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Chapter 2: Problem 38

Use proof by cases to prove that \(|x+y| \leq|x|+|y|\) for all real numbers \(x\) and \(y\).

Short Answer

Expert verified
Yes, the inequality \(|x+y| \leq |x|+|y|\) holds true for all real numbers \(x\) and \(y\), as it has been confirmed by considering all possible cases of their signs.

Step by step solution

01

Case when \(x \geq 0\) and \(y \geq 0\)

Since both \(x\) and \(y\) are nonnegative, the sum \(x+y\) is nonnegative as well. Thus, \(|x+y|=x+y\). Also, by definition, \(|x|=x\) and \(|y|=y\). Therefore, it is clear that \(|x+y| \leq |x|+|y|\), since \(x+y\) is the same as \(x+y\).
02

Case when \(x < 0\) and \(y < 0\)

In this scenario, both \(x\) and \(y\) are negative, which means that their sum \(x+y\) is also negative. Hence, \(|x+y|=-(x+y)\). By the definition of absolute value, \(|x|=-x\) and \(|y|=-y\). So, it is evident that \(|x+y| \leq |x|+|y|\), because \(-x-y\) equals \(-x-y\).
03

Case when \(x\) and \(y\) have opposite signs

Without loss of generality, let's assume that \(x\) is nonnegative and \(y\) is negative. This means that \(|x|=x\) and \(|y|=-y\). Because the sum \(x+y\) could be either nonnegative or negative, we have \(|x+y|=x+y\) if \(x+y \geq 0\), and \(|x+y|=-(x+y)\) if \(x+y<0\). In both cases, we can see that \(|x+y| \leq |x|+|y|\), because \(x-y\) is always greater than or equal to \(x+y\) or \(-x-y\). The argument is symmetric for when \(x\) is negative and \(y\) is nonnegative.

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Key Concepts

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

Absolute Value Inequality
Absolute value inequalities, like \(|x+y| \leq |x| + |y|\), are tools used to express relationships between absolute values of sums and individual absolute values. The absolute value of a number is essentially the distance of that number from zero on the number line, without regard to direction.
When tackling absolute value inequalities, we are often interested in finding the largest or smallest possible range for a scenario.
This inequality, \(|x+y| \leq |x| + |y|\), is known as the triangle inequality. It reflects the shortest distance between two points, which is useful in a variety of mathematical contexts, like geometry and algebra.
Notice that when applying such inequalities, it鈥檚 crucial to understand and break down cases where the absolute values equal, or differ in sign. This understanding helps dissect a larger problem into manageable parts.
Real Numbers
Real numbers encompass the broadest range of numbers found in everyday mathematics, including all the points on a number line. They consist of both rational numbers, like fractions and integers, as well as irrational numbers, such as square roots and pi.
The significance of real numbers is substantial when discussing problems such as \(|x+y| \leq |x| + |y|\), as we consider all possible numeric inputs.
Real numbers don't follow simple binary categorizations like positive or negative. They span continuous values, meaning when you craft proofs or solve inequalities like the above, any scenario can present itself:
  • Both numbers can be positive.
  • Both numbers can be negative.
  • They can differ in sign.
Understanding real numbers is fundamental as they allow us to account for an array of possibilities.
Step-by-Step Proof
Proof by cases is an effective method used in proving inequalities or other mathematical statements, specifically by categorically covering all possible scenarios.
This approach dissects each scenario where the main statement might behave differently.
In our exercise, we were asked to demonstrate \(|x+y| \leq |x| + |y|\), across different circumstances:
  • When both \(x\) and \(y\) are positive.
  • When both are negative.
  • When one is positive and the other is negative.
Breaking down the problem into these cases makes it easier to illustrate why the inequality holds under every condition.
A step-by-step proof walks you through each possibility, showing that regardless of what numbers you plug into the equation, the inequality remains true. Each step ensures complete coverage and eliminates any doubt about hidden or overlooked exceptions. Being meticulous in each case, provides a clear vision of the truth within the complexity.

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Most popular questions from this chapter

Find the quotient \(q\) and remainder \(r\) as in Theorem 2.5 .6 when \(n\) is divided by \(d\). $$n=47, d=9$$

The Egyptians of antiquity expressed a fraction as a sum of fractions whose numerators were \(1 .\) For example, \(5 / 6\) might be expressed as $$\frac{5}{6}=\frac{1}{2}+\frac{1}{3}$$ We say that a fraction \(p / q,\) where \(p\) and \(q\) are positive integers, is in Egyptian form if $$\frac{p}{q}=\frac{1}{n_{1}}+\frac{1}{n_{2}}+\cdots+\frac{1}{n_{k}}$$ where \(n_{1}, n_{2}, \ldots, n_{k}\) are positive integers satisfying \(n_{1}

Find the quotient \(q\) and remainder \(r\) as in Theorem 2.5 .6 when \(n\) is divided by \(d\). $$n=0, d=9$$

What is wrong with the following "proof" that $$ \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1} \neq \frac{n^{2}}{n+1} $$ for all \(n \geq 2 ?\) Suppose by way of contradiction that $$ \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}=\frac{n^{2}}{n+1} $$ Then also $$ \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}+\frac{n+1}{n+2}=\frac{(n+1)^{2}}{n+2} $$ We could prove statement ( 2.4 .14 ) by induction. In particular, the Inductive Step would give $$ \left(\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}\right)+\frac{n+1}{n+2}=\frac{n^{2}}{n+1}+\frac{n+1}{n+2} $$ Therefore, $$ \frac{n^{2}}{n+1}+\frac{n+1}{n+2}=\frac{(n+1)^{2}}{n+2} $$ Multiplying each side of this last equation by \((n+1)(n+2)\) gives $$ n^{2}(n+2)+(n+1)^{2}=(n+1)^{3} $$ This last equation can be rewritten as $$ n^{3}+2 n^{2}+n^{2}+2 n+1=n^{3}+3 n^{2}+3 n+1 $$ Or $$ n^{3}+3 n^{2}+2 n+1=n^{3}+3 n^{2}+3 n+1 $$ which is a contradiction. Therefore, $$ \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1} \neq \frac{n^{2}}{n+1} $$ as claimed.

Deal with plane convex sets. \(A\) plane convex set, subsequently abbreviated to "convex set," is a nonempty set \(X\) in the plane having the property that if \(x\) and \(y\) are any two points in \(X,\) the straight-line segment from \(x\) to \(y\) is also in \(X .\) The following figures illustrate. Suppose that \(n \geq 3\) points in the plane have the property that each three of them are contained in a circle of radius 1 . Prove that there is a circle of radius 1 that contains all of the points.
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