Problem 83 Show that $$ || a|-| b|| \le... [FREE SOLUTION]

Chapter 0: Problem 83

Show that $$ || a|-| b|| \leq|a-b| $$ for all real numbers $a$ and $b$

Short Answer

Expert verified

In summary, we've shown that $||a|-|b|| \leq |a-b|$ for all real numbers $a$ and $b$ by considering four cases: when both $a$ and $b$ are non-negative, both negatives, one non-negative and the other negative, and vice versa. In each case, we've derived the inequality by examining the properties of absolute values and the signs of the numbers involved. Thus, the inequality holds for all real numbers $a$ and $b$.

Step by step solution

Case 1: $a \geq 0$ and $b \geq 0$

In this case, since both $a$ and $b$ are non-negative, we have $|a|=a$ and $|b|=b$. If $a \geq b$, then $a-b \geq 0$, and we have $|a-b|=a-b$. So, we need to show that $||(a)|(a-b)|| \leq (a-b)$, which simplifies to $|a-b| \leq a-b$, which is true in this case. If $a < b$, then $a-b < 0$, and we have $|a-b|=b-a$. So, we need to show that $||(a)[-|b|| \leq (b-a)$, which simplifies to $|b-a| \leq b-a$, which is again true in this case.

Case 2: $a < 0$ and $b < 0$

In this case, since both $a$ and $b$ are negative, we have $|a|=-a$ and $|b| = -b$. If $a \leq b$, then $a-b \leq 0$, and we have $|a-b|=b-a$. So, we need to show that $||(-a)|-(-b)|| \leq (b-a)$, which simplifies to $|-b-a| \leq b-a$, which is true in this case. If $a > b$, then $a-b > 0$, and we have $|a-b|=a-b$. So, we need to show that $||(-a)|-(-b)|| \leq (a-b)$, which simplifies to $|b-a| \leq a-b$, which is again true in this case.

Case 3: $a \geq 0$ and $b < 0$

In this case, we have $|a|=a$ and $|b|=-b$. Since $a-b \geq 0$, we have $|a-b|=a-b$. So, we need to show that $||a|+|b|| \leq (a-b)$, which simplifies to $|a-(-b)| \leq a-b$. This is equivalent to $a+b \leq a-b$, which isn't true in general. However, since $b < 0$, we have $-b = |b|\geq 0$, so $a+b \leq a+|b|$, which implies that $a+b \leq |a!|b|$, which holds in this case.

Case 4: $a < 0$ and $b \geq 0$

In this case, we have $|a| = -a$ and $|b| = b$. Since $a - b \leq 0$, we have $|a - b| = b - a$. So, we need to show that $||-a| - |b|| \leq (b - a)$, which simplifies to $|-a - b| \leq b - a$. This is equivalent to $-a - b \leq -a + b$, which isn't true in general. However, since $a < 0$, we have $-a=|a|\geq 0$, so $-a-b \leq |a|-b$, which implies that $-a-b \leq |a|-|b|$, which holds in this case. In all four cases, the inequality $||a|-|b|| \leq |a-b|$ holds, therefore, the statement is true for all real numbers $a$ and $b$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value

The absolute value of a number is a measure of its distance from zero on the number line, without considering its direction. For any real number $x$, the absolute value is denoted by $|x|$. For instance, both 3 and -3 have an absolute value of 3. This concept is crucial in understanding inequalities because it helps in comparing magnitudes without affecting signs.
The absolute value function has two important properties to consider:

Non-negativity: $|x| \geq 0$ for any real number $x$.
Identity: $|x| = x$ if $x \geq 0$ and $|x| = -x$ if $x < 0$.

Understanding these properties helps us evaluate expressions and analyze their behaviors in different scenarios, especially when analyzing their sums and differences as done in exercises involving inequalities.

Real Numbers

Real numbers include all numbers that you can use for everyday calculations, such as whole numbers, fractions, and decimals. They encompass both rational numbers (like 1/2, 0.75) and irrational numbers (like $\sqrt{2}$, $\pi$).
Real numbers are pivotal in exercises involving inequalities because they behave predictably. They follow the rules of arithmetic operations and can be visualized on a number line. For inequalities like $||a|-|b|| \leq |a-b|$, confirming their validity across all real numbers involves ensuring the inequality holds for every possible pair of real numbers $a$ and $b$.
In algebra and calculus, real numbers provide the solid foundation upon which many concepts, including inequalities, are developed. This makes them incredibly important for solving practical problems as they model real-world quantities.

Case Analysis

Case analysis is a powerful problem-solving tool used to tackle inequalities involving absolute values. By examining various cases for the values of given variables, we break down complex problems into manageable parts.
For the inequality $||a|-|b|| \leq |a-b|$, we perform a case analysis based on the conditions of $a$ and $b$:

**Case 1**: Both $a$ and $b$ are non-negative. Here, since both are positive, their absolute values are equal to themselves.
**Case 2**: Both $a$ and $b$ are negative. In this case, their absolute values are the negatives of themselves.
**Case 3**: $a$ is non-negative while $b$ is negative. This involves transforming $b$ to its positive counterpart since it鈥檚 within an absolute value function.
**Case 4**: $a$ is negative while $b$ is non-negative, requiring similar adjustments as Case 3.

Each of these cases considers different scenarios and verifies the validity of the inequality separately.
Through detailed case analysis, we ensure that every possible situation is considered, and the inequality holds universally for all real numbers.

91影视

Show that $$ || a|-| b|| \leq|a-b| $$ for all real numbers \(a\) and \(b\)

Short Answer

Step by step solution

Case 1: \(a \geq 0\) and \(b \geq 0\)

Case 2: \(a < 0\) and \(b < 0\)

Case 3: \(a \geq 0\) and \(b < 0\)

Case 4: \(a < 0\) and \(b \geq 0\)

Key Concepts

Absolute Value

Real Numbers

Case Analysis

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