Question

Show that \(\|a|-| b|\| \leq| a-b | \text { for all constants } a \text { and } b\). (Hint: Write \(|a|=|(a-b)+b|\) and apply the triangle inequality to \(|(a-b)+b| .)\)

Short answer

Short Answer: Applying the triangle inequality and rearranging, we find that \(|a| - |b| \leq |a-b|\) and \(|b| - |a| \leq |b-a|\). Upon combining these inequalities, we obtain \(||a|-| b|| \leq| a-b |\), proving the result for all constants a and b.

Step-by-step solution

Write |a| as a sum of |(a-b)| and |b|

We are given the hint to write \(|a|=|(a-b)+b|\). Let's substitute a with (a-b) + b: $$|a|=|(a-b)+b|.$$

Apply the triangle inequality

Now, we will apply the triangle inequality to \(|(a-b)+b|\): $$|(a-b)+b| \leq |a-b| + |b|.$$

Rearrange to obtain \(|a| - |b|\)

We want to obtain an inequality concerning \(||a|-| b||\). So, let's rearrange the inequality we have: $$|a| - |b| \leq |a-b|.$$

Apply the triangle inequality to obtain \(|b| - |a|\)

To obtain an inequality with \(|b|-|a|\), we will follow the same approach with a slight modification in the hint. Write \(|b|=|(b-a) + a|\) and apply the triangle inequality to \(|(b-a)+a|\): $$|b|=|(b-a) + a| \leq |b - a| + |a|$$ $$|b| - |a| \leq |b - a|.$$

Combining the inequalities

Now, we have two inequalities, one with \(|a| - |b|\) and one with \(|b|-| a|\). Let's combine them: $$||a| - |b|| = |(a-b)+|- |(b-a)| \leq |a - b|.$$ Thus, we have proven that \(||a|-| b|| \leq| a-b |\) for all constants a and b.

Key concepts

Inequalities

Inequalities are fundamental concepts in mathematics that help us compare two values. They allow us to understand relationships such as greater than, less than, or equal to. In the context of our original exercise, we are dealing with an inequality that involves absolute values. Here's why inequalities are useful:

• Inequalities express a range of possibilities rather than fixed values. This is especially helpful in problem-solving and analysis.
• They allow for flexibility in solutions, indicating that a result can be above, below, or equal to a given threshold.
• When combined with absolute values, inequalities help in establishing bounds, as seen with the triangle inequality.

By using the given strategies in the solution, such as substituting and rearranging terms, we find that \(||a|-|b|| \leq |a-b|\). This inequality tells us that the absolute difference between the absolute values of two numbers is never more than the absolute value of their difference.

Absolute Value

Absolute value is a concept that measures how far a number is from zero without considering its direction. So, the absolute value of any number is always non-negative.

• The absolute value of a real number \(a\) is denoted by \(|a|\).
• It is defined as:\[|a| =\begin{cases} a, & \text{if } a \geq 0\-a, & \text{if } a < 0\end{cases}\]
• This is crucial when working with inequalities since it influences how values are compared and combined.

In the exercise, absolute values were pivotal for its successful resolution. By recognizing that \(|a|=|(a-b)+b|\), we could effectively apply the triangle inequality to establish our proof. The handling and rearrangement of absolute values are key skills for solving similar problems.

Properties of Real Numbers

Real numbers are an essential part of mathematics, encompassing both rational and irrational numbers. They have various properties that are used to solve equations and inequalities.

• They follow the Commutative, Associative, and Distributive Laws.
• Real numbers are ordered, which means you can arrange them in a sequence from the smallest to the largest.
• They respect the properties of inequalities, such as if \( a \leq b \) and \( c \leq d \), then \( a+c \leq b+d \).

In the context of the exercise, the properties of real numbers allowed us to manipulate expressions and apply inequalities successfully. By using the properties of addition and the ordering of real numbers, a key aspect was the understanding that modifying terms like \(|b-a|\) is legitimate provided the overall inequality constraints remain correct. This underpins how we can explore and enjoy the broad spectrum of problem-solving in mathematics.