Real Ordering is not Compatible with Subtraction
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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
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 $\S 1.8 \ (4) \ \text{(i)}$