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Mathematics 26 Online
OpenStudy (anonymous):

Prove the following is symmetric. Let a ~ b on "all rational numbers" iff a-b=m for some integer m.

OpenStudy (anonymous):

I know it is set up as If a~b, then b~ a. But I am stuck on setting up the proof.

OpenStudy (kainui):

Hmm maybe I am misunderstanding, since it seems like this should be antisymmetric because: a~b iff a-b=m but then we can factor out a negative sign to get: (a-b)=-(b-a) so we have b-a=-m so it seems like a~b = -(b~a)

OpenStudy (anonymous):

I fairly sure it is symmetric because i cannot determine any counter examples

OpenStudy (anonymous):

I just don't remember how to set up proofs to determine symmetry.

OpenStudy (bobo-i-bo):

You have misunderstood. We do not require that b-a = m = a - b for b~a when a~b since we only require that b-a=n, for some integer n. Your confusion is that you think that n has to equal to m.

OpenStudy (bobo-i-bo):

To start your proof: Let a~b. Then there exists an integer m such that a-b=m. (Now you have to prove that b-a is an integer and so therefore b~a).

OpenStudy (anonymous):

OK...

OpenStudy (anonymous):

Sorry totally lost on the process. I am OK on reflexive and transitive, but writing proofs for symmetry has me stumped.

OpenStudy (anonymous):

Since a and b are integers then b-a is also an integer.

OpenStudy (anonymous):

Therefore b ~ a

OpenStudy (anonymous):

Would this be correct?

OpenStudy (bobo-i-bo):

No, since a and b are not necessarily integers, as they could be rational numbers

OpenStudy (anonymous):

Open eyes and now I see...the problem is it is Not symmetric and i can find a counter example. thanks for your help.

OpenStudy (bobo-i-bo):

Nono, a~b DOES imply b~a :P

OpenStudy (bobo-i-bo):

So, premise is that: a~b From a~b you must prove that b~a. But to prove that b~a you must prove that b-a is an integer... so basically you have to prove that if a~b, then b-a is an integer.... does that make sense? :P

OpenStudy (anonymous):

Yep..let's try this Let b-a = n for some integer n. Then b= n+a then a-(n+a) =m. Then a-n-a =m then -n = m, thus n is an integer,

OpenStudy (anonymous):

or should I add |-n| = m, then n is an integer.

OpenStudy (bobo-i-bo):

Yes, that is basically right! But you could rephrase/organise your proof a bit better... you used an assumption that n is an integer

OpenStudy (bobo-i-bo):

So if you start from the premise: a~b, then that implies that a-b=m, and therefore b-a=??? :P

OpenStudy (anonymous):

b-a = -m

OpenStudy (anonymous):

Since - m is an integer then b~a.

OpenStudy (anonymous):

Think I make things harder than they need to be.

OpenStudy (bobo-i-bo):

It's just knowing what the premises are and what conclusion you need to reach. But you got there ^_^

OpenStudy (anonymous):

Thank you for your help.

OpenStudy (bobo-i-bo):

Np :)

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