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Critique for proof is always encouriged I suppose, I'm notthe best expert on proving things in math usually... I wonder if more rigorous or clever technique/style could be used in this problem.

task

if there's set of numbers $\{x_1, x_2, ... x_n\}$ they have a mean of $\hat{x}$

prove that $ \sum_{i=1}^{n} (x_i - \hat{x})=0$

answer

Because original form of sum is always a finite sum it can be written in the form as follows $(x_1 - \hat{x}) + (x_2 - \hat{x}) + (x_3 - \hat{x}) +... + (x_n - \hat{x}) $

separate into two sums and add them together $\sum_{i=1}^{n} (x_i ) + \sum_{i=1}^{n} (\hat{-x} ) = \sum_{i=1}^{n} (x_i - \hat{x}) $

we know from the common arithmetic mean formula such that $\hat{x} = \sum_{i=1}^{n} (x_i /n) \leftrightarrow \hat{x} *n = \sum_{i=1}^{n} (x_i ) $

then we can write $\sum_{i=1}^{n} (x_i - \hat{x})= \sum_{i=1}^{n} (x_i ) + \sum_{i=1}^{n} (\hat{-x} ) $ $= \hat{x}*n + \sum_{i=1}^{n} (\hat{-x}) $

on the right hand side sum we have just constant value being summed repeatedly, which means repeated addition is multiplication $= \hat{x}*n - \hat{x}*n =0 $

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  • $\begingroup$ You have the right idea. Summation is commutative and associative $\sum (a_i+b_i) = \sum a_i + \sum b_i.$ By definition $\hat x = \frac {1}{n}\sum_\limits{i=1}^n x_i.$ And, $\hat x$ is a constant so $\sum_\limits{i=1}^n \hat x = n\hat x$ $\endgroup$ – Doug M Apr 5 '18 at 21:58
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You do have the idea. Depending on how much detail you want or need to include, you could make your proof more concise. For example, write: $\sum_{i=1}^{n} (x_i - \hat{x})=\sum_{i=1}^{n}x_i - \sum_{i=1}^{n}\hat{x}=n\cdot \frac{\sum_{i=1}^{n}x_i}{n}-n \cdot \hat{x} = n \cdot \hat{x} - n \cdot \hat{x} = 0$. It looks like you essentially explained each step.

Since you asked about rigor, one thing to note is that here, we used properties of the summation here (For example $\sum_{i=1}^{n}(a_i+b_i)=\sum_{i=1}^{n}a_i+\sum_{i=1}^{n}b_i$). Technically, you define summations formally and use that definition to prove its properties (for summations, you use induction). Then for proofs like this, you use the results without proof of its properties.

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