Do there exist $n$ rational numbers, $n>2$, such that their sum is $1$, difference is $0$?

Question

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.

Answer 1

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 $a - \sum_i a_i = \frac{1}{2} - \frac{1}{2} =0$