# Number Theory Question

1. Nov 17, 2006

### Astro

I have a follow-up question to the "Calculating Averages" thread.
Please see the "Followup_Question" attachment.

Any help is, as always, most appreciated.

(See this link for the original question: https://www.physicsforums.com/showthread.php?t=143759)

#### Attached Files:

• ###### Followup_Question.pdf
File size:
11.2 KB
Views:
60
2. Nov 21, 2006

### driscoll79

Hey Astro,

You cannot do what you are suggesting for a pretty simple reason. Think of the one-dimensional number line. Each positive number x on the number line represents the distance between that number and 0. So, any line segment on the number line with length x lies on a closed interval $$[0 + \alpha, x + \alpha]$$ where $$\alpha$$ is an arbitrary constant in the real numbers. So, when you average the three numbers 1, 3, and 8, you are in fact averaging the lengths of three intervals [0,1], [0,3], [0,8] (here I'm taking $$\alpha$$ to be 0, but that's really inconsequential).

Your method posits taking half the length of the intervals [1,3] and [3,8] and averaging those values, which to be honest doesn't make any sense when you think of this geometrically. It also doesn't make sense logically, since you eliminated an interval in the process. However, you do have the germ of an idea here. If we consider three subintervals that lie on the interval [0,8]: [0,1], [1,3], and [3,8] and if we apply a version of your method:

$$\frac{1-0}{2} + \frac{3-1}{2} + \frac{8-3}{2}$$,

then we arrive at:

$$\frac{1 + 2 + 5}{2} = 4$$

...which is the average of 1, 3, and 8. I'm not certain if I answered your question as thoroughly as you'd hoped, but I hope that this helps you in some way.

Best,