Solving compound inequalities

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 $$0 and $$0

3 Answers

No we can't, we have three cases

• $$b<1-b \implies b<\frac12$$

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

• $$b>1-b \implies b>\frac12$$

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

• $$b=1-b \implies b=\frac12$$

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

• thanks for the answer. What if it is also known that $0<a<0.5$ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ? Commented Nov 21, 2018 at 20:58
• @user3222 In the first case we have $\frac{a}{1-a} <b \implies a<b-ab \implies a<\frac{b}{1+b}$
– user
Commented Nov 21, 2018 at 21:02

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

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. $$ac < bd\\ a+c < b+d$$