Real Number Ordering is Compatible with Addition

Theorem

$\forall a, b, c \in \R: a < b \implies a + c < b + c$

where $\R$ is the set of real numbers.

Proof

From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.

By definition of ordered integral domain, the usual ordering $\le$ is compatible with ring addition.

$\blacksquare$

Examples

$15 + 3$ Greater than $12 + 3$

We have that:

$15 > 12$

so by Real Number Ordering is Compatible with Addition:

$15 + 3 > 12 + 3$

That is:

$18 > 15$

$15 - 3$ Greater than $12 - 3$

We have that:

$15 > 12$

so by Real Number Ordering is Compatible with Addition:

$15 - 3 > 12 - 3$

That is:

$12 > 9$

Sources

• 1971: Wilfred Kaplan and Donald J. Lewis: Calculus and Linear Algebra ... (previous) ... (next): Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-2}$: Inequalities
• 1972: Murray R. Spiegel and R.W. Boxer: Theory and Problems of Statistics (SI ed.) ... (previous) ... (next): Chapter $1$: Inequalities
• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.4$: Inequalities: $\text{(II)}$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): inequality
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): inequality: 2.
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): order properties (of real numbers): $(3)$