Real Number Inequalities can be Added/Proof 1
Theorem
Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Then:
- $a + c > b + d$
Proof
| \(\ds a\) | \(>\) | \(\ds b\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a + c\) | \(>\) | \(\ds b + c\) | Real Number Ordering is Compatible with Addition |
| \(\ds c\) | \(>\) | \(\ds d\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b + c\) | \(>\) | \(\ds b + d\) | Real Number Ordering is Compatible with Addition |
Finally:
| \(\ds a + c\) | \(>\) | \(\ds b + c\) | ||||||||||||
| \(\ds b + c\) | \(>\) | \(\ds b + d\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a + c\) | \(>\) | \(\ds b + d\) | Trichotomy Law for Real Numbers |
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.8 \ (3)$