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

Question

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 $\frac{b}{b-1} \in [0,\infty)$.

Answer 1

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

Answer 2

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

Answer 3

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.