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?)

$\frac{1}{x-{a}_{1}}+\frac{1}{x-{a}_{2}}+\cdots +\frac{1}{x-{a}_{n}}=0$

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?)