A graph on irrationals where p is adjacent to q if p^q or q^p is rational.

The answers are that $|E(G)|=2^{aleph_0}$ and that every vertex of $G$ has degree $aleph_0$.

Proof:
The positive real solutions to $x^y=2$ form a curve of cardinality $2^{aleph_0}$, and at most $aleph_0$ of these have $x$ or $y$ rational, so $|E(G)|=2^{aleph_0}$. (It cannot be larger, because this is also the number of pairs of vertices.)

For $g in G$ with $g>1$, the set of positive real numbers $x$ such that $g^x$ is a prime number contains at most one rational number since if there were two, then we would obtain an equation $p^q=p'$ with $p,p'$ distinct primes and $q$ rational, which is impossible by unique factorization. Thus $g$ has infinite degree. On the other hand, the degree of $g$ is at most countable since $g^x = q$ and $x^g=q$ have at most one solution $x$ each for each $q in mathbf{Q}$. Finally, the same arguments apply when $g<1$, with reciprocals of primes in place of primes.

Remark: The same statements hold for $G^*$ except that one should exclude the vertex $1$.

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