Solve Spivak Inequality for x: 0<x<1

[Von Neumann]

Question:

Find all numbers x for which \frac{1}{x}+\frac{1}{1-x}&gt;0.

Solution:

If \frac{1}{x}+\frac{1}{1-x}&gt;0,

then \frac{1-x}{x(1-x)}+\frac{x}{x(1-x)}&gt;0;

hence \frac{1}{x(1-x)}&gt;0.

Now we note that

\frac{1}{x(1-x)} \rightarrow ∞ as x \rightarrow 0

and \frac{1}{x(1-x)} \rightarrow 0 as x \rightarrow 1.

Thus, 0&lt;x&lt;1.

Notes:

Not quite sure if that's the sort of solution Spivak is looking for in Ch.1.

 

[Infrared]

A non-zero number and its reciprocal will always have the same sign so \frac{1}{x(1-x)} will be positive where x(1-x) is

 

[Von Neumann]

Ah, I see. Don't know how I didn't see that.