0
$\begingroup$

Let $a,b,c$ be three integers greater than $0$, and assume there is a real number $t$ such that $ \frac{a}{a+b}=\frac{\left\lfloor t\right\rfloor}{\left\lfloor t\right\rfloor+1}. $ Is there a way to show that $c$ is bounded in the interval $[a-d,a+d]$ for some $d$ constant, if we would say that $\frac{t}{t+1}=\frac{c}{c+b}$? Is there other stuff needed to be verifies first?

Thanks.

2 Answers 2

1

Let $\lfloor t \rfloor = n$. Note that $\frac{a}{a+b} = \frac{a/b}{a/b+1}$. Since the function $ t \to \frac{t}{t+1}$ is increasing on $(-1,\infty)$, if $\frac{a}{a+b} = \frac{n}{n+1}$ and $\frac{c}{c+b} = \frac{t}{t+1}$ we must have $a/b = n$ and $c/b = t$. But $n \le t < n+1$, so $a = b n \le b t = c < b (n+1) = a + b$.

0

You need to be clear about what is input and what is output. It looks like you are given $a$ and $b$ and hope to find a $t$ such that $\frac{a}{a+b}=\frac{\lfloor t \rfloor}{\lfloor t+1 \rfloor}$. First, are you sure this is possible? For example, let $a=2,b=5$ and there is no such $t$. If you take off the floor function, you can solve it with $t=0.4$. But if you take off the floor, $t$ is a fixed number and you just find $c=a$. So what is the question?

  • 0
    It will be true only for $b=1$ if $\frac{a}{a+b}$ is in lowest terms. In which case, $t\in [a,a+1)$. You can certainly then solve $\frac{t}{t+1}=\frac{c}{c+1}$, but it will give $c=t$.2011-04-24