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Do there exist $n$ rational numbers $a,a_2,...,a_n\;$, $n>2$, such that their difference (in order) is $0$ and their sum is $1$?

That is, are their any rational solutions $a_1,a_2,...,a_n$ to:

$a+a_2+a_3$ $=$ $1$, $a-a_2-a_3$ $=$ $0$

$a+a_2+a_3+a_4$ $=$ $1$, $a-a_2-a_3-a_4$ $=$ $0$

.........

$a+a_2+a_3+...+a_n$ $=$ $1$, $a-a_2-a_3-...-a_n$ $=$ $0$

(The order of the numbers is important)

If not, is there a simple proof to show that the specified conditions can't be satisfied?

Thanks for help.

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  • $\begingroup$ Do you have $a_i\le a_{i+1}$? also can you have $a_i=a_{i+1}$? $\endgroup$ – Elliot G Mar 7 '17 at 6:41
  • $\begingroup$ what about $.5, .25, .25$ ? $\endgroup$ – Badam Baplan Mar 7 '17 at 6:42
  • $\begingroup$ @BadamBaplan How would a pair of 4 rational numbers work? I see how you got the pair of 3 rational numbers, thanks for that. $\endgroup$ – J. Linne Mar 7 '17 at 6:43
  • $\begingroup$ Your conditions imply that the first number is $1/2$. Then take any set of positive rational numbers adding to $1/2$, and you're done $\endgroup$ – Badam Baplan Mar 7 '17 at 6:44
  • $\begingroup$ @BadamBaplan so the first value $a$ must be $1/2$, otherwise no solution will work? $\endgroup$ – J. Linne Mar 7 '17 at 6:45
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The difference relation gives you $a = \sum_i a_i$, and then plugging that into the sum relation gives $a + a = 1$, so that $a = \frac{1}{2}$.

Now you can quickly check that any set of rational numbers $\{a_i\}$ with $\sum_i a_i = \frac{1}{2}$ has $a + \sum_i a_i = \frac{1}{2} + \frac{1}{2} = 1$ and likewise $a - \sum_i a_i = \frac{1}{2} - \frac{1}{2} =0$

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