<s>[INST] Where is the error in this apparent contradiction arising from the function $f(\pm n)=\left(\frac{n}{n+1}\right)^{\pm 1}$? We define a function $ f \left( \pm n \right) =\left( \frac{ n }{ n + 1 } \right) ^ { \pm 1 } $.
That is, $$ f \left( m \right) = \begin{cases}\frac{ m }{ m + 1 } &\text{ if } m \ge 0 \\ \frac{ -m + 1 }{ -m } & \text{ if }m \lt 0 \end{cases}. $$
From this, $ f \left( - a \right) = \frac{ a + 1 }{ a } = \frac{ a }{ a } + \frac{ 1 }{ a }= 1 + \frac{ 1 }{ a } $
However, we can also do this:
$ f \left( \frac{ \alpha }{ \beta } > 0 \right) = \frac{ \alpha }{ \beta } \div \left( \frac{ \alpha }{ \beta } + 1 \right) = \frac{ \alpha }{ \beta } \div \left( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \beta } \right) = \\ = \frac{ \alpha }{ \beta } \div \frac{ \alpha + \beta }{ \beta } = \frac{ \alpha \beta }{ \alpha \beta + \beta ^ 2 } $
Since $ \frac{ \alpha \beta + \beta ^ 2 }{ \alpha \beta } = 1 + \frac{ \beta ^ 2 }{ \alpha \beta } = 1 + \frac{ \beta }{ \alpha } = \frac{ \alpha + \beta }{ \alpha } $, $ \frac{ \alpha \beta }{ \alpha \beta + \beta ^ 2 } =  \frac{ \alpha }{ \alpha + \beta } $.
Since $ f \left( k \right) = \frac{ 1 }{ f \left( - k \right)} $, $ f \left( - \frac{ \gamma }{ \delta } \right) = \frac{ \gamma }{ \gamma + \delta } $.
But this conflicts with the previous definition!
It says that $f \left( -2 \right) = \frac{ 1 }{ 2 }$ AND $= 1 + \frac{ 1 }{ 2 }$!
Where is the problem?
 [/INST] you have , where you used alpha and beta
$\frac{a}{b}\div \frac{a + b}{b} = \frac{a}{a +b}$
but you've ended up with $b^2$ etc which is right, but you could cancel b
then you say 'since' and you don't cancel b out properly
</s>