Real Number Inequalities can be Added

Theorem

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

Then:

$a + c > b + d$

Proof 1

$\ds a$

$>$

$\ds b$

$\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:

$\ds a + c$

$>$

$\ds b + c$

$>$

$\ds b + d$

$\ds \leadsto \ \ $

$\ds a + c$

$>$

$\ds b + d$

Trichotomy Law for Real Numbers

$\blacksquare$

Proof 2

$\ds a$

$>$

$\ds b$

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

$\blacksquare$

Sources

1990: Edwin E. Moise: Elementary Geometry from an Advanced Standpoint (3rd ed.): Chapter $1$ / Section $1.4$: "The Ordering of the Real Numbers."
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(a)}$