# How rounding real numbers affect mean

First of all, I am not used to construct mathematical demonstration, so I will discuss my point in a argument based fashion. This is why I am asking to mathematician a more formal and correct way to prove my point.

I often have to compute mean of rounded numbers, and I would make sure that rounding number does not affect too much the mean.

Having three functions, $\mathrm{floor}$, $\mathrm{round}$ and $\mathrm{ceil}$ which respectively take the previous, the closest, the next integer to $x_i$, the following relations hold: $$\mathrm{floor}(x_i) \leq \mathrm{round}(x_i) \leq \mathrm{ceil}(x_i)$$

And also:

$$\mathrm{floor}(x_i) \leq x_i \leq \mathrm{ceil}(x_i)$$

I have the intuition that rounding number cannot change the mean more than a unit. Thus I would like to prove than:

$$|\bar{x} - \bar{x}^*| \leq 1$$

Where:

$$\bar{x} = \frac{1}{n}\sum_i^n{x_i} \,,\quad \bar{x}^* = \frac{1}{n}\sum_i^n{\mathrm{round}(x_i)}$$

My reasoning is the following:

I may state that:

$$\frac{1}{n}\sum_i^n{\mathrm{floor}(x_i)} \leq \bar{x}^* \leq \frac{1}{n}\sum_i^n{\mathrm{ceil}(x_i)}$$

And:

$$\frac{1}{n}\sum_i^n{\mathrm{floor}(x_i)} \leq \bar{x} \leq \frac{1}{n}\sum_i^n{\mathrm{ceil}(x_i)}$$

Because of the definition of those functions and linearity of inequalities operations (there are only addition and multiplication by positive numbers).

We also may observe than:

$$\mathrm{ceil}(x_i) \leq x_i + 1$$

And:

$$\mathrm{floor}(x_i) \geq x_i - 1$$

By expanding summation, it leads to:

$$\bar{x} - 1 \leq \bar{x}^* \leq \bar{x} + 1$$

Which is what I want to prove. I am right doing this reasoning. How could I make it more formal?

• I don't see a problem with this proof, but I guess that a way to maybe make it more simple is using the fact that $floor(x_i) + 1 =ceil(x_i)$ from the begging – mathstu15 Nov 11 '15 at 10:23