Proof

Let \(P(n)\) be the statement, “it takes \(n-1\) breaks to reduce a \(n\)-square chocolate bar to single squares.”

Base case: Consider \(P(2)\text{.}\) The squares must be arranged into a \(1\times 2\) rectangle, and we require \(2-1 = 1\) breaks to reduce this to single squares.

Inductive case: Fix an arbitrary \(n\ge 2\) and assume \(P(k)\) is true for all \(k \lt n\text{.}\) Consider a \(n\)-square rectangular chocolate bar. Break the bar once along any row or column. This results in two chocolate bars, say of sizes \(a\) and \(b\text{.}\) That is, we have an \(a\)-square rectangular chocolate bar, a \(b\)-square rectangular chocolate bar, and \(a+b = n\text{.}\)

We also know that \(a \lt n\) and \(b \lt n\text{,}\) so by our inductive hypothesis, \(P(a)\) and \(P(b)\) are true. To reduce the \(a\)-square bar to single squares takes \(a-1\) breaks; to reduce the \(b\)-square bar to single squares takes \(b-1\) breaks. Doing this results in our original bar being reduced to single squares. All together it took the initial break, plus the \(a-1\) and \(b-1\) breaks, for a total of \(1+a-1+b-1 = a+b-1 = n-1\) breaks. Thus \(P(n)\) is true.

Therefore, by strong induction, \(P(n)\) is true for all \(n \ge 2\text{.}\)