Assume that a chocolate bar consists of n squares arranged in a rectangular patt

Bevan Mcdonald

Bevan Mcdonald

Answered question

2021-10-11

Assume that a chocolate bar consists of n squares arranged in a rectangular pattern. The entire bar, a smaller rectangular piece of the bar, can be broken along a vertical or a horizontal line separating the squares. Assuming that only one piece can be broken at a time, determine how many breaks you must SUccessively make to break the bar into n separate squares. Use strong induction to prove your answer.

Answer & Explanation

Asma Vang

Asma Vang

Skilled2021-10-12Added 93 answers

Step 1
Given: A chocolate bar that consists of n squares arranged in a rectangle.
To proof: We make n1 breaks to break a chocolate bar.
PROOF BY STRONG INDUCTION
Let P(n) be "We make n1 breaks to break a chocolate bar."
Basis step n=1
Let us consider a chocolate bar that consists of 1 square. The chocolate bar is then already completely broken into pieces and thus we require 0=11=n1 breaks to completely break the chocolate bar.
Thus P(1) is true.
Inductive step We assume that P(1), P(2), .... , P(k) are all true, thus we make i breaks to break a chocolate bar consisting of i squares (i=1,2,,k).
We then need to prove that P(k+1) is true. Let us consider a chocolate bar with k+1 squares.
Let m and q be positive integers such that k+1=mg and such that the squares of the chocolate bar form a rectangle with m rows and q columns.
Let the first break be breaking off the last column of the rectangle. Then we divide the chocolate bar in a (m1)×q rectangle and in a 1×q rectangle.
Since P((m1)q) is true, we require (m1)g1 breaks to break the (m1)×q rectangle.
Since P(q) is true, we require q1 breaks to break the 1×q rectangle.
We then note that we require (m1)q1+(q1)+1 break to complete break the chocolate bar consisting of k+1 squares.
Step 2
(m1)q1+(q1)+1=mqq1=mq1=(k+1)1
Thus we require (k+1)1 breaks to break the chocolate bar with k+1 squares and thus P(k+1) is true.
Conclusion By the principle of strong induction, P(n) is true for all positive integers n.
Note: The inductive hypothesis in this proof contains more true statements than a proof using (not-strong) mathematical induction.
Answer
We make n1 breaks to break a chocolate bar.

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