Prove that the set of all integers forms a group under addition.
Answers
Answer:
To show that the integers, Z together with usual addition form a group, you just need to check that the 4 properties* (or axioms, if you like) of a group are satisfied:
There exists an identity element in your group that fixes every element under the given binary operation.
Yes, the number 0 is the identity since 0+k=k+0=k, ∀k∈Z
Closure: given any two elements a and b in the set, we need to show that a+b stays in the set.
Yes, given any two integers a and b, their sum a+b is again an integer.
Associativity of the given operation.
Yes, addition in Z is associative i.e. (a+b)+c=a+(b+c)
Existence of inverses: Given any element a in the set, we need to find another element (call it "a inverse") such that a and its inverse commute and their operation together give the identity element.
Yes, given a∈Z, then −a is the inverse since a+(−a)=(−a)+a=0.
Since Z with addition satisfies 1-4 above, it is a group. Done.
*These 4 can be economized into just 3 properties, depending on which book you may be using.
In general, if you want to prove that some other structure is what you think it should be (e.g. a ring, a topological space, a σ-algebra etc.) then the usual method of proof is to simply show that it satisfies each of the defining properties. Just work through and check off a list, just as I've done above.
Hope this helps.