<s>[INST] bijection from $(3,4)$ to $ (7,8) - \{ 7+\frac{1}{2}, 7+\frac{1}{3}, 7+\frac{1}{4} ,\cdots \}$ Find a bijection from $ (3,4) $ to $ (7,8) - \{ 7+\frac{1}{2}, 7+\frac{1}{3}, 7+\frac{1}{4} ,\cdots \}$
Any ideas? I know it may be easy but I can't think anything.
Thank you!
 [/INST] Without loss of generality, suppose the first interval is also $(7,8)$.  (you can just apply the function $x+4$ to go inbetween the two.)  If we tried the function $f(x)=x$, that would give a bijection between $(7,8)$ and $(7,8)$.  The problem is, how do we avoid the points $7+\frac{1}{2}$ etc..?  
Consider the points of the form $7+\frac{n-1}{n}$ for $n=3,4,\dots$.  Lets change the function $f(x)=x$ only on the numbers of the form $7+\frac{n-1}{n}$ or $7+\frac{1}{n}$.  Send $7+\frac{1}{2}$ to $7+\frac{2}{3}$, and $7+\frac{2}{3}$ to $7+\frac{3}{4}$.  Then send $7+\frac{1}{3}$ to $7+\frac{4}{5}$ and $7+\frac{3}{4}$ to $7+\frac{5}{6}$.  Continuing in this way, send $7+\frac{1}{n}$ to $7+\frac{2n-2}{2n-1}$ and $7+\frac{n-1}{n}$ to $7+\frac{2n-3}{2n-2}$.  This gives the bijection.
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