Solving compound inequalities

Question

Consider the following two inequalities,

$\frac{a}{1-a} < b$

and

$\frac{a}{1-a}< (1-b)$

Is it correct to substitute the first into the second, and write,

$b<(1-b)$

to derive $b < \frac{1}{2}$ ?

EDIT:

It is also known that $0<a<0.5$ and $0<b<1$

Answer 1

No we can't, we have three cases

• $b<1-b \implies b<\frac12$

$$\frac{a}{1-a} <b< 1-b$$

• $b>1-b \implies b>\frac12$

$$\frac{a}{1-a} <1-b<b$$

• $b=1-b \implies b=\frac12$

$$\frac{a}{1-a} <\frac12$$

Answer 2

If you have two inequalities, you can combine to get $\frac{a}{1-a}< \min\{ b, (1-b)\}$

Inequalities which can be combined are $$a < b\\ b < c$$to get $a < c$ and you can also add and multiply, i.e. $$a < b\\ c < d$$ gives e.g. $$ac < bd\\ a+c < b+d$$

Answer 3

Surely not. We may also have $${a\over 1-a}<1-b<b$$for example $$a=0.002\\b={2\over 3}$$