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Would it be possible to prove that there is an equation that includes a number N of unknown numbers that are all equal, between 0 and 1 and whose sum is equal to 1 ? And to find this equation ?

I don't know if the problem is well defined and if it's possible. It works with A - B = 0 with A = B = 0.5 but I'm struggling to prove and extend it. I'm not sure that it would be "legal" to do any multiplication or division.

Regards

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  • $\begingroup$ Why does $A-B=0$ force $A=B=\frac 12$? That is one solution, but there are many more that do not have $A+B=1$ $\endgroup$ – Ross Millikan May 14 at 15:23
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You can certainly do $$\left(a-\frac 13\right)^2+\left(b-\frac 13\right)^2+\left(c-\frac 13\right)^2=0$$ with the obvious generalization to $n$ variables.

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  • $\begingroup$ Thanks for your answer, why do we need to square members ? How to prove that this is working with n variables? $\endgroup$ – A.Ben May 15 at 5:01
  • $\begingroup$ We square them because squares are always greater than or equal to zero. If a sum of squares is zero, each term must be zero. That is how I get away from the problem of having one equation in more than one variable-really I have $n$ equations because each term must be zero. Can you see the connection between the fact that I have three terms and the fractions are $\frac 13$? That shows how to do it with $n$ variables. $\endgroup$ – Ross Millikan May 15 at 5:53
  • $\begingroup$ Ok thanks. Yes I see the connection. But is it mathematically "legal" to just replace the 1/3 by 1/n without proving that it is true ? I may prove it with recursion but I don't know if it is enough to stand as a proof $\endgroup$ – A.Ben May 15 at 7:05
  • $\begingroup$ Ok I got it, I guess the sentence "If a sum of squares is zero, each term must be zero" can be sufficient to prove this. Thanks a lot $\endgroup$ – A.Ben May 15 at 11:08

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