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 $\S 1.8 \ (3)$