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July 30th, 2016, 03:49 AM   #11
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What I was saying is that I think the set of all exponentiations

$\displaystyle {q^r}$ : $\displaystyle q \in Q$

is countably infinite if $\displaystyle r \in Q$

but uncountable if $\displaystyle r \in R$
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July 30th, 2016, 05:10 AM   #12
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Originally Posted by Loren View Post
What about the possible cardinality for the set of operations with elements numbering as N=2^Q, as studiot may be suggesting?
If all arguments to the operation are rational, the number of possibke outputs of the operation is at most countably infinite. This is because each set of arguments produces a single output and the set of arguments is the set of $n$-tuples $(q_1,q_2,\ldots,q_n)$.

In fact, an operator can be viewed as a single-valued function $f(q_1,q_2,\ldots,q_n)$.
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July 30th, 2016, 12:33 PM   #13
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Thanks, both. I think I understand.
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