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Zaara
Group Title
Group Theory:
any one pls guide me with a link to find the definition of these.........
1.(Zm,+)
2.(Zm,*)
 one year ago
 one year ago
Zaara Group Title
Group Theory: any one pls guide me with a link to find the definition of these......... 1.(Zm,+) 2.(Zm,*)
 one year ago
 one year ago

This Question is Closed

experimentX Group TitleBest ResponseYou've already chosen the best response.1
what is Zm?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
they prob mean \(Z_m=\{0,1,2,\ldots,m1\}\)
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
the set of integers mod \(m\)
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
Yeah!!!! i need complete sentences to explain it... do u have any stuff???
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
show that it satisfies the property to be called a group.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
Exactly!!! can i get that proof or its details... i mean any link???
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let 'a' and 'b' be two elements of Zm, show that it is closed in +, ie, \( (a+b) \mod b \in Z_m \)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
addition operation is associative, and find the identity element of Zm
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Edit:: \[ (a+b) \mod m \in Z_m \]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this is always true becuse \( (a+b) \mod m = c < m, c \in Z\m \)
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
What i have written below is correct right?? or should there be any change????? Definitions: Let Zm be the set of non negative integers less than m: {0,1, ., m−1} The operation +m is defined as a +m b = (a + b) mod m. This is addition modulo m. The operation ∙m is defined as a ∙m b = (a + b) mod m. This is multiplication modulo m. Using these operations is said to be doing arithmetic modulo m. Example: Find 7 +11 9 and 7 ·11 9. Solution: Using the definitions above: – 7 +11 9 = (7 + 9) mod11 = 16 mod11 = 5 – 7 ·11 9 = (7 ∙ 9) mod11 = 63 mod 11 = 8
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
huh!! what exactly is the question?
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
I wanna define it with examples!
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
aren't those two different distinct groups?
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
u mean the addition modulo and the other???
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yep!! just this one .. (Zm,+), this is a group.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
Thats it...what is confusing u??
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
you are mixing two things ... doing both of them at same time.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
where did i mix them?? i have defined Zm,then defined addition and multiplication modulo and have given example 4 each...
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let's work out one at each ... we do with (Zm,+) first.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
ohhh... u mean i have not seperately done it..ryt??? ok got it.... i have jst given it together for my ease....
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
but why do you want to give example?
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
definition with example is most expected... so that one can easily understand it....
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
choose m=10, or 5 for sake of ease. then {0,1,2,3,4} are the elements of group!! show that it remains closed, and is associative and has identity '0', and has inverse.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
then is my answer wrong????
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
no i don't think so ... you put stuff in one place, and it hurt my eyes.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
is it ok now??? Let Zm be the set of nonnegative integers less than m: {0,1, ., m−1} 1)The operation +m is defined as a +m b = (a + b) mod m. This is addition modulo m.(Zm,+) Example: Find 7 +11 9 Solution: Using the definitions above: 7 +11 9 = (7 + 9) mod11 = 16 mod11 = 5 2)The operation ∙m is defined as a ∙m b = (a + b) mod m. This is multiplication modulo m. Example: Find 7 ·11 9. Solution: Using the definitions above: 7 ·11 9 = (7 ∙ 9) mod11 = 63 mod 11 = 8
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
@experimentX
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
sorry ... i was busy!!
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
if possible make some alterations and input the needed stuffs... and edit my answer...pls:)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
okay .. you showed that it is closed. you need to test for these http://en.wikipedia.org/wiki/Group_(mathematics)#Definition
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
shud i show associativity,identity element and inverse element????
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
@experimentX
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yes!! you need to show them!!
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
bt in the link u gave... they have shown for multiplication modulo only.... what can i do for addition modulo???????
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
same ... show that (a+b) mod m is in the set!!.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
instead of multiplying, we add ,,, since operation is +.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Never seen a Group Theory question before... this should be fun... :D
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Once you've shown closure, next step is to show that there is an element e of Zm (a nonnegative integer less than m) such that for any k in Zm e + k = k + e = k
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
similar to that??? r u sure???
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
+ here is + modulo m, ok
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Quite sure.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Okay, I'll do this (because it's easier :P) and you do the "existence of inverse" part. Let k be in Zm Consider 0. 0 is a nonnegative integer. Since m is positive, 0 < m So 0 is in Zm 0 + k = 0 + k (mod m) = k(mod m) = k, since k is in Zm (and is therefore a nonnegative integer less than m). Similarly, k + 0 = k + 0 (mod m) = k(mod m) = k. Thus, we have shown that 0 + k = k + 0 = k Hence, 0 is the identity element in <Zm, +>.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
what abt inverse???
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
. Here's the gist of it. Consider an arbitrary element of Zm, and show that there is something that you can add (modulo m) to it so that the sum is 0 (the additive identity).
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
show that for a, ma is the it's inverse, and ma is also in the set.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
@experimentX A spoiler alert, next time? LOL JK
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
well .. sorry!!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Never mind, lol... Anyway, let a be in Zm. Your task is to show that an element ma exists, and ma is in Zm, and a + (m  a) = 0, where "+" here is + (mod m)
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
why ma??? i dnt get it...
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Well, remember that in the world of Zm, every multiple of m is basically zero. So, when I said "what do you need to add to a so that you get zero", I really mean what do you need to add to a to get m...
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Okay, first of all, the inverse of the additive identity is itself... 0 + 0 = 0, so, yeah... Now, let a be in Zm, where a is not equal to zero. Consider ma. since a is in Zm, a < m a  a < m  a m  a > 0, meaning ma is nonnegative, (positive, even) m < m + a m  a < m + a  a m  a < m, meaning ma is less than m. Thus, ma is nonnegative and is less than m, therefore ma is in Zm. Consider a + (ma) = a + m  a = m = 0(mod m) Thus, a + (ma) = 0(mod m) a + (ma) = 0, in Zm (ma) is the inverse of a. Thus we have shown that for any element a in Zm, there is an element (ma) such that a + (ma) = 0, and therefore, all elements in <Zm, +> have inverses :D
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
And there you go, we've shown closure, associativity, existence of an identity element and existence of inverses, we can conclude that <Zm, +(mod m)> forms a group. (An Abelian group, too, but more on some other time, I guess)
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.0
great!!!!!!!!!!!!!!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
I have to go now, work on the other one: <Zm, *(mod m)> But since we're having spoilers anyway, it's not a group :P The reason for that is for you to find... *cough*oneisthemultiplicativeidentity*cough*theresnothingyoucanmultiplytozerotogetone *cough*zerohasnomultiplicativeinverse *cough* LOL anyway Terence out :D
 one year ago
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