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

\(\displaystyle x < y\) \(\leadstoandfrom\) \(\displaystyle y > x\) Definition of Dual Ordering
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle y + \paren {-x} > x + \paren {-x}\) Real Number Ordering is Compatible with Addition
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle y + \paren {-x} > 0\) Real Number Axioms: $\R \text A 4$: Inverse Elements
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle y - x > 0\) Definition of Field Subtraction

Hence the result.

$\blacksquare$


Note

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