I am having this equation: 1 x − a 1 + 1 x − a 2 + ⋯ + 1 x − a n = 0where a 1 &lt; a 2 &lt; ⋯ &lt; 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 &lt; a 2 &lt; ⋯ &lt; 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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I am having this equation:      1          x      −              a        1              +      1          x      −              a        2              +  ⋯  +      1          x      −              a        n              =  0where       a    1    &amp;lt;      a    2    &amp;lt;  ⋯  &amp;lt;      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    &amp;lt;      a    2    &amp;lt;  ⋯  &amp;lt;      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?)

emerhelienapj · Accepted Answer

Let&#039;s call your function   f. We&#039;ll look at it moving left to right along the line. When   x  &amp;lt;      a    1  ,   f  (  x  )  &amp;lt;  0, because all the summands are negative. So there are no roots there.Now let&#039;s look at   (      a    1    ,      a    2    ). In particular let&#039;s look at   x  →      a    1    +  . All but the first term is negative, but the first term is going to   +  ∞ while the others are remaining finite. So       lim          x      →              a        1        +              f  (  x  )  =  +  ∞. In particular,   f is positive somewhere on   (      a    1    ,      a    2    ). Similarly we get that   f is negative somewhere on   (      a    1    ,      a    2    ) by considering   x  →      a    2    −  .You can take it from here.

Kade Reese · Accepted Answer

We can use the intermediate value theorem with this beginning.On   (      a    1    ,      a    2    ), approaching       a    1   gives us       x    1   with   f  (      x    1    )  &amp;gt;  0. Approaching       a    2   gives us       x    2   with   f  (      x    2    )  &amp;lt;  0. Then the intermediate value theorem gives us       x    3    ∈  (      x    1    ,      x    2    ) with   f  (      x    3    )  =  0.

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.

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