Here are the three:
Fill in details.
Hi!
I have given av relation ~ by where:
~
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
Symmetric
Transative if
Thanks in advance for any help/tips!
*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
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 ;-)
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
To continue, how is this proof actually set;
Function is defined as followed:
⊕:
by
⊕
Show that,
~ and ~
then
⊕ ~ ⊕
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!
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.