbers closed under addition?

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


What is closed and not-closed under addition?

The set of even numbers is closed under addition, the set of odd numbers is not.


Are irrational addition numbers closed under the closure property?

No. For example, the square root of two plus (minus the square root of two) = 0, which is not an irrational number.


Why are odd integers closed under multiplication but not under addition?

The numbers are not closed under addition because whole numbers, even integers, and natural numbers are closed.


Which of the following is an example of why irrational numbers are not closed under addition?

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.


What operations are irrational numbers closed under?

None.


Is the set of irrational numbers closed under mulriplication?

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


When are complex numbers closed under addition?

Quite simply, they are closed under addition. No "when".


Is the set of irrational numbers closed under subtraction?

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.