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