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

This page or section has statements made on it that ought to be extracted and proved in a Theorem page. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed. To discuss this page in more detail, feel free to use the talk page.

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