# Picking socks probability proof

I'm trying to follow the proof from question 1b out of the book "50 Challenging Problems in Probability" by Mosteller. The problem states:

A drawer contains red and black socks. When two socks are drawn at random, the probability that both are red is $1/2$. How small can the number of socks in the drawer be if the number of black socks is even?

The answer given starts as follows:

Let there be $r$ red and $b$ black socks. The probability of the first socks' being red is $\frac{r}{r+b}$ and if the first sock is red, the probability of the second's being red now that a red has been removed is $\frac{r-1}{r+b-1}$. Then we require the probability that both are red to be $\frac{1}{2}$, or $$\frac{r}{r+b} * \frac{r-1}{r+b-1} = \frac{1}{2}$$ Notice that $$\frac{r}{r+b} > \frac{r-1}{r+b-1}, \text{for b} >0$$ Therefore we can create the inequalities $$(\frac{r}{r+b})^2 > \frac{1}{2} > (\frac{r-1}{r+b-1})^2$$

This is where I'm confused. Why is it that $(\frac{r}{r+b})^2 > \frac{1}{2}$? If $r=1, b=100$ then obviously $(\frac{1}{101})^2 < \frac{1}{2}$. Am I missing some obvious assumption?

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$$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2}$$
Given that is true, and that one of the fractions is larger than the other, it's clear that their squares must lie either side of $\frac{1}{2}$
Note that in your example with $r=1, b=100$, it is NOT the case that $\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2}$ - if you ignore the condition, the result wouldn't be expected to hold.