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If 0=(x-a)(x-b)(x-c)...(x-x)..=0. So it's a product sum that we write with pi instead of sigma but how? There should be indexes but I'm not convinced that I understand what notation to use.

$$\prod_{ z=a }^{ y }{\left(x-z\right)} = 0$$

I don't think the expression above is correct but it's my attempt to illustrate the 0. I'm not trying to solve a specific problem. I want to learn how to formalize it if you can tell how.

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Is it supposed to say, $x-x$? Because then there is no need for Pi notation, the whole thing is $0$.... – TylerHG Aug 29 '14 at 1:29
Yes the whole thing is zero. It's a "math joke" to see that the multiplication will be zero from (x-a)*(x-b)*(x-c)...(x-x).. if you only view (x-a)(x-b)(a-c)... then it's what some engineer tricked me with ("Can you know what the answer is: ( (x-a)(x-b)(x-c)... – Programmer 400 Aug 30 '14 at 4:28

You could try defining a family like: $$ \mathcal F = \{a, b, c, \ldots, x, \ldots, z\} $$ and then doing: $$ \prod_{\alpha \in \mathcal F} (x - \alpha) = 0 $$

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If $a$ is one of the elements of the set, you may not want to use $a$ as your dummy variable. – GEdgar Aug 29 '14 at 1:32
@GEdgar My dummy variable is supposed to be alpha, or $\alpha$. I guess it's hard to see. Perhaps I should use something like $\lambda$ instead? – Adriano Aug 29 '14 at 1:33

Seems like you're using $x$ in multiple ways. Integer subscripts are a wonderful way of avoiding this.

I think the statement you want is $0 = (x-a_1) (x- a_2) \ldots (x- a_n)$, which can also be written as $\prod_{i=1}^n (x-a_i) = 0$.

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This is actually a classic 'trick' question hinging on the poor notation. See or, for example. – Alex Zorn Aug 29 '14 at 1:07
@AlexZorn Yes, an electronics engineer tested it like that for me on his phone display. – Programmer 400 Aug 30 '14 at 4:30

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