Is the set of integers Z={0,+-1,+-2, } .

  • Thread starter Thread starter florenti
  • Start date Start date
  • Tags Tags
    Integers Set
florenti
Messages
7
Reaction score
0
Is the set of integers Z={0,+-1,+-2,...}...

Hi,

Can anybody help me.

I know that integers under Addition is a group, but

Is the set of Integers Z={0,+-1,+-2,...} together with the operation of subtraction a (noncommutative) group.

Thanks a lot

Flor
 
Physics news on Phys.org


Well, have you tried anything at all?

In order to be a group, a set with an operation must satisfy these rules:

1) There is an "identity". Here that means there must be some integer, e, such that x- e= x and e- x= x for any integer x. Is there such a number?

2) If there is an identity, e, then every integer must have an "inverse", "-x", such that x- (-x)= e. Is there such a number?

3) The operation must be associative. For any three integers a, b, c, (a-b)- c must be equal to a-(b-c). Is that true?
 
Welcome to PF!

florenti said:
I know that integers under Addition is a group, but

Is the set of Integers Z={0,+-1,+-2,...} together with the operation of subtraction a (noncommutative) group.

Hi Flor! Welcome to PF! :smile:

A group must have an inverse of every element, so the non-negative integers are not a group (I think :redface: they're a "semi-group").

The integers Z={0,+-1,+-2,...} are a commutative group …

commutative just means that, for example, 7 + 11 = 11 + 7. :smile:

(and you don't need to mention the operation of subtraction … subtraction is the inverse of addition, so it has to be in the group anyway.)
 


Thanks a lot,

So the set of integers with operation of subtraction is not a group because the subtraction of integers is not associative.So a-(b-c) is not equal to (a-b)-c.

Thanks a lot,

florent
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top