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.
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Show that set of Irrational Numbers does not form a group w.r.t multiplication.
By definition, the two sets do not overlap. This is because the irrationals are defined as the set of real numbers that are not members of the rationals.
Factional exponents, in general, are not rational. For example, the length of the diagonal of a unit square, which is sqrt(2).
Negative integers, integers, negative rationals, rationals, negative reals, reals, complex numbers are some sets with specific names. There are lots more test without specific names to which -10 belongs.
Is { 0, 20 } closed under multiplication
Yes, every Cauchy sequence of real numbers is convergent. In other words, the real numbers contain all real limits and are therefore continuous, and yes the integers are discrete in that the set of integers only contains (very very few, with respect to the set of rationals) rational numbers, i.e. their values can always be accurately displayed unlike the set of reals which is dense with irrational numbers. It's so dense with irrationals in fact, that by comparison, the set of rationals can be called a null set, however that is a different topic.