In order to be a group with respect to addition, the integers must satisfy the following axioms: 1) Closure under addition
2) Associativity of addition
3) Contains the additive identity
4) Contains the additive inverses 1) The integers are closed under addition since the sum of any two integers is an integer. 2) The integers are associative with respect to addition since (a+b)+c = a+(b+c) for any integers a, b, and c. 3) The integer 0 is the additive identity since z+0 = 0+z = z for any integer z. 4) Each integer n has an additive inverse, namely -n since n+(-n) = -n+n = 0.
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0 belongs to the reals. It is a member of the irrationals, the rationals. It is also a member of the integers; It is a member (the identity) of the group of even integers, 3*integers, 4*integers etc with respect to addition.
no
The set of positive integers does not contain the additive inverses of all but the identity. It is, therefore, not a group.
Because the set is not closed under addition. If x and y are odd, then x + y is not odd.
Integer Subsets: Group 1 = Negative integers: {... -3, -2, -1} Group 2 = neither negative nor positive integer: {0} Group 3 = Positive integers: {1, 2, 3 ...} Group 4 = Whole numbers: {0, 1, 2, 3 ...} Group 5 = Natural (counting) numbers: {1, 2, 3 ...} Note: Integers = {... -3, -2, -1, 0, 1, 2, 3 ...} In addition, there are other (infinitely (uncountable infinity) many) other subsets. For example, there is the set of even integers. There is also the subset {5,7}.
0 belongs to the reals. It is a member of the irrationals, the rationals. It is also a member of the integers; It is a member (the identity) of the group of even integers, 3*integers, 4*integers etc with respect to addition.
No. It is not a group.
Is the set of negative interferes a group under addition? Explain,
no
The set of positive integers does not contain the additive inverses of all but the identity. It is, therefore, not a group.
Yes, with respect to multiplication but not with respect to addition.
The set of integers, under addition.
Assuming that the question is in the context of the operation "addition", The set of odd numbers is not closed under addition. That is to say, if x and y are members of the set (x and y are odd) then x+y not odd and so not a member of the set. There is no identity element in the group such that x+i = i+x = x for all x in the group. The identity element under addition of integers is zero which is not a member of the set of odd numbers.
Because the set is not closed under addition. If x and y are odd, then x + y is not odd.
A cyclic group, by definition, has only one generator. An example of an infinite cyclic group is the integers with addition. This group is generated by 1.
Integer Subsets: Group 1 = Negative integers: {... -3, -2, -1} Group 2 = neither negative nor positive integer: {0} Group 3 = Positive integers: {1, 2, 3 ...} Group 4 = Whole numbers: {0, 1, 2, 3 ...} Group 5 = Natural (counting) numbers: {1, 2, 3 ...} Note: Integers = {... -3, -2, -1, 0, 1, 2, 3 ...} In addition, there are other (infinitely (uncountable infinity) many) other subsets. For example, there is the set of even integers. There is also the subset {5,7}.
Say we have a group G, and some subgroup H. The number of cosets of H in G is called the index of H in G. This is written [G:H].If G and H are finite, [G:H] is just |G|/|H|.What if they are infinite? Here is an example. Let G be the integers under addition. Let H be the even integers under addition, a subgroup. The cosets of H in G are H and H+1. H+1 is the set of all even integers + 1, so the set of all odd integers. Here we have partitioned the integers into two cosets, even and odd integers. So [G:H] is 2.