Reciprocal of Strictly Positive Real Number is Strictly Positive

Theorem

$\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$

Proof

Let $x > 0$.

Aiming for a contradiction, suppose $\dfrac 1 x < 0$.

Then:

\(\ds x\)

\(>\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x \times \dfrac 1 x\)

\(<\)

\(\ds 0 \times 0\)

Real Number Ordering is Compatible with Multiplication: Negative Factor

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(<\)

\(\ds 0\)

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

But from Real Zero is Less than Real One:

$1 > 0$

Therefore by Proof by Contradiction:

$\dfrac 1 x > 0$

$\blacksquare$

Also see

• Reciprocal of Strictly Negative Real Number is Strictly Negative

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.5$: Example
• 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(i)}$