# Let $b \in [0,1)$. Prove that $\frac{b}{1-b} \in [0,\infty)$

Can someone check my solution for this problem? It seems to me that it’s incomplete, and I’m not sure.

Problem: Let $$b \in [0,1)$$. Prove that $$\frac{b}{1-b} \in [0,\infty)$$.

Solution: We know that $$b \in [0,1)$$, so $$0 \leq b < 1$$. From here we can also deduce that $$0 < 1-b \leq 1$$. So $$\frac{1}{1-b} \geq 1$$. Multiplying by $$b$$ we obtain that $$\frac{b}{1-b} \geq b$$. Since $$b \geq 0$$ we conclude that $$\frac{b}{1-b} \geq 0$$. Therefore $$\frac{b}{b-1} \in [0,\infty)$$.

• looks good to me Commented Jul 16, 2020 at 23:05
• @MushuNrek thank you for the feedback! Commented Jul 16, 2020 at 23:06
• This is a neat argument
– user675768
Commented Jul 16, 2020 at 23:17

Another method is to note that $$$$\dfrac{b}{1-b}=\sum_{i=1}^\infty b^i$$$$ Since each term is non-negative, therefore, sum of this series is also non-negative.

• That’s a really clever way to deal with this problem! Thank you for the suggestion, I really appreciate it! Commented Jul 17, 2020 at 2:10

Define the function $$f$$ from $$[0,1)$$ to $$\Bbb R$$ by

$$(\forall x\in[0,1))\;\; f(x)=\frac{x}{1-x}$$

$$f$$ is continuous at $$[0,1)$$.

$$f$$ is differentiable at $$[0,1)$$ and

$$(\forall x\in[0,1))\;\; f'(x)=\frac{1-x+x}{(1-x)^2}>0$$ $$f$$ is then strictly increasing at $$[0,1)$$.

Thus, $$f$$ is a bijection from $$[0,1)$$ to $$f([0,1))=[f(0),\lim_{x\to 1^-}f(x))=[0,+\infty)$$

we conclude that $$(\forall b\in[0,1))\;\; f(b)=\frac{b}{1-b}\ge 0$$

Remark:

You can simply say $$0\le b<1\; \implies$$ $$b\ge 0 \text{ and } 1-b>0 \;\implies$$ $$\frac{b}{1-b}\ge 0\; \implies$$ $$\frac{b}{1-b}\in [0,+\infty)$$

• That’s really a good and rigorous way to solve this problem! Thank you for that amazing answer! Commented Jul 16, 2020 at 23:38

I think your proof is giving too much (unnecessary) detail. The only thing that needs to be proved is $$\dfrac{b}{1-b}\geq0$$, which follows from the fact $$b\geq0$$ and $$1-b>0$$ (since $$b<1$$). Done.

• I think this is the cleanest answer -- it uses the fewest assumptions and takes the fewest steps. Commented Jul 17, 2020 at 15:02