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Show that if \(n\) and \(k\) are positive integers, then \(\lceil n / k\rceil=\) \(\lfloor(n-1) / k\rfloor+ 1\)

Short Answer

Expert verified
True, \(\lceil \frac{n}{k} \rceil = \lfloor \frac{n-1}{k} \rfloor + 1\).

Step by step solution

01

- Understand Ceiling Function

The ceiling function \(\lceil x \rceil\) of a real number \(x\) is the smallest integer greater than or equal to \(x\).
02

- Understand Floor Function

The floor function \(\lfloor x \rfloor\) of a real number \(x\) is the largest integer less than or equal to \(x\).
03

- Change Variable

Let \(y = \frac{n}{k}\). We need to show that \(\lceil y \rceil = \lfloor (y-\frac{1}{k}) \rfloor + 1\).
04

- Apply Floor Function on \(y-\frac{1}{k}\)

Consider \(y-\frac{1}{k} = \frac{n}{k} - \frac{1}{k} = \frac{n-1}{k}\). Applying the floor function, we obtain \(\lfloor \frac{n-1}{k} \rfloor\).
05

- Add 1

According to the requirement, add 1 to the result from Step 4: \(\lfloor \frac{n-1}{k} \rfloor + 1\).
06

- Compare with Ceiling Function

Notice that \(\lceil \frac{n}{k} \rceil\) by definition is the smallest integer greater than or equal to \(\frac{n}{k}\). Meanwhile, \(\lfloor \frac{n-1}{k} \rfloor + 1\) adjusts for any fractional part ensuring it aligns with \(\lceil \frac{n}{k} \rceil\). Thus both expressions yield the same result.

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Key Concepts

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

Ceiling Function
The ceiling function, denoted as \(\backslash lceil x \rceil\), returns the smallest integer greater than or equal to a given real number \(x\). For example, \(\backslash lceil 2.3 \rceil = 3\) and \(\backslash lceil 5 \rceil = 5\). It's useful when you need to 'round up' a number to its nearest whole integer.
Floor Function
The floor function, denoted as \(\backslash lfloor x \rfloor\), gives you the largest integer less than or equal to a given real number \(x\). For instance, \(\backslash lfloor 4.7 \rfloor = 4\) and \(\backslash lfloor 3 \rfloor = 3\). This operation 'rounds down' a number to its nearest whole integer. In mathematical expressions, the floor function assesses the integer part of a number.
Proof in Discrete Mathematics
Proofs in discrete mathematics often involve demonstrating the equality of two mathematical expressions or showing that a condition holds for all elements in a specific set. In the provided exercise, we are proving the relationship between the ceiling and floor functions. We began by setting a variable \(\backslash y = \backslash frac{n}{k}\). Using definitions, we showed that \(\backslash lceil \backslash frac{n}{k} \rceil = \lfloor \backslash frac{n-1}{k} \rfloor + 1\), by breaking down the problem and applying the properties of ceiling and floor functions. These methods are foundational tools in learning how to logically, and rigorously, prove mathematical statements.

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Most popular questions from this chapter

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