No. You can well multiply two irrational numbers and get a result that is not an irrational number.
The main subgroup is the rational numbers. The set of irrational numbers is not closed with regard to addition basic arithmetical operations and so does not form a group.
Yes, the set is closed.
No. A number cannot be closed under addition: only a set can be closed. The set of rational numbers is closed under addition.
If you mean the set of non-negative integers ("whole numbers" is a bit ambiguous in this sense), it is closed under addition and multiplication. If you mean "integers", the set is closed under addition, subtraction, multiplication.
no it is not
Hennd
No. The set of rational numbers is closed under addition (and multiplication).
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.
Rational and irrational numbers are part of the set of real numbers. There are an infinite number of rational numbers and an infinite number of irrational numbers. But rational numbers are countable infinite, while irrational are uncountable. You can search for these terms for more information. Basically, countable means that you could arrange them in such a way as to count each and every one (though you'd never count them all since there is an infinite number of them). I guess another similarity is: the set of rational numbers is closed for addition and subtraction; the set of irrational numbers is closed for addition and subtraction.
The set of even numbers is closed under addition, the set of odd numbers is not.
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.
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.
The main subgroup is the rational numbers. The set of irrational numbers is not closed with regard to addition basic arithmetical operations and so does not form a group.
Yes. The set of real numbers is closed under addition, subtraction, multiplication. The set of real numbers without zero is closed under division.
Yes, the set is closed.