Final part of determining an equivalence relation

[kathrynag]

Homework Statement

For a,b elements of the real numbers, define a~b if \left|a-b\right|\leq
1. Determine if we have an equivalence relation.

Homework Equations

The Attempt at a Solution

I've already done the first two parts of determining. it's only the last part that I'm having trouble with, so i will omit the first two parts.
We bwant to determine if for all a,b,c elements of S, if (a,b) is an element of R and (b,c) is an element of R, then (a,c) is an element of R.
We know \left|a-b\right|\leq1 and \left|b-c\right|\leq1.
This means a-b\leq1,b-a\leq1, b-c\leq1, and c-b\leq1.
Furthermore, a\leq1+b, b\leq1+a, b\leq1+c, c\leqb+1.
I know from this, I need to be able to show a-c\leq1 and c-a\leq1 in order to have an equivalence relation.
That's what I don't see right off hand is how to get that.

 

[CompuChip]

How about a = 0, b = 1, c = 2?

 

[kathrynag]

But that's not how to determine an equivalence relation. I can't just pick numbers because it has to work for all a, b and c.

 

[CompuChip]

it has to work for all a, b and c.

Exactly.

You might want to carefully re-read the question :)

 

[kathrynag]

Could I do something like this:
Ok if c\leq1+b and a\leq1+b.
We have c\leqa or c-a\leq0.
If a\leq1+b and c\leq1+b, we can say a\leqc or a-c\leq0.
Could I do something like that since that implies \left|a-c\right|\leq0, but we wanted \left|a-c\right|\leq1. Thus not an equivalence relation.

 

[CompuChip]

Why do you want to do it the hard way?
If it is an equivalence relation, |a - b| <= 1 and |b - c| <= 1 should imply |a - c| <= 1 for all a, b and c.
So if you can find just one pair (a, b, c) for which it's not true, you are done, right?

 

[kathrynag]

Why do you want to do it the hard way?
If it is an equivalence relation, |a - b| <= 1 and |b - c| <= 1 should imply |a - c| <= 1 for all a, b and c.
So if you can find just one pair (a, b, c) for which it's not true, you are done, right?

Well, I know my professor would not want to just see a pair since he is a very strict grader.

 

[tiny-tim]

hi kathrynag!

(have a ≤)

Well, I know my professor would not want to just see a pair since he is a very strict grader.

honestly, compuchip is right …

your professor asked you to determine whether it is an equivalence relation …

if it is, then of course you have to prove it strictly … as you say, you can't just pick numbers because it has to work for all a, b and c

but if it isn't, then you prove that simply by finding one a b and c for which it doesn't work

 

[CompuChip]

Did you read this Wikipedia page I linked you to earlier (actually, the first paragraph of the introduction will do)?