# Why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$

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!

• How is this algebraic number theory? – Asaf Karagila Feb 18 '11 at 18:18
• @Asaf: It isn't. You should feel free to do tag edits when you see any that are so off (you have more than enough reputation for it). – Arturo Magidin Feb 18 '11 at 18:19
• Sorry for the mistag. On a second thought, this was so easy there's only one correct answer - don't do maths on a friday night. – Milosz Wielondek Feb 18 '11 at 18:19
• Arturo, I usually do. From time to time I use my iPhone to actively use the site, at which case I just tend to be a jerk about it. :-) – Asaf Karagila Feb 18 '11 at 18:25
• 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. – shuhalo Feb 19 '11 at 4:44

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

• When $x \neq 0$. – I. J. Kennedy Feb 18 '11 at 22:45
• ..and when $x\neq -1$ – InsideOut Feb 6 at 17:59

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 $x \neq 0$.

• #Matthew: It is correct as an identity of rational functions, i.e. polynomial fractions. – Gone Feb 18 '11 at 19:19
• 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. – Matthew Conroy Feb 18 '11 at 22:39

Divide both the numerator and the denominator by $x$.