I need some help understanding why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$.
I would be grateful if someone could explain. Be explicit.
Thank you!
I need some help understanding why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$.
I would be grateful if someone could explain. Be explicit.
Thank you!
$\frac{x}{x+1} = \frac{x \cdot 1}{x \cdot (1+\frac{1}{x})}= \frac{x}{x} \cdot \frac{1}{1+\frac{1}{x}} = 1 \cdot \frac{1}{1+\frac{1}{x}} = \frac{1}{1+x^{-1}}$
Your equation is not correct. Note that the expression on the left-hand side is defined for $x=0$, but the right-hand expression is not. Thus, the two expressions are not equivalent. However, assuming $x\neq 0$, we may transform the left-hand expression into the right-hand expression as tpv has shown.
In particular to tpv's explanation, note that $\frac{x}{x}=1$ only if $x \neq 0$.
Divide both the numerator and the denominator by $x$.