Prove by contradiction that: $A∩B=B∩A$.
My Solution: Assume to the contrary that
$A∩B ≠ B∩A$ .... (i)
Consequently, $B∩A≠ A∩B$ ...... (ii).
Since for all sets $A$ and $B$ , $A=B → B=A$.
Then, (i) and (ii) as a whole suggest $A∩B≠A∩B$.
The foregoing statement contradicts the fact that for every set $A$ , $A = A$.
By such an antilogy we've shown that $B∩A = A∩B$.
Recently enrolled for math major. I try and devising my proving skills before the first day.
On
Do you need a proof by contradiction? It's fairly easy to do it like this:
Let $x\in A\cap B$. Then of course $x\in B$, and $x\in A$, ie, $x\in B\cap A$. Thus $A\cap B\subset B\cap A$
Now let $x\in B\cap A$. Then $x\in A$, and $x\in B$, so $x\in A\cap B$. Thus $B\cap A\subset A\cap B$
$A\cap B\subset B\cap A$ and $B\cap A \subset A\cap B$, so $A\cap B = B\cap A$
You could modify this quite easily to do a proof by contradiction, it would just be an odd way I'd think.
In general the easiest way to prove equality between 2 sets is to show that each is contained in the other, through the use of an arbitrary element.
$$\begin{align} x \in A \cap B &\iff (x \in A) \land (x \in B)\\ &\iff (x\in B) \land (x \in A)\\ &\iff x \in B \cap A, \end{align}$$
where I used that the logical "AND" $\land$ is commutative. Thus we conclude $A\cap B = B \cap A.$