My Math Forum Cardinality of all operations on rational numbers

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 July 30th, 2016, 03:49 AM #11 Senior Member   Joined: Jun 2015 From: England Posts: 905 Thanks: 271 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 $\displaystyle r \in R$
July 30th, 2016, 05:10 AM   #12
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Quote:
 Originally Posted by Loren 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 $f(q_1,q_2,\ldots,q_n)$.

 July 30th, 2016, 12:33 PM #13 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory Thanks, both. I think I understand.

 Tags cardinality, numbers, operations, rational

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