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Zaara
Group Theory: any one pls guide me with a link to find the definition of these......... 1.(Zm,+) 2.(Zm,*)
they prob mean \(Z_m=\{0,1,2,\ldots,m-1\}\)
the set of integers mod \(m\)
Yeah!!!! i need complete sentences to explain it... do u have any stuff???
show that it satisfies the property to be called a group.
Exactly!!! can i get that proof or its details... i mean any link???
let 'a' and 'b' be two elements of Zm, show that it is closed in +, ie, \( (a+b) \mod b \in Z_m \)
addition operation is associative, and find the identity element of Zm
Edit:: \[ (a+b) \mod m \in Z_m \]
this is always true becuse \( (a+b) \mod m = c < m, c \in Z\m \)
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
huh!! what exactly is the question?
I wanna define it with examples!
aren't those two different distinct groups?
u mean the addition modulo and the other???
yep!! just this one .. (Zm,+), this is a group.
Thats it...what is confusing u??
you are mixing two things ... doing both of them at same time.
where did i mix them?? i have defined Zm,then defined addition and multiplication modulo and have given example 4 each...
let's work out one at each ... we do with (Zm,+) first.
ohhh... u mean i have not seperately done it..ryt??? ok got it.... i have jst given it together for my ease....
but why do you want to give example?
definition with example is most expected... so that one can easily understand it....
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.
then is my answer wrong????
no i don't think so ... you put stuff in one place, and it hurt my eyes.
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
sorry ... i was busy!!
if possible make some alterations and input the needed stuffs... and edit my answer...pls:)
okay .. you showed that it is closed. you need to test for these http://en.wikipedia.org/wiki/Group_(mathematics)#Definition
shud i show associativity,identity element and inverse element????
yes!! you need to show them!!
bt in the link u gave... they have shown for multiplication modulo only.... what can i do for addition modulo???????
same ... show that (a+b) mod m is in the set!!.
instead of multiplying, we add ,,, since operation is +.
Never seen a Group Theory question before... this should be fun... :D
Once you've shown closure, next step is to show that there is an element e of Zm (a non-negative integer less than m) such that for any k in Zm e + k = k + e = k
similar to that??? r u sure???
+ here is + modulo m, ok
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 non-negative 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, +>.
-.- 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).
show that for a, m-a is the it's inverse, and m-a is also in the set.
@experimentX A spoiler alert, next time? LOL JK
Never mind, lol... Anyway, let a be in Zm. Your task is to show that an element m-a exists, and m-a is in Zm, and a + (m - a) = 0, where "+" here is + (mod m)
why m-a??? i dnt get it...
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...
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 m-a. since a is in Zm, a < m a - a < m - a m - a > 0, meaning m-a is non-negative, (positive, even) m < m + a m - a < m + a - a m - a < m, meaning m-a is less than m. Thus, m-a is non-negative and is less than m, therefore m-a is in Zm. Consider a + (m-a) = a + m - a = m = 0(mod m) Thus, a + (m-a) = 0(mod m) a + (m-a) = 0, in Zm (m-a) is the inverse of a. Thus we have shown that for any element a in Zm, there is an element (m-a) such that a + (m-a) = 0, and therefore, all elements in <Zm, +> have inverses :D
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)
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