Show that this is an equivalence relation

**jokke22** · April 26th 2009, 04:40 AM

Hi!

I have given av relation ~ by $\mathbb{N}^2$ where:
$(n,m)$ ~ $(k,l) <=> n+l = m+k$

How would you show that ~ is an equivalence relation?

Need something to get me started, though I understand the relation, not just "how" to prove it.
In other words, show that the realtion is reflexive, symmetric and transitive.
Reflexive $a \mathcal{R} a$
Symmetric $a \mathcal{R} b \implies b \mathcal{R} a$
Transative if $a \mathcal{R} b \mbox{ and } b \mathcal{R}c \implies a \mathcal{R}c$

Thanks in advance for any help/tips!

**Plato** · April 26th 2009, 08:53 AM

Here are the three:
$\begin{gathered} x + y = y + x \hfill \\ a + d = b + c\, \Leftrightarrow \,c + b = d + a \hfill \\ a + d = b + c\,\& \,c + f = d + e\, \Leftrightarrow \,a + f = b + e \hfill \\ \end{gathered}$

Fill in details.

**jokke22** · April 26th 2009, 10:45 AM

Originally Posted by Plato

Here are the three:
$\begin{gathered} x + y = y + x \hfill \\ a + d = b + c\, \Leftrightarrow \,c + b = d + a \hfill \\ a + d = b + c\,\& \,c + f = d + e\, \Leftrightarrow \,a + f = b + e \hfill \\ \end{gathered}$

Fill in details.

Thank you for your reply!

So I could rewrite this as:

"Edited for new approach..."

**jokke22** · April 26th 2009, 11:03 AM

Hmmm... I can't seem to get it...

Could you elaborate a tiny bit...? :P

**Jhevon** · April 26th 2009, 11:19 AM

Originally Posted by jokke22

Hmmm... I can't seem to get it...

Could you elaborate a tiny bit...? :P

what do you need elaborated? Plato listed why the relation fulfills the three properties. the first equation shows reflexivity, the second set show symmetry, and the third, transitivity

**jokke22** · April 26th 2009, 11:29 AM

Originally Posted by Jhevon

what do you need elaborated? Plato listed why the relation fulfills the three properties. the first equation shows reflexivity, the second set show symmetry, and the third, transitivity

It's getting late over here... I just didn't see the connection at first!

Thank you yet again!

**spearfish** · April 26th 2009, 12:08 PM

For REFLEXIVE:

Suppose aRa --> (n,m) R (n,m)
Then n + m = m + n

n + m = n + m

((n+ m) = (n +m)) / (n+m) Divide both sides by n+m

1 = 1

Therefore, R is reflexive

**Jhevon** · April 26th 2009, 12:13 PM

Originally Posted by spearfish

For REFLEXIVE:

Suppose aRa --> (n,m) R (n,m)
Then n + m = m + n

n + m = n + m

((n+ m) = (n +m)) / (n+m) Divide both sides by n+m

1 = 1

Therefore, R is reflexive

*sigh*, no, for reflexive, you need to show that (n,m) R (n,m) (that is, all points relate to themselves under this relation). Since n + m = m + n (by the commutativity of addition on the integers), we have reflexivity.

now do the others.

it's getting really late there, huh?

Plato did most of the work for you, you only need to explain his equations. don't try to make new ones

**jokke22** · April 26th 2009, 12:24 PM

Just to get this right, as its not given any values and on how the relation stands, the answer Plato has given is the "general" solution to all similar relations to mine?

First timer on this subject, pain with some new terms etc.

Haha! Going past 11pm now... Could be that I just suck really bad at this subject as well ;-)

**spearfish** · April 26th 2009, 12:27 PM

I think you were referring to jokke22, lol. I was just helping. Anyways, here are the others:

SYMMETRIC:
Suppose (n,m)R(k,l)
Then n + l = m +k
n = m + k -l

Suppose (k,l) R (n,m)
Then k + m = l + n
k + m = l + (m+k-l)
k+ m = m+ k

Therefore, R is Symmetric

TRANSITIVE: a = (n,m), b = (k,l), c = (o,p)
Suppose (n,m)R(k,l)
Then n + l = m + k
l = m + k - n

Suppose (k,l)R(o,p)
Then k + p = l + o
k + p = (m + k-n) +o
k -k + p = m -n + o
p + n = m + o
aRc -->(n,m)R(o,p)
Therefore, R is Transitive

**jokke22** · April 26th 2009, 12:33 PM

Originally Posted by spearfish

I think you were referring to jokke22, lol. I was just helping. Anyways, here are the others:

SYMMETRIC:
Suppose (n,m)R(k,l)
Then n + l = m +k
n = m + k -l

Suppose (k,l) R (n,m)
Then k + m = l + n
k + m = l + (m+k-l)
k+ m = m+ k

Therefore, R is Symmetric

TRANSITIVE: a = (n,m), b = (k,l), c = (o,p)
Suppose (n,m)R(k,l)
Then n + l = m + k
l = m + k - n

Suppose (k,l)R(o,p)
Then k + p = l + o
k + p = (m + k-n) +o
k -k + p = m -n + o
p + n = m + o
aRc -->(n,m)R(o,p)
Therefore, R is Transitive

I just see a whole bunch of letters now...

I will look everything over tomorrow and see if I get this!

Thank you though! Appriciate all the help I can get! Kudos to all the math geniuses here!

**jokke22** · April 27th 2009, 09:54 AM

To continue, how is this proof actually set;

Function is defined as followed:
⊕: $\mathbb{N}^2 * \mathbb{N}\Rightarrow\mathbb{N}$
by
$(n,m)$ ⊕ $(k,l) = (n+k, m+l)$

Show that,
$(n,m)$ ~ $(n_1,m_1)$ and $(k,l)$ ~ $(k_1,l_1)$
then
$(n,m)$ ⊕ $(k,l)$ ~ $(n_1,m_1)$ ⊕ $(k_1,l_1)$

Finding it hard to "prove", though I understand it to some level.

Hmmm...
Will post if I find a reasonable answer, so any pointers would be greatly appriciated!

**Danneedshelp** · April 27th 2009, 11:15 AM

Let (n,m),(k,l),and (x,y) be arbitrary elements of N^2. Now, suppose (n,m)~(k,l) and (k,l)~(x,y). By our definition this implies n+l=m+k and k+y=l+x. So, n+l=m+k=k+y=l+x ⇒ n=m=y=x ⇒ n+m=x+y. So, (n,m)~(x,y). Therefore, we have shown the relation ~ is transitive.

**jokke22** · April 27th 2009, 11:29 AM

Originally Posted by Danneedshelp

Let (n,m),(k,l),and (x,y) be arbitrary elements of N^2. Now, suppose (n,m)~(k,l) and (k,l)~(x,y). By our definition this implies n+l=m+k and k+y=l+x. So, n+l=m+k=k+y=l+x ⇒ n=m=y=x ⇒ n+m=x+y. So, (n,m)~(x,y). Therefore, we have shown the relation ~ is transitive.

**jokke22** · April 27th 2009, 12:05 PM

Originally Posted by Danneedshelp

Let (n,m),(k,l),and (x,y) be arbitrary elements of N^2. Now, suppose (n,m)~(k,l) and (k,l)~(x,y). By our definition this implies n+l=m+k and k+y=l+x. So, n+l=m+k=k+y=l+x ⇒ n=m=y=x ⇒ n+m=x+y. So, (n,m)~(x,y). Therefore, we have shown the relation ~ is transitive.

This shows that this

~

relation is transitive...or?

Confused here...

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