Question

do intergers form a group? explain

Answer 1

no u can't find any integer b for an integer a such that a*b=1
except for 1

Answer 2

thats it?

Answer 3

I can not write you the formal proof. But you just need to make sure that integers fit all the axioms of a group.
Namely: \[\forall a,b,c \in Z\]
a*b is in Z
(a*b)*c=a*(b*c)
An identity element e such that e*a=a*e=a
An inverse element such that a*b=b*a=1 (or the identity for that group)

Answer 4

Where * denotes an operation (typically addition or multiplication)

Answer 5

malevolence19 .....has written correctly the definition of a group but the 3dr condition doers not hold for inverse so it can't be a group....
if a,b both are integers and the condition a*b=b*a=1 holds then b=1/a...which is a fraction...for a>1...and not belongs to the integer group....

Answer 6

Exactly. :)

Answer 7

The integers are not a group under multiplication but they are under addition. For instance:

a+b is always an integer
(a+b)+c=a+(b+c)
0+a=0+a=a
a+b=b+a=0 (i.e., b=-a)

So it does form a group under addition.

Answer 8

Dip correct me if i'm wrong D:

Answer 9

yes....u did it under addition but i did it under multiplication...thats y our ans. differs.

Answer 10

I was just making sure I wasn't telling him wrong! :D

Answer 11

but i was also correct from my point of view....

Answer 12

Yes :P

Answer 13

Both correct just different contexts