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Chapter 2: Problem 10

(a) If \(a

Short Answer

Expert verified
The first inequality \(a+c<b+d\) is proven by demonstrating that adding less to less results in a lesser sum. The second inequality \(0 \leq a c \leq b d\) is proven by showing that multiplying smaller numbers produces a smaller or equal product.

Step by step solution

01

Analyze the first inequality

We are given that \(a<b\) and \(c \leq d\). This means that b is greater than a, and d is greater than or equal to c. The inequality manifests because this means there's more 'b' than there is 'a', and at least as much 'd' as there is 'c'.
02

Prove the first inequality

Adding these inequalities side by side will result in \(a + c < b + d\), because adding less to less will still be less.
03

Analyze the second inequality

We are given that \(0<a<b\) and \(0 \leq c \leq d\). This means that a is greater than 0 but less than b, and c is greater than or equal to 0, but less than or equal to d. The inequality manifests because this means b is definitely bigger than a, and 'd' will always be at least as big as 'c'.
04

Prove the second inequality

Multiplying these inequalities side by side will result in \(0 \leq a c \leq b d\), because the product of smaller numbers results in a smaller product, and a larger number multiplied with a larger or equal number results in a larger or equal product.

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Most popular questions from this chapter

Show that if \(A\) and \(B\) are bounded subsets of \(\mathbb{R}\), then \(A \cup B\) is a bounded set. Show that \(\sup (A \cup B)=\sup \\{\sup A, \sup B\\}\)

Sketch the graph of the equation \(y=|x|-|x-1|\)

(a) Show that if \(x, y\) are rational numbers, then \(x+y\) and \(x y\) are rational numbers. (b) Prove that if \(x\) is a rational number and \(y\) is an irrational number, then \(x+y\) is an irrational number. If, in addition, \(x \neq 0\), then show that \(x y\) is an irrational number.

If \(S:=\\{1 / n-1 / m: n, m \in \mathbb{N}\\}\), find inf \(S\) and \(\sup S\).

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