# Proving that an equality is false.

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I have two sets of 64 numbers (1.1 to 7.4). Both number sets are created using the same equation for values of i from 0 to 63.

m = 1.1 + ( i * 0.1 )

n = 1.1 + ( i * 0.1 )

I am trying to understand if the following equality is false in all cases except when the terms in each expression are equal (e.g. m^-12 = n^-12 and so on).

m^-12 + m^-11 + m^-10 + m^-9 + m^-8 + m^-7 + m^-6 + m^-5 + m^-4 + m^-3 + m^-2 = n^-12 + n^-11 + n^-10 + n^-9 + n^-8 + n^-7 + n^-6 + n^-5 + n^-4 + n^-3 + n^-2

There are no restrictions regarding utilization of the numbers in the number sets: the same number may be used repeatedly in the expression (e.g. 0.1^-12 + 0.1^-12 and so on).

Alternatively, I need to identify when the equality will be true in cases where the terms are different (e.g. m^-12 not equal to n^-12 and so on).

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An equality is proven to be false. The solution is detailed and well presented.

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I have two sets of 64 numbers (1.1 to 7.4). Both number sets are created using the same equation for values of i from 0 to 63.

m = 1.1 + ( i * 0.1 )

n = 1.1 + ( i * 0.1 )

I am trying to understand if the following equality ...

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- BSc , Wuhan Univ. China
- MA, Shandong Univ.

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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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