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Is the set of composite number closed under multiplication?
Is { 0, 20 } closed under multiplication
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.
Why the set of rationals does not form a group wrt multiplication?
All the elements in a group must be invertible with respect to the operation. The element 0, which belongs to the set does not have an inverse wrt multiplication.
Why are the rational numbers under the operation of multiplication not a group?
I believe it is because 0 does not have an inverse element.
If you multiply the numerator and the denominator of a fraction by the same number the result will be a fraction?
Yes, unless the number used for multiplication is 0. In that case you will have 0/0 which is not defined.
