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Chapter 5: Problem 81

Show that a three-dimensional \(2^{n} \times 2^{n} \times 2^{n}\) checkerboard with one \(1 \times 1 \times 1\) cube missing can be completely covered by \(2 \times 2 \times 2\) cubes with one \(1 \times 1 \times 1\) cube removed.

Short Answer

Expert verified
By induction, a \(2^n \times 2^n \times 2^n\) checkerboard with one missing cube can be covered by \(2 \times 2 \times 2\) cubes with one cube removed.

Step by step solution

01

Understand the Problem

We need to show that a 3D checkerboard of size \(2^n \times 2^n \times 2^n\), with one cube missing, can be completely covered using \(2 \times 2 \times 2\) cubes with one cube removed.
02

Base Case for n=1

Consider the base case where \(n=1\). The 3D checkerboard is of size \(2^1 \times 2^1 \times 2^1 = 2 \times 2 \times 2\). Removing one \(1 \times 1 \times 1\) cube results in 7 cubes remaining, which can be covered by one \(2 \times 2 \times 2\) cube with one cube removed.
03

Inductive Hypothesis

Assume that the statement holds true for \(k=n\), i.e., a \(2^n \times 2^n \times 2^n\) cube with one cube missing can be completely covered with \(2 \times 2 \times 2\) cubes with one cube removed.
04

Inductive Step

Now, consider the \(k=n+1\) case, i.e., a \(2^{n+1} \times 2^{n+1} \times 2^{n+1}\) checkerboard with one cube missing. Divide this larger cube into 8 smaller \(2^n \times 2^n \times 2^n\) cubes. By the inductive hypothesis, each of these smaller cubes (with one missing cube) can be covered with \(2 \times 2 \times 2\) cubes with a single cube missing.
05

Applying the Hypothesis

When only one \(2^n \times 2^n \times 2^n\) block has the missing cube, the hypothesis ensures that each block can be covered by the \(2 \times 2 \times 2\) cubes with one \(1 \times 1 \times 1\) cube removed. Thus, the entire \(2^{n+1} \times 2^{n+1} \times 2^{n+1}\) structure can be covered.
06

Conclusion

Since the base case holds and the inductive step succeeds, by mathematical induction, a \(2^n \times 2^n \times 2^n\) checkerboard with one cube missing can be completely covered by \(2 \times 2 \times 2\) cubes with one cube removed for any \(n \ge 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

3D Checkerboard
Let's start by imagining a three-dimensional checkerboard. This isn't like a regular 2D game board鈥攊t's more like a Rubik's cube.
For this problem, we consider a large cube divided into smaller unit cubes. Specifically, we deal with a cube of size \(2^n \times 2^n \times 2^n\), where \(n\) is a positive integer.
Now, taking out one of these tiny unit cubes results in a slightly imperfect 3D checkerboard. Our ultimate goal is to cover this imperfect 3D checkerboard using only \(2 \times 2 \times 2\) cubes that also have one unit cube missing.
This is a classic example of a problem in higher dimensions, emphasizing spatial thinking and pattern recognition.
Base Case
In any induction proof, we start with what is called the 'base case.' It's the simplest version of the problem.
For \(n=1\), our 3D checkerboard becomes \(2^1 \times 2^1 \times 2^1 = 2 \times 2 \times 2\). This is an 8-unit cube.
Remove one unit cube, and you have 7 remaining. We need to check if these 7 remaining cubes can be covered with one \(2 \times 2 \times 2\) cube that also has one unit cube removed.
It's clear that they can be covered, as there is exactly one \(2 \times 2 \times 2\) cube left with one missing unit. This satisfies our base case.
Inductive Hypothesis
With the base case confirmed, we move to the inductive hypothesis. This step is like saying, 'Assume the rule works for some number \(k\).'
Here, we assume that any \(2^k \times 2^k \times 2^k\) checkerboard with one missing unit cube can be covered by \(2 \times 2 \times 2\) cubes missing one cube each.
This assumption helps us bridge the gap to prove it for the next number, \(k+1\).
Inductive Step
Next is the inductive step, which essentially says, 'If our assumption works for \(k\), it should work for \(k+1\) as well.'
Consider the larger \(2^{k+1} \times 2^{k+1} \times 2^{k+1}\) cube. We can break this larger cube into 8 smaller \(2^k \times 2^k \times 2^k\) cubes.
By our inductive hypothesis, each of these smaller cubes, with one cube missing, can be covered by \(2 \times 2 \times 2\) cubes missing one unit each.
Therefore, the entire larger cube with only one missing unit can be covered.
This completes the inductive step, ensuring that our covering rule works for any \(n\).
In conclusion, thanks to mathematical induction, we confirm that a \(2^n \times 2^n \times 2^n\) checkerboard with one cube missing can indeed be covered by \(2 \times 2 \times 2\) cubes with one cube removed, for any \(n \geq 1\).

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