# Csc 236 H1f Lecture Summary for Week 2 Fall 2015

• Published 2015

#### Abstract

Conjecture: For all n ≥ 1, every chocolate bar with n squares can be split up (into individual squares) with n− 1 breaks. Let’s prove the conjecture using complete induction. P (n) : All chocolate bars with n squares can be split up with n− 1 breaks. Base case: A chocolate bar with only one square is already split up and needs 0 = 1− 1 breaks. So P (1) holds. Induction step: Assuming i ≥ 2 and for all 1 ≤ j < i, P (j) is true (IH), we want to prove that P (i) is also true. The first break splits the chocolate bar in two pieces A and B, with a and b squares, respectively. Note that 1 ≤ a < n, 1 ≤ b < n, and n = a + b. By the IH, A can be split up with a− 1 breaks and B can be split up with b− 1 breaks. Hence, the chocolate bar can be entirely split up with 1 + a− 1 + b− 1 = a + b− 1 = n− 1 breaks. So, P (i) holds. Conclusion: By complete induction, ∀n ≥ 1, P (n) (i.e., for all n ≥ 1, every chocolate bar with n squares can be split up with n− 1 breaks).

### Cite this paper

@inproceedings{2015Csc2H, title={Csc 236 H1f Lecture Summary for Week 2 Fall 2015}, author={}, year={2015} }