Problem 55 - Discrete Mathematics with Applications

Question

Prove each, where \(x \in \mathbb{R}\) and \(n \in \mathbf{Z}.\) \(\left\lfloor\frac{n^{2}}{4}\right\rfloor=\frac{n^{2}-1}{4}\) if \(n\) is odd

Step-by-Step Solution

Answer

For \(n\) being an odd integer, we can express \(n\) as \(n = 2k + 1\), where \(k\) is an integer. Substituting this expression in the given equation and simplifying, we can prove that the equation \(\left\lfloor\frac{n^{2}}{4}\right\rfloor=\frac{n^{2}-1}{4}\) holds true for odd integers \(n\). This is because the simplified equation \(\left\lfloor k^2+k\right\rfloor=k^2+k\) represents an integer value in both sides.

Step 1: (Step 1: Expressing \(n\) as an odd integer)

Since \(n\) is an odd integer, we can express it as \(n = 2k + 1\), where \(k\) is an integer.

Step 2: (Step 2: Substituting \(n\) in the given equation)

Let's substitute the expression for \(n\) in the given equation. So, we will have: \[\left\lfloor\frac{(2k+1)^{2}}{4}\right\rfloor=\frac{(2k+1)^{2}-1}{4} \]

Step 3: (Step 3: Expanding and simplifying the equation)

Now, we will expand the equation and simplify it wherever possible: \[\left\lfloor\frac{4k^2+4k+1}{4}\right\rfloor=\frac{4k^2+4k+1-1}{4} \] \[\left\lfloor\frac{4k^2+4k+1}{4}\right\rfloor=\frac{4k^2+4k}{4} \] \[\left\lfloor\frac{4k^2+4k}{4}\right\rfloor=\frac{4k^2+4k}{4} \]

Step 4: (Step 4: Factoring the equation)

We can factor out "4" from the equation: \[\left\lfloor\frac{4(k^2+k)}{4}\right\rfloor=\frac{4(k^2+k)}{4} \]

Step 5: (Step 5: Canceling out terms)

Now, we can cancel out the '4' terms in both sides of the equation: \[\left\lfloor k^2+k\right\rfloor=k^2+k\] This is true, as \(k^2 + k\) will always be an integer. Therefore, we have proved that for \(n\) being an odd integer, the given equation holds true: \[\left\lfloor\frac{n^{2}}{4}\right\rfloor=\frac{n^{2}-1}{4}\]

Key Concepts

Floor Function

The floor function, denoted by \( \left\lfloor x \right\rfloor \), maps a real number to the largest integer less than or equal to that number. This function is quite straightforward, yet powerful in mathematics for working with integers. Think of it as rounding down whatever real number it is applied to.Here are a few properties to remember:

• \( \left\lfloor x \right\rfloor \leq x < \left\lfloor x \right\rfloor + 1 \) for every real number \( x \).
• \( \left\lfloor x \right\rfloor = x \) when \( x \) is an integer.
• For any real number \( x \) and integer \( n \), \( \left\lfloor x + n \right\rfloor = \left\lfloor x \right\rfloor + n \).

An interesting use of the floor function is evident in integer manipulations, like the one from our exercise, for mapping values to discrete outcomes.

Odd Integers

Odd integers are integers that cannot be divided evenly by 2. They have the general form \( n = 2k + 1 \), where \( k \) is any integer. The sequence of odd numbers starts as 1, 3, 5, 7, 9, and continues indefinitely by adding 2 each time.A quick check to determine odd integers:

• If the last digit in base 10 is 1, 3, 5, 7, or 9, it's odd.
• If you divide it by 2, you get a remainder of 1.

Odd integers are vital in various mathematical proofs because they introduce unique properties in numerical identities. In the problem we're examining, using the form \( n = 2k + 1 \) simplifies the work by avoiding even numbers.

Integer Algebra

Integer algebra focuses on operations with integers, including addition, subtraction, multiplication, and division.It incorporates various properties that make calculations systematic and understandable. Understand these key properties:

• Commutative Property: The order of addition or multiplication does not change the outcome, e.g., \( a + b = b + a \).
• Associative Property: Grouping of numbers doesn't affect addition or multiplication, e.g., \( (a + b) + c = a + (b + c) \).
• Distributive Property: Multiplying a number by a group of numbers added together is the same as doing each multiplication separately, e.g., \( a(b + c) = ab + ac \).
• Identity Elements: Zero is the identity for addition, and one is the identity for multiplication.

The problem explores integer algebra through simplifying and manipulating expressions using properties, emphasizing that any arithmetic involving integers results in another integer.

Mathematical Induction

Mathematical induction is a powerful method of mathematical proof typically used to establish statements binding to all natural numbers.Here's a simple breakdown of how it works:

• Base Case: First, show that the statement holds for the initial value (often \( n = 0 \) or \( n = 1 \)).
• Inductive Step: Assume the statement holds for some integer \( n \). Then prove it holds for \( n+1 \).

When both the base case and the inductive step are confirmed, the statement is true for all natural numbers. While this exercise doesn't use induction, understanding this concept allows for extension to more complex proofs.By observing a repetitive pattern as shown in proof, induction strengthens our grasp on the reliability of statements tied to sequences or series.