Solve Spivak Inequality for x: 0<x<1

[Von Neumann]

Question:

Find all numbers $x$ for which $\frac{1}{x}+\frac{1}{1-x}>0$.

Solution:

If $\frac{1}{x}+\frac{1}{1-x}>0$,

then $\frac{1-x}{x(1-x)}+\frac{x}{x(1-x)}>0$;

hence $\frac{1}{x(1-x)}>0$.

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, $0<x<1$.

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.