Problem 65 - Precalculus

Question

(As background for this exercise, you might want to work Exercise \(23 .)\) Prove that $$ \max (a, b)=\frac{a+b+|a-b|}{2} $$ Hint: Consider three separate cases: \(a=b ; a>b ;\) and \(b>a\).

Step-by-Step Solution

Answer

The formula is accurate for all cases: \(a = b\), \(a > b\), and \(b > a\).

Step 1: Consider the Case a = b

In this situation, if \(a = b\), then \(|a-b| = 0\). So, plugging these values into the expression, we have: \\[\frac{a+b+|a-b|}{2} = \frac{a+a}{2} = \frac{2a}{2} = a\] \Since \(a = b\), this means \(\max(a, b) = a = b\). Therefore, the formula holds for this case.

Step 2: Consider the Case a > b

When \(a > b\), the maximum value between \(a\) and \(b\) is \(a\). In this case, \(|a-b| = a-b\). Substitute these into the expression: \\[\frac{a+b+|a-b|}{2} = \frac{a+b+(a-b)}{2} = \frac{2a}{2} = a\] \Since \(\max(a, b) = a\) and the expression evaluates to \(a\), the formula is true for this scenario.

Step 3: Consider the Case b > a

For \(b > a\), the maximum value between \(a\) and \(b\) is \(b\). Here, \(|a-b| = b-a\). Substitute into the expression: \\[\frac{a+b+|a-b|}{2} = \frac{a+b+(b-a)}{2} = \frac{2b}{2} = b\] \Here, \(\max(a, b) = b\) and the formula outputs \(b\) as expected, proving correctness in this case.

Step 4: Conclusion: Formula Verification

We considered all possible scenarios for \(a\) and \(b\): when \(a = b\), \(a > b\), and \(b > a\). In each case, the expression \(\frac{a+b+|a-b|}{2}\) equates to the maximum value of \(a\) and \(b\). This demonstrates that the given formula is universally accurate.

Key Concepts

Maximum Function

The maximum function is a simple yet powerful concept in mathematics. It helps in determining the largest of two or more numbers. When dealing with just two numbers, say \(a\) and \(b\), the function is written as \(\max(a, b)\). This function outputs the greater value among the two.
For example, if \(a = 7\) and \(b = 5\), then \(\max(7, 5) = 7\). Its simplicity is deceptive because its applications span across various fields such as computer science, economics, and statistics.
Understanding the maximum function involves recognizing how it can be described algebraically. The formula \(\max(a,b)=\frac{a+b+|a-b|}{2}\) finds the larger of two numbers without using any conditional statements. Instead, it uses the absolute value to dynamically adjust the sum, ensuring the maximum is always selected.

• For \(a = b\), the formula simplifies to \(a\) because \(|a-b| = 0\).
• For \(a > b\), it simplifies to \(a\) because \(|a-b| = a-b\).
• For \(b > a\), it simplifies to \(b\) as \(|a-b| = b-a\).

This formulation is not only mathematically elegant but also computationally efficient.

Inequalities

Inequalities are a fundamental concept that allows comparing values to determine which is larger or smaller. Writing this relationship succinctly, \(a > b\) means \(a\) is greater than \(b\), while \(a < b\) implies the opposite.
In the maximum function formula, inequalities determine the outcome of the expression \(|a-b|\), which is pivotal for correctly identifying the max element.
For example:

• In the case \(a > b\), we use \(|a-b| = a-b\), thus confirming \(a\) as the maximum.
• In the case \(b > a\), \(|a-b| = b-a\), which confirms \(b\) as the maximum.

The interplay of inequalities and absolute values ensures the correct element is selected without needing explicit if-else logic. It's a direct application of logical reasoning combined with algebraic manipulation.
It's important to grasp this concept as it lays a foundation for more complex mathematical theories and proofs.

Proof Techniques

Speaking of proofs, proving a mathematical statement or formula is an essential part of mathematical studies. A common technique is proof by cases, where you divide the problem into exhaustive individual cases and prove each one separately.
This approach was used in our maximum function formula. Breaking down into three cases: when \(a = b\), \(a > b\), and \(b > a\), provides clarity and guarantees all possibilities are covered.

• Case 1: \(a = b\) shows the outcome is simply \(a\) since both inputs are the same.
• Case 2: \(a > b\) results in \(a\) as \(|a-b| = a-b\), and the expression simplifies to show \(a\) is greater.
• Case 3: \(b > a\) leads to \(b\) as \(|a-b| = b-a\), highlighting \(b\) as the greater value.

Proof by cases is powerful because it breaks complex problems into manageable parts. It reveals both the effectiveness and robustness of mathematical logic. Whether in competitions, coursework, or real-world applications, mastering proof techniques is invaluable for any mathematician.