Real Number Inequalities can be Added/Proof 2

Theorem

Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.


Then:

$a + c > b + d$


Proof

\(\ds a\) \(>\) \(\ds b\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds a + c\) \(>\) \(\ds b + c\) Real Number Ordering is Compatible with Addition
\(\ds c\) \(>\) \(\ds d\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds b + c\) \(>\) \(\ds b + d\) Real Number Ordering is Compatible with Addition
\(\ds \leadsto \ \ \) \(\ds a + c\) \(>\) \(\ds b + d\) Transitive Law

$\blacksquare$

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