Real Ordering Incompatible with Division

Theorem

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

Then it does not necessarily hold that:

$\dfrac a c > \dfrac b d$

Proof

Proof by Counterexample:

For example, set $a = 5, b = 3, c = 4, d = 1$

Then $\dfrac a c = 1 \dfrac 1 4$ while $\dfrac b d = 3$.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.8 \ (4) \ \text{(ii)}$