# Help with proofs in general

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I'm new to proofs and I'm not sure from which assumptions one has to start with in a proof. I'm trying to prove the generalized associative law for groups and if I start with the axioms of a group as the assumptions then I already have the proof.

From what basic assumptions should one start with in proving something? I'm thinking of starting with the definition of an equivalence relation. Would that even help with proving something like (a R b) R c = a R (b R c) for all a,b,c $\in \mathbb{G}$?

fresh_42
Mentor
2021 Award
I'm new to proofs and I'm not sure from which assumptions one has to start with in a proof. I'm trying to prove the generalized associative law for groups and if I start with the axioms of a group as the assumptions then I already have the proof.

From what basic assumptions should one start with in proving something? I'm thinking of starting with the definition of an equivalence relation. Would that even help with proving something like (a R b) R c = a R (b R c) for all a,b,c $\in \mathbb{G}$?
What is "the generalized associative law" and how is "group" defined? There are basically two different approaches: as an enumeration of properties of the group operation (neutral element, inverse element and so on) or as the solutions of certain equations like ##xa=b##.

This means: it is always helpful to state exactly what is known (or given) and what has to be shown. The more precise these statements are the easier it is to find a proof. Of course not always, but often. In any case it helps reading a proof and for sure the debate on a platform like PF.

Well a group is defined as an ordered pair of a binary relation R and the set $\mathbb{G}$ as (R,$\mathbb{G}$) and the generalized associative law is that $a_1 ~R~ a_2 ~R ~a_3~ R ...R~ a_n \in \mathbb{G}$ this is independent of how you place brackets around the elements.

If I can prove this for a few elements in the group then I can do this with all of them through induction right? Without using the definition of a group (having to be associative, there must exist an identity element, and there must exist an inverse for each element) how can I do this?

Actually, It is quite easy to do this if we assume the relation is equivalent (since symmetry means commutativity and proving associativeness follows very quickly) but under the definition of a group it doesn't have to be equivalent so I have to find another way to prove the associative property darn

fresh_42
Mentor
2021 Award
Well, you aren't allowed to use commutativity since non Abelian groups also obey the associative law. But induction sounds good.
In this case it is really more an exercise in precise wording than it is to find the proof.

chiro
For proof's I'd recommend looking at it from an intuitive view before trying to formalize things.

Language is often suited to ones intuition and mathematics is no exception to that rule.

If you can get the intuition (which may draw from all sorts of visual examples and physical intuition) then the formality will get easier.

The formalities are necessary to make everything precise and consistent - but the intuition of ideas, concepts and information should be a primary point - not a secondary one in my opinion.

Also - what mathematicians (and those who use mathematics) do is to learn how to combine information to get to the resolution they are intending to make (not so much the answer they are expecting - but the resolution in regard to that answer).

Learning that takes a bit of practice - but it can be done and has been done by many who have studied and worked at the subject of mathematics and its applications.

Stephen Tashi
If I can prove this for a few elements in the group then I can do this with all of them through induction right?

The technique wouldn't be called "induction".