Problem 53 - Calculus: One and Several Variables

Question

Show that ||\(a|-| b|| \leq|a-b|\) for all real numbers \(a\) and \(b\). HINT: Calculate ||\(a|-| b||^{2}\).

Step-by-Step Solution

Answer

Indeed, for any two real numbers \(a\) and \(b\), the absolute difference between their absolute values is always less than or equal to the absolute difference between the numbers. This has been confirmed by direct mathematical proof.

Step 1: Observe the given inequality

The given inequality is \(||a|-| b|| \leq |a-b|\). We know that for all real numbers, the inequality between them squared holds if the inequality is non-negative.

Step 2: Square both sides of the inequality

Let's square the left and the right sides of the inequality. Squaring both sides we get: \[(|a|-|b|)^2 \leq (a-b)^2\]

Step 3: Expand the left and right side of inequality

On the left we get \(|a|^2 - 2|a||b| + |b|^2\) which simplifies to \(a^2 - 2|a||b| + b^2\). On the right we get \(a^2 - 2ab + b^2\). Setting these equal we find: \[a^2 - 2|a||b| + b^2 \leq a^2 - 2ab + b^2\]

Step 4: Simplify the inequality

Cancelling the \(a^2\) and \(b^2\) from both sides, we can simplify the inequality to \[-2|a||b| \leq -2ab\].

Step 5: Final conclusion

Divide both sides by -2 to get \(|a||b| \geq ab\). This is true because the absolute value of a product is less than or equal to the product of absolute values. Therefore the original inequality is true.

Key Concepts

Absolute Value

The concept of absolute value involves measuring the distance of a number from zero on a number line, regardless of its direction. It's denoted by two vertical bars surrounding the number: \(|x|\). For example, both \(|3|\) and \(-3|\) equal 3.
Understanding absolute value is crucial in solving inequalities as it treats negative numbers as their positive counterparts, effectively removing the direction indicator and focusing purely on magnitude.
An important property of absolute values used in inequalities is:

• \(|a| \leq |b|\) implies that the magnitude of \(a\) is less than or equal to the magnitude of \(b\).
• The reverse is \(|a| \geq |b|\) which implies the magnitude of \(a\) is greater than or equal to the magnitude of \(b\).

These properties are essential when working through exercises in inequalities, especially when determining relationships between numbers and their absolute values.

Squaring Inequalities

Squaring inequalities is a powerful method used to simplify and solve inequalities that involve square terms. When squaring both sides of an inequality \(|x| \leq |y|\), one must ensure that the inequality remains true while simplifying it.
A key idea to remember is, squaring maintains the order of an inequality if both sides are non-negative, which is a safe assumption with absolute values since they are always non-negative.
Consider the process of squaring \(||a|-|b|| \leq |a-b|\):

• Square \(||a|-|b||\) to get \((|a|-|b|)^2\).
• Square \(|a-b|\) to get \((a-b)^2\).

This results in new relations that can be expanded and simplified to aid in solving the inequality. Keep in mind to carefully reduce and ensure all manipulations respect the inequality direction.

Real Numbers

Real numbers encompass both rational and irrational numbers, forming the set of all magnitudes that can be plotted on a number line.
When solving inequalities, understanding that \(a\) and \(b\) are real numbers implies that they can possess any value, positive or negative, whole or fractional.
This comprehensive range ensures that considerations about squares, roots, and absolute values applied to real numbers remain broadly applicable and must account for every possible a and b:

• Operations such as finding the absolute value or squaring must be handled with knowledge that real numbers include negative values, affecting the inequality resolve path.
• When demonstrating inequalities among real numbers, one must ensure clarity in addressing both the magnitude and the sign of the elements involved in the inequality.

Recognizing that real numbers cover such an extensive range aids in successfully proving inequalities like the one in our exercise.