Understanding Symmetry Arguments

This is an idea used in the solution of Problem 4 in Siklos' book.

What is the average value of the integers greater than or equal to 0 and less than 1000 that aren't divisible by 2 or 5?

Can someone explain why it is simply 1000/2 (which is 'obvious by symmetry' :wink:

? I found the average using a pairing argument...

(edited 6 years ago)

Reply 1

atsruser

Original post by FXLander

This is an idea used in the solution of Problem 4 in Siklos' book.

What is the average value of the integers greater than or equal to 0 and less than 1000?

Can someone explain why it is simply 1000/2 (which is 'obvious by symmetry' :wink:

? I found the average using a pairing argument...

It's not clear to me which symmetry he's thinking of. To my mind, the "obvious" symmetry to exploit is the geometrical one, whereby if we take a copy of a 1000 integer "staircase" function, turn it upside down, and put it on top of the first, then the average is "obviously" 1000, since all the columns are now 1000 high. Hence the average of a total 1/2 that size is 1000/2.

Maybe it's something else though.

Reply 2

Zacken

Another way (which is just the algebraic version of atsusers geometric perspective) is that the sum of the first n+1 numbers, starting from 0 is n(n+1)/2 using the symmetry on either side of 500; Gauss's method.

So the average is n/2

Reply 3

Pangol

Original post by FXLander

? I found the average using a pairing argument...

I think that he means the same thing as you. I presume that you mean that you paired 1 with 999, 2 with 998, etc. I would say that this is an application of symmetry - the set of numbers you are considering is, in this sense, symmetrical about 500, in that they are equally spaced out on either side of this middle value.

Reply 4

atsruser

Original post by Zacken

Wasn't Gauss's method the pairing one?

https://nrich.maths.org/2478

[edit: sorry, yes you're describing the pairing method, but my method is a doubling one - it's slightly different]

(edited 6 years ago)

Reply 5

artful_lounger

Universities Forum Helper

I'm fairly sure your pairing argument is the same as his symmetry one - he just summed the pairs, so to speak.

Reply 6

FXLander

Yeah I think I understand what he means now... In the problem we are considering only those numbers that aren't divisible by 2 or 5. So if I had a list with the integers from 0 to 1000, I would be crossing off multiples of 2 and 5. And since these multiples are symmetric about 500, the numbers 'left over' must be symmetric about 500 as well (i.e. 500 will still be the middle value). Therefore the average is simply that after I have crossed off all the multiples.

Does that make sense?

Thanks for all the great explanations btw especially the geometrical interpretation, never thought of that before :smile:

(edited 6 years ago)

Reply 7

RogerOxon

Original post by FXLander

What is the average value of the integers greater than or equal to 0 and less than 1000?

Can someone explain why it is simply 1000/2 (which is 'obvious by symmetry' :wink:

? I found the average using a pairing argument...

It isn't 1000/2. Did you mis-quote the question?

You asked for the average of 0..999. The average of 0..1000 or 1..999 is 500. If you add the extra 0 in, and divide by 1000 not 999, then you slightly reduce the average.

(edited 6 years ago)

Reply 8

FXLander

Original post by RogerOxon

Sorry I misquoted it. It's asking for the average of the integers greater than 0 and less than a 1000 that are not divisible by 2 or 5.

Reply 9

RogerOxon

Original post by FXLander

Sorry I misquoted it. It's asking for the average of the integers greater than 0 and less than a 1000 that are not divisible by 2 or 5.

That makes more sense, but adds an extra step.

When you look at the pairs that add to 1000, you can see that you will either have both (x, 1000-x), or neither, when removing numbers divisible by 2 or 5. There's no difference if the 500 (the only non-pairing number) is or isn't eliminated (it is though).