Let $a,b$ cardinal numbers. If $a \cdot b=1 \iff a=1 \land b=1$

Question

$(\implies)$

Let $A,B$ sets such that $\#(A)=a,\#(B)=b$ and $1=\#(\{x\})$ with $x\in \mathcal{U}$.

Since $\#(A\times B)=1$ then exist a bijective function $f$ between $A\times B$ and $\{x\}$, but $f$ is biyective then exist the function $g$ such that $g(x)=(u,v); (u,v)\in A\times B$.

Then, as $g$ is a bijective function and $ran(g)=dom(f)$ then $A\times B=\{(u,v) \}$.By definition of $A\times B$ then $u\in A \land v\in B\implies \{u\}=A \land \{v\}=B\implies \#(A)=1 \land \#(B)=1\implies a=1 \land b=1$

$(\Longleftarrow)$

Let $A=\{u\}$ and $B=\{v\}$ such that $\#(A)=a=1\land \#(B)=b=1$. By definition of $A\times B\implies u\in A\land v\in B\implies A\times B=\{(u,v)\}\implies \#(A\times B)=1\implies a\cdot b=1 $

It's okay my proof?

Answer 1

Mathematically, your proof is almost solid. I only see two problems, and they're small problems:

By definition of $A\times B$ then $u\in A \land v\in B\implies \{u\}=A \land \{v\}=B\implies \#(A)=1 \land \#(B)=1\implies a=1 \land b=1$

The fact that $u \in A$ does not imply that $\{u\} = A$. How do you know that $A$ doesn't contain other elements besides $u$? See if you can come up with a proof for that part.

The second problem is even smaller. You write:

By definition of $A\times B\implies u\in A\land v\in B\implies A\times B=\{(u,v)\}$

But the fact that $u\in A\land v\in B$ does not imply that $A\times B=\{(u,v)\}$, because $A$ and $B$ could contain other elements besides $u$ and $v$. But your reader already knows that $A = \{u\}$ and $B = \{v\}$, so just leave out the part about $u\in A\land v\in B$, and write this:

By the definition of $A\times B$, we can find that $A\times B=\{(u,v)\}$

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You have some English language mistakes in your write-up of the proof. Here's a revised version:

$(\implies)$

Let $A,B$ be sets such that $\#(A)=a,\#(B)=b$ and $1=\#(\{x\})$ with $x\in \mathcal{U}$.

Since $\#(A\times B)=1$, there exists a bijective function $f$ between $A\times B$ and $\{x\}$, but since $f$ is bijective, there exists a function $g$ such that $g(x)=(u,v)$, where $(u,v)\in A\times B$.

Then, as $g$ is a bijective function and $ran(g)=dom(f)$, we have $A\times B=\{(u,v) \}$. By the definition of $A\times B$, we have $u\in A \land v\in B\implies \{u\}=A \land \{v\}=B\implies \#(A)=1 \land \#(B)=1\implies a=1 \land b=1$

$(\Longleftarrow)$

Let $A=\{u\}$ and $B=\{v\}$ such that $\#(A)=a=1\land \#(B)=b=1$. By the definition of $A\times B$, we can find that $u\in A\land v\in B\implies A\times B=\{(u,v)\}\implies \#(A\times B)=1\implies a\cdot b=1 $