Consider the following two inequalities,

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


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

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


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


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


3 Answers 3


No we can't, we have three cases

  • $b<1-b \implies b<\frac12$

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

  • $b>1-b \implies b>\frac12$

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

  • $b=1-b \implies b=\frac12$

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

  • $\begingroup$ 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 ? $\endgroup$
    – rranjik
    Commented Nov 21, 2018 at 20:58
  • 1
    $\begingroup$ @user3222 In the first case we have $\frac{a}{1-a} <b \implies a<b-ab \implies a<\frac{b}{1+b}$ $\endgroup$
    – user
    Commented Nov 21, 2018 at 21:02

Surely not. We may also have $${a\over 1-a}<1-b<b$$for 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. $$ac < bd\\ a+c < b+d$$


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