Problem 55 - Discrete Mathematics with Applications

Question

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

Step-by-Step Solution

Answer

For any odd integer n, we can define it as n = 2k + 1, where k is an integer. We found that \(n^2\) for an odd integer can be expressed as \(4k^2 + 4k + 1\), and \(\frac{n^2}{4}\) is simplified to \(k^2+k+\frac{1}{4}\). We also found that \(n^2 - 1 = 4k^2 + 4k\), which when divided by 4 gives us \(k^2 + k\). We can see that both expressions are equal, as both expressions are k² + k. Therefore, we have proven that for any odd integer n, \(\left|\frac{n^2}{4}\right|=\frac{n^2 - 1}{4}\).

Step 1: Define Odd and Even Integers

For any odd integer n, we can define it as n = 2k + 1, where k is an integer. This is because odd integers are not divisible by 2, but one more than an even integer, represented by 2k where k is an integer.

Step 2: Substitute the Definition of an Odd Integer

Now, we will substitute the definition for an odd integer, n = 2k + 1, into the function we need to prove: \(\left|\frac{n^2}{4}\right|\).

Step 3: Calculate n^2 for an Odd Integer

First, we need to calculate n^2, so we square the definition of an odd integer, (2k + 1)². Using the square of a binomial formula, \((a + b)^2 = a^2 + 2ab + b^2\), we get: \((2k + 1)^2 = (2k)^2 + 2(2k)(1) + (1)^2 = 4k^2 + 4k + 1\)

Step 4: Substitute n^2 into Function

Next, we substitute the expression for \(n^2\), i.e., \(4k^2 + 4k + 1\), into the function we need to prove: \(\left|\frac{n^2}{4}\right|\). So, we have: \(\left|\frac{4k^2 + 4k + 1}{4}\right|\)

Step 5: Simplify the Expression

Now, we can simplify the expression inside the absolute value brackets: \(\frac{4k^2 + 4k + 1}{4} = k^2 + k + \frac{1}{4}\) Since k and k^2 are integers, we know that the sum k^2 + k is an integer because the sum of integers is also an integer. The expression k^2 + k will always be non-negative since k^2 will always be non-negative and k will be non-negative for half of the cases (when k is non-negative). So, in either case, the absolute value sign won't change the expression: \(\left|k^2 + k + \frac{1}{4}\right| = k^2 + k + \frac{1}{4}\)

Step 6: Derive the Expression for n^2 - 1

Now, we need to find the expression for \(n^2 - 1\) for an odd integer n. Using the expression for n^2 that we found before, we have: \(n^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k\)

Step 7: Prove the Equality

Lastly, we need to show that the expression we derived for the absolute value function is equal to the given expression, \(\frac{n^2 - 1}{4}\). So we will substitute the expression for \(n^2 - 1\) that we found above: \(\frac{n^2 - 1}{4} = \frac{4k^2 + 4k}{4} = k^2 + k\) Comparing this final expression with what we found in Step 5, we can see that they are equal, as both expressions are k² + k: \(k^2 + k = k^2 + k\) Thus, the proof is complete. For any odd integer n, we have: \(\left|\frac{n^2}{4}\right| = \frac{n^2 - 1}{4}\)

Key Concepts

Odd and Even Integers

Understanding the difference between odd and even integers is essential in various areas of mathematics. An even integer is any number that can be divided by two without leaving a remainder, making it a multiple of two. In contrast, an odd integer will have a remainder of one when divided by two. It cannot be evenly divided by two, thus always one more (or less) than an even integer.

For example, the numbers 2, 4, and 6 are even because they result in whole numbers when divided by 2. On the other hand, the numbers 1, 3, and 5 are odd. They cannot be halved to give a whole number. An important characteristic of odd numbers is their representation in mathematical terms as n = 2k + 1, where k is an integer. This notation is very useful when proving mathematical propositions involving odd integers, as you can substitute n with 2k + 1 to simplify and solve problems as seen in the exercise.

Absolute Value

The absolute value of a number essentially refers to its distance from zero on the number line, irrespective of direction. It is denoted by two vertical lines, |n|. The absolute value of both positive and negative numbers is always non-negative. For instance, the absolute value of both -5 and 5 is 5.

The crux is that when dealing with absolute value in equations and inequalities, you may be considering both the negative and positive scenarios of a variable. However, if you can ascertain that the variable, in any case, is non-negative, the absolute value bars have no effect, as the number is already positive. This becomes particularly significant in proofs where the expression inside the absolute value can be simplified, and its sign can be determined, as was the case in the textbook exercise.

Square of a Binomial

The square of a binomial is a fundamental concept in algebra that deals with squaring a two-term algebraic expression. A binomial takes the form (a + b) and its square is noted as (a + b)^2. This is expanded using the formula (a + b)^2 = a^2 + 2ab + b^2, which comes from applying the distributive property, also known as FOIL (First, Outer, Inner, Last).

This expansion is of great use when simplifying algebraic expressions and performing proofs, as it allows you to break down the squared terms into more manageable parts. In our example, (2k + 1)^2 was expanded to 4k^2 + 4k + 1, subsequently facilitating the process of proving the given expression involving the absolute value of a squared odd integer.