BMAD 64 Posted March 18, 2016 Report Share Posted March 18, 2016 Consider the following sequence where (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) ... now clearly this is the same as (0) + (0) + (0) + (0) + (0) + (0) ... = 0 however if I apply the associative property of addition to this series I get... 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) ... which clearly equals 1 + (0) + (0) + (0) + (0) + (0) ... = 1 But 1 does not 0, is the associative property wrong? Quote Link to post Share on other sites

0 Solution bonanova 84 Posted March 19, 2016 Solution Report Share Posted March 19, 2016 Association holds for finite series. Infinite series converge only under certain conditions, which fail in this case. So it's not proper to give the series a value at all. But let's ignore that and call the series S anyway. Then what is S? Clearly it's 1/2: Spoiler S = 1-1+1-1+1-1+1-... 0+ S = 0+1-1+1-1+1-1+1-... -------------------------- 0+2S = 1+0+0+0+0+0+... = 1 2S=1; S=1/2. Quote Link to post Share on other sites

0 harey 8 Posted March 23, 2016 Report Share Posted March 23, 2016 My favorite one: Take the sum of all the integers. Call it S. Take the sum of even integers. That sum is S/2. Because S is infinite, S/2=S, their difference is zero. So the sum of the odd integers is zero. Quote Link to post Share on other sites

## Question

## BMAD 64

Consider the following sequence where

(1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) ... now clearly this is the same as (0) + (0) + (0) + (0) + (0) + (0) ... = 0

however if I apply the associative property of addition to this series I get...

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) ... which clearly equals 1 + (0) + (0) + (0) + (0) + (0) ... = 1

But 1 does not 0, is the associative property wrong?

## Link to post

## Share on other sites

## 2 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.