Real Ordering is not Compatible with Subtraction

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:

$a - c > b - d$

That is, the usual ordering is not compatible with subtraction.

Proof

Proof by Counterexample:

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

Then $a - c = 1$ while $b - d = 2$.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.8 \ (4) \ \text{(i)}$