Inequality iff Difference is Positive

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Theorem

Let $x, y \in \R$.

Then the following are equivalent:

$(1): \quad x < y$

$(2): \quad y - x > 0$

Proof

$\ds x < y$

$\leadstoandfrom$

$\ds y > x$

Definition of Dual Ordering

$\ds $

$\leadstoandfrom$

$\ds y + \paren {-x} > x + \paren {-x}$

Real Number Ordering is Compatible with Addition

$\ds $

$\leadstoandfrom$

$\ds y + \paren {-x} > 0$

Real Number Axiom $\R \text A4$: Inverses for Addition

$\ds $

$\leadstoandfrom$

$\ds y - x > 0$

Definition of Field Subtraction

Hence the result.

$\blacksquare$

Note

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If the notion of an ordering on $\R$ has not already been defined rigorously, this is often taken to be the definition of $x < y$.

Sources

1964: William K. Smith: Limits and Continuity ... (previous) ... (next): $\S 2.3$: Inequalities