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Chapter 0: Problem 42

Show that if \(a

Short Answer

Expert verified
We are given two inequalities, \(a<b\) and \(c\leq d\). To prove the desired inequality, \(a+c<b+d\), we combine these inequalities as follows: Adding \(a < b\) to \(c\leq d\) results in \(a+c<b+c\leq b+d\). Thus, we can conclude that \(a+c<b+d\).

Step by step solution

01

Understand the given inequalities

We have to deal with two inequalities: 1. \(a < b\): It means that 'a' is less than 'b'. 2. \(c \leq d\): It means that 'c' is less than or equal to 'd'.
02

Add the given inequalities

We can add the given inequalities to have: \(a < b\) and \(c \leq d\) Adding both inequalities, we get: \(a + c \leq b + d\) However, to be consistent with what we need to prove, we should recall that \(a < b\), which means \(a + c < b + c\). Now, since \(c \leq d\), this means that \(b + c \leq b + d\).
03

Combine the inequalities

From the inequalities in Step 2: \(a + c < b + c \leq b + d\) So, we can conclude that \(a + c < b + d\). This completes the proof.

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