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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!

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    It might be helpful to note, that there is a blow-up at -1, as expected while at 0 the right expression is formally undefined, but it makes sense give it the value 0 (by continouos extension, if you know this). Plotting this function might help you.2011-02-19

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$\frac{x}{x+1} = \frac{x \cdot 1}{x \cdot (1+\frac{1}{x})}= \frac{x}{x} \cdot \frac{1}{1+\frac{1}{x}} = 1 \cdot \frac{1}{1+\frac{1}{x}} = \frac{1}{1+x^{-1}}$

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    When $x \neq 0$.2011-02-18
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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 $x \neq 0$.

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    I'm not sure what you mean, Bill. Can you elaborate? For instance, the denominator of the rhs is not a polynomial, so I'm not clear what you intend by saying this is an identity of polynomial fractions.2011-02-18
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Divide both the numerator and the denominator by $x$.