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

This Question is Closed

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0they prob mean \(Z_m=\{0,1,2,\ldots,m1\}\)

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0the set of integers mod \(m\)

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah!!!! i need complete sentences to explain it... do u have any stuff???

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1show that it satisfies the property to be called a group.

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0Exactly!!! can i get that proof or its details... i mean any link???

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1addition operation is associative, and find the identity element of Zm

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Edit:: \[ (a+b) \mod m \in Z_m \]

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

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0What 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1huh!! what exactly is the question?

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0I wanna define it with examples!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1aren't those two different distinct groups?

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0u mean the addition modulo and the other???

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

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0Thats it...what is confusing u??

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1you are mixing two things ... doing both of them at same time.

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

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1but why do you want to give example?

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1choose 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.

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0then is my answer wrong????

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

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0is 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry ... i was busy!!

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0if possible make some alterations and input the needed stuffs... and edit my answer...pls:)

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

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0shud i show associativity,identity element and inverse element????

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yes!! you need to show them!!

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1same ... show that (a+b) mod m is in the set!!.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1instead of multiplying, we add ,,, since operation is +.

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Once 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

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0similar to that??? r u sure???

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1+ here is + modulo m, ok

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Okay, 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, +>.

terenzreignz
 3 years ago
Best 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).

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1show that for a, ma is the it's inverse, and ma is also in the set.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1@experimentX A spoiler alert, next time? LOL JK

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Never 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)

Zaara
 3 years ago
Best ResponseYou've already chosen the best response.0why ma??? i dnt get it...

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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...

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Okay, 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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1And 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)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1I 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
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