suppose $f:Q→R$, how we can find cardinality of set of all possible functions $f$,I know it's probably the same as cardinality of real numbers but I can't think of any bijection for that or any other way to prove that.
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1$\mathbb{R}$ has the same cardinality as $2^{\mathbb{N}}$ so $\mathbb{R}^{\mathbb{Q}}$ has the same cardinality as $2^{\mathbb{N} \times \mathbb{N}} \cong 2^{\mathbb{N}}$. – Qiaochu Yuan Dec 30 '20 at 20:58
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@QiaochuYuan but does $(a^M)^K = a^{M\times K}$. Do we know that? (Of course once we think in those terms it's probably easy to come up with the bijection.) – fleablood Dec 30 '20 at 21:03
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Yes. This is currying: https://en.wikipedia.org/wiki/Currying – Qiaochu Yuan Dec 30 '20 at 21:03
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@fleablood: Yes, it’s easy. – Brian M. Scott Dec 30 '20 at 21:35
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Yes, I was asking rhetorically for the sake of the OP. – fleablood Dec 30 '20 at 21:38