Question

Prove the following statements using either direct or contrapositive proof. Let \(a, b, c \in Z\) and \(n \in \mathbb{N}\). If \(a \equiv b(\bmod n)\) and \(a \equiv c(\bmod n)\), then \(c \equiv b(\bmod n)\).

Short answer

By direct proof, from the given congruences \(a \equiv b (\bmod n)\) and \(a \equiv c (\bmod n)\), it's proven that \(c \equiv b (\bmod n)\).

Step-by-step solution

Restate the given congruences

The congruences \(a \equiv b \bmod n\) and \(a \equiv c \bmod n\) mean that \(n\) divides \((a-b)\) and \((a-c)\) respectively, which can be restated as there exist integers \(i\) and \(j\) such that \(a-b=ni\) and \(a-c=nj\).

Rewrite the equations

The above two equations can be rewritten as \(a = ni+b\) and \(a = nj+c\). Set the two right-hand sides equal to get \(ni+b = nj+c\).

Manipulate the equality to prove the target statement

Subtract \(nj\) and \(b\) from both sides of the equation obtained in Step 2: \(ni - nj = c - b\).\ This can be rewritten as \(n(i-j) = c-b\), which means \(n\) divides \((c-b)\). So, by the definition of modulo, we have \(c \equiv b (\bmod n)\).