Discrete structures exercise I have this exercise in my worksheet I am a beginner.
Prove or disprove that if $A,B$ and $C$ are sets such that $A\times B = A \times C$ then $B = C$.
 A: The statement is true if $A\neq \emptyset$. If $x \in A$ and $y \in B$, then $(x,y)\in A\times B=A\times C \Rightarrow y\in C$, so $B\subset C$. By a similar way, we can show that $C\subset B$, so $B=C$.
A: Your new question is also wrong (note that it is considered better style not to change the question, but accept a valid answer and write a new question linking this one).
To your question: If $A = \emptyset$, then $A \times B = \emptyset$ for any $B$, so $B = \{0\}$ and $C = \emptyset$ gives a counterexample.
If we add the condition $A \ne \emptyset$, then your claim is true, for: Let $B, C$ be sets such that $A \times B = A \times C$, choose $a \in A$ (here we need $A \ne \emptyset$). Let $b \in B$, then $(a,b) \in A \times B = A\times C$, so $b \in C$. For $c \in C$, we argue along the same line, we have $(a,c) \in A \times C = A \times B$, so $c \in B$. This gives $B = C$.
A: Now that another answer also has a complete proof, here is how I would do this: simplify the equation $\;A \times B = A \times C\;$ by expanding the definitions and simplifying the resulting predicate logic expression.
In other words,
\begin{align}
& A \times B = A \times C \\
\equiv & \;\;\;\;\;\text{"extensionality; definition of $\;\times\;$, twice"} \\
& \langle \forall x,y :: x \in A \land y \in B \;\equiv\; x \in A \land y \in C \rangle \\
\equiv & \;\;\;\;\;\text{"extract common part out of $\;\equiv\;$"} \\
& \langle \forall x,y :: x \in A \Rightarrow (y \in B \equiv y \in C) \rangle \\
\equiv & \;\;\;\;\;\text{"expand $\;\Rightarrow\;$; split into two independent quantifications"} \\
& \langle \forall x :: x \not\in A \rangle \;\lor\; \langle \forall y :: y \in B \equiv y \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\emptyset\;$; extensionality"} \\
& A = \emptyset \;\lor\; B = C \\
\end{align}
So the fact that $\;A = \emptyset\;$ is a special case automatically follows from the calculation.
A: consider all US citizens and subtract lets say all brazilians.
now consider again all US citizens and subtract lets say all chinese. 
what do you get? and do all brazilians equal all chinese citizens?
A: Take $A = \{1,2,3\}$, $B = \{3,4\}$ and $C = \{3,5\}$. Then $A \backslash B = A \backslash C$ but $B \neq C$.
