Irrational numbers are not closed under multiplication, then why are real and complex numbers considered rings? Aren't real and complex numbers supposed to have irrational numbers?
*Made the mistake of thinking if subsets were to be invalid, the higher sets would be invalid too. Thanks to @Zachary Selk for clarify it!
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Yes, real numbers do indeed contain irrational numbers. However, the only thing irrationals can "escape to" are rationals, i.e. reals. So reals are a ring.
As a matter of fact, every nonzero ring contains a subset that is not a ring.
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Well, the irrationals are not closed under multiplication, e.g., $\sqrt 2 \cdot \sqrt 2 = 2$. And what would be the zero and unit elements?
However, what is interesting is that one can define a ring structure on the set ${\Bbb I}$ of irrational numbers by using (any) bijection $\phi:{\Bbb I}\rightarrow{\Bbb R}$.
The irrationals are not closed under multiplication, but the reals and complex numbers are. Why should the irrationals not being closed matter to whether the reals and complex numbers are rings? $\{2\}$ is not closed under multiplication (or addition), but it is a subset of the integers, which form a ring.