Question

I'm supposed to prove if the set Z^- (set of integers under subtraction) is :
1. Commutative
2. Associative
3. An identity
4.Are there any inversions?
5. if so what is the inverse of an arbitrary element
Please help! medal to person who helps

Answer 1

commutative means x - y = y - x

Answer 2

associative means (x - y) - z = x - (y -z)

Answer 3

how would i do it with Z^- though. My professor did Z^+ in class.. like for commutative it was: a, b is an element of Z^+, Suppose a^b = b^a is true. If a = 1, b =2, 1^2=1 does not equal 2^1 = 2. Thats all he had for commutative

Answer 4

Your note confused me.

for Z^+ 1+2 = 3 and 2+1=3 so 1+2= 2+1 is true.

for Z^- 1-2 = -1 and 2 -1 = 1 so 1-2 ≠ 2 -1 --> not commutative.

Answer 5

Ok I see that but what about exponents? How would you do a^b = b ^a in Z^-

Answer 6

you have to show both for commutative right

Answer 7

For your \( Z^+\) example, this does NOT mean integers being added ?

you are doing exponentiation ?
If so, I am do not understand exactly what is going on

Answer 8

Well he said prove for the set Z^- which is Integers under Subtraction. So I see how you prove it with the 1-2 = 2-1 part but dont you have to check the same thing with a^b = b^a for commutative? And I thought since Z^- was under subtraction that Z^+ would be addition

Answer 9

I am interpreting a ^ b to mean a operator b (not exponentiation)

and that \(Z^-\) means the integers are being operated on by -

but if that is the case, \(Z^+\) would mean integers being operated on by +, which we know will be commutative and associative. But your example seems to show that \(Z^+\) is not commutative under +

Answer 10

yes that is correct it isnt

Answer 11

he showed us in class that it wasn't. but how would i do the exponentiation for commutative under Z^-

Answer 12

I don't know what is going on here.

Answer 13

I am getting the idea that ^ is the operator, and Z are the integers. what exactly does
Z^+ mean ? what does the + mean ?

Answer 14

yes, but what does it mean ?

Answer 15

do you have a link to any background on this question ?

Answer 16

no it was done in class. Z+ means set of posititve integers

Answer 17

but you said
Z^- (set of integers under subtraction)

which is different from Z- means set of negative integers

Answer 18

how would subtraction look then. Maybe my professor messed up on the symbols

Answer 19

do you have any more notes on what he did for Z^+ ?

Answer 20

yes

Answer 21

if you post them, maybe I can make sense of them ?

Answer 22

Ex: State properties of the following binary compositions:
-Exponentiation on the set of positive integers.
Exponentiation on Z^+
a*b = a^b
1. Commutative: a * b = b * a , a^b = b^a
Counter example: a,b are elements of Z^+
Suppose a^b = b^a is true.
If a = 1, b = 2, then 1^2 = 1 does not equal 2^1 = 2 so its not commutative
2. Associative?
Not associative:
COunter-example: (a^b)^c = a^(b^c)
Let a = 2, b = 3, c = 2
2^(3)^2 = 2(3^2)
128 does not equal 512

3. Is it an identity?
Is a = a^e = e^a ?
e = 1
a = a^1 does not equal 1^a
Therefore no identity

And he never showed how to do inverse

Answer 23

ok, that makes more sense. So I think you should have posted your problem this way:

State properties of the following binary compositions:
-Exponentiation on the set of negative integers.

Answer 24

Though (just possibly) the the problem is
State properties of the following binary compositions:
-Subtraction on the set of integers.

Answer 25

can you show me how to do it both ways?

Answer 26

it makes much more sense to do
set of integers under subtraction (which means use all integers and - as the operator)
if a and b are in Z, a - b will also be in Z, so at least it is closed. we can now ask, is this operator commutative, associative, etc.

Answer 27

Exponentiation on negative integers has problems.
for example
-2^-2 means 1/(-2)^2 = 1/4 which is neither negative, nor an integer
so it is not closed... and it does not make sense to check for the other properties.

but if we did... -1^-2 ≠ -2^(-1) so not commutative

Answer 28

for Integers under the operation subtraction you should find
1) not commutative 1 -2 ≠ 2 -1
2) not associative (1 - 2) -3 ≠ 1 - (2 - 3)
3) has an identity: a - 0 = a for all a
4) inverse: a - a = 0 the inverse of a is itself

Answer 29

Let me know how this turns out.

Answer 30

Thanks Phi i get everything but 3 and 4. For 3, did u choose a-0 because identity is 0 and - because its Z^-? Also for 4) how is the inverse of a itself?

Answer 31

I am assuming the operator is subtraction
so given two members "a" and "b" of Z (the integers), we compute a - b
the identity element 0 means a - 0 = a
(though I think a true identity commutes: 0 - a = a should also be true... so I am not totally sure if 0 qualifies as an identity)

Answer 32

Inverse is defined as
For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element.

in this case, if we have 0 be the identity element, then for a - b =0
b must be a
in other words a is its own inverse.

Answer 33

see https://en.wikipedia.org/wiki/Group_(mathematics)#Definition
for how these definitions are used to define a GROUP

Answer 34

ok thank you so much! this helped a lot