# I am having this equation: 1/(x−a_1)+1/(x−a_2)+⋯+1(/x−a_n)=0 where a_1<a_2<⋯<an are real numbers. Now I want to prove with the intermediate value theorem that this equation has n−1 solutions in the real numbers.

Ethen Frey 2022-07-22 Answered
I am having this equation:
$\frac{1}{x-{a}_{1}}+\frac{1}{x-{a}_{2}}+\cdots +\frac{1}{x-{a}_{n}}=0$
where ${a}_{1}<{a}_{2}<\cdots <{a}_{n}$ are real numbers.
Now I want to prove with the intermediate value theorem that this equation has $n-1$ solutions in the real numbers.
My thoughts:
With ${a}_{1}<{a}_{2}<\cdots <{a}_{n}$, you can see that every summand gets smaller than the summand before.
My other thought was that about the $n$-summands, with the intermediate value theorem you know that every zero (point) is in the interval and is located between the $n$-summands. So there are $n-1$ solutions for this equation!
Questions:
How can I prove my thoughts in a formal correct way? (Are my thoughts generally correct?)
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emerhelienapj
Let's call your function $f$. We'll look at it moving left to right along the line. When $x<{a}_{1}$, $f\left(x\right)<0$, because all the summands are negative. So there are no roots there.
Now let's look at $\left({a}_{1},{a}_{2}\right)$. In particular let's look at $x\to {a}_{1}^{+}$. All but the first term is negative, but the first term is going to $+\mathrm{\infty }$ while the others are remaining finite. So $\underset{x\to {a}_{1}^{+}}{lim}f\left(x\right)=+\mathrm{\infty }$. In particular, $f$ is positive somewhere on $\left({a}_{1},{a}_{2}\right)$. Similarly we get that $f$ is negative somewhere on $\left({a}_{1},{a}_{2}\right)$ by considering $x\to {a}_{2}^{-}$.
You can take it from here.

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We can use the intermediate value theorem with this beginning.
On $\left({a}_{1},{a}_{2}\right)$, approaching ${a}_{1}$ gives us ${x}_{1}$ with $f\left({x}_{1}\right)>0$. Approaching ${a}_{2}$ gives us ${x}_{2}$ with $f\left({x}_{2}\right)<0$. Then the intermediate value theorem gives us ${x}_{3}\in \left({x}_{1},{x}_{2}\right)$ with $f\left({x}_{3}\right)=0$.

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