Suppose that $A$ and $B$ are sets of real numbers and $A \cong B$ (isomorphic). Let $2A = \{2a|a \in A\}$ and $3B = \{3b|b \in B\}$. Prove that $A \times 2A \cong B \times 3B$ by finding a bijective function between the sets.
So prove that $A \times 2A$ is a bijective to $B \times 3B$. I am stuck on how to the bijection is injective and surjective. I don't know how to use the given information about the cardinality between $A$ and $B$.
Hint: First, "A x 2A is a bijection" and "B x 3B is a bijection" does not make any sense here, these things are sets and not functions. Be sure to know the difference before going on. Now, if $f:A \to B$ is a bijection between $A$ and $B$, define $\varphi: A \times 2A \to B \times 3B$ by $$\varphi(a, 2a') = (f(a), 3f(a')).$$Check that $\varphi$ is bijective using that $f$ is, by hypothesis. Good luck!