anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
commutative means x - y = y - x
phi
  • phi
associative means (x - y) - z = x - (y -z)
anonymous
  • anonymous
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

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phi
  • phi
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.
anonymous
  • anonymous
Ok I see that but what about exponents? How would you do a^b = b ^a in Z^-
anonymous
  • anonymous
you have to show both for commutative right
phi
  • phi
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
anonymous
  • anonymous
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
phi
  • phi
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 +
anonymous
  • anonymous
yes that is correct it isnt
anonymous
  • anonymous
he showed us in class that it wasn't. but how would i do the exponentiation for commutative under Z^-
phi
  • phi
I don't know what is going on here.
phi
  • phi
I am getting the idea that ^ is the operator, and Z are the integers. what exactly does Z^+ mean ? what does the + mean ?
anonymous
  • anonymous
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phi
  • phi
yes, but what does it mean ?
phi
  • phi
do you have a link to any background on this question ?
anonymous
  • anonymous
no it was done in class. Z+ means set of posititve integers
phi
  • phi
but you said Z^- (set of integers under subtraction) which is different from Z- means set of negative integers
anonymous
  • anonymous
how would subtraction look then. Maybe my professor messed up on the symbols
phi
  • phi
do you have any more notes on what he did for Z^+ ?
anonymous
  • anonymous
yes
phi
  • phi
if you post them, maybe I can make sense of them ?
anonymous
  • anonymous
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
phi
  • phi
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.
phi
  • phi
Though (just possibly) the the problem is State properties of the following binary compositions: -Subtraction on the set of integers.
anonymous
  • anonymous
can you show me how to do it both ways?
phi
  • phi
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.
phi
  • phi
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
phi
  • phi
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
phi
  • phi
Let me know how this turns out.
anonymous
  • anonymous
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?
phi
  • phi
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)
phi
  • phi
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.
phi
  • phi
see https://en.wikipedia.org/wiki/Group_(mathematics)#Definition for how these definitions are used to define a GROUP
anonymous
  • anonymous
ok thank you so much! this helped a lot

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