walters
  • walters
Abstract Algebra help
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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walters
  • walters
\[T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 },f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\}\] show that T is a group or not i've shown that 0 is associative on T can u pls help on showing the identity and the inverse
walters
  • walters
0 is a binary operation not zero
walters
  • walters
@experimentX

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experimentX
  • experimentX
use \\ in latex for new line
walters
  • walters
\[f _{6}(x)=\frac{ 1 }{ x }\]} this is the lst element
experimentX
  • experimentX
\[ T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 }, \\ f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\} \]
experimentX
  • experimentX
hmm ... f1 is the identity element.
walters
  • walters
let \[f _{7}(x) \in T\] be the identity with respect to 0,then \[(f _{7}0f _{1})(x)=(f _{1}0f _{7})(x)=f _{1}(x)\]
experimentX
  • experimentX
unfortunately, the last element ... has inverse as itself.
walters
  • walters
from there i did this \[(f _{7}0f _{1})(x)=f _{1}(x)\] and \[(f _{1} 0f _{7})(x)=f _{1}(x)\]
walters
  • walters
on the first part i got \[f _{7}(x)=x\] but the second part i don't know how to do it
experimentX
  • experimentX
huh?? how did you do that??
walters
  • walters
\[(f _{7}0f _{1})(x)=f _{1}(x)\] then \[f _{7}(f _{1}(x))=f _{1}(x)\] \[f _{7}(x)=x\]
experimentX
  • experimentX
hmm .. what is f7, where is f7 ??
walters
  • walters
i let it to be element of T because by the definition of identity we have e*a=a*e=a
experimentX
  • experimentX
isn't |dw:1362683941681:dw|
walters
  • walters
i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer
walters
  • walters
and we know that identity element is unique
experimentX
  • experimentX
there isn't any other identity element on the group.
walters
  • walters
if i am using f7 and f2 i have |dw:1362684306098:dw| how can i simplify it until i get x as my identity
experimentX
  • experimentX
I think you are not getting the meaning|dw:1362684677725:dw| of identity
experimentX
  • experimentX
|dw:1362684807109:dw|
experimentX
  • experimentX
hell ... the page refreshed and my work is gone.
experimentX
  • experimentX
|dw:1362685396384:dw|
experimentX
  • experimentX
looks like i made some mistake.
walters
  • walters
so if is like this how wat am i gonna say about the inverse
walters
  • walters
actually i mean how to write them
experimentX
  • experimentX
could you check those stuffs?
experimentX
  • experimentX
if inverse exist, then it must be unique.
walters
  • walters
the inverse not always is unique but the identity must always be unique
experimentX
  • experimentX
the identity is unique on this case ... f(x) = x is the only identity element.
walters
  • walters
the inverse is given by a*b=b*a=e
walters
  • walters
where e is the identity we have found
walters
  • walters
i think this is not a group
experimentX
  • experimentX
why do you think it's not a group?
walters
  • walters
the only way to get the inverse of a certain element u must multiply two elements and this does not clearly tell us the inverse of a function.As we know that for every function wen we use the operation combining the function and its inverse we must get the identity
experimentX
  • experimentX
here, the operation is not multiplication ... but i think there is no unique inverser. But I am not sure ... let me do this seriously on copy.
walters
  • walters
ok
experimentX
  • experimentX
I am getting the same result as before have you verified associativity??
walters
  • walters
yes it holds
walters
  • walters
i think wat u did is similar wen trying to show that 0 is a binary operation ,because wat u have as the inverse i have them at my fisrt condition that one of checking indeed the our operation hold
experimentX
  • experimentX
what a fool i've been ... f3, f5, f6 <-- they all are self inverse.
experimentX
  • experimentX
this is a group.
walters
  • walters
can u show them pls
experimentX
  • experimentX
assuming associativity holds for all operations define your idenity by \( f_1(x) = x \) ... in any case you won't get \( f_x f_y = f_x \) if y \( \neq 1\)
experimentX
  • experimentX
for every operation this is closed. earliser I showed, f2 and f4 are inverses of each other.|dw:1362690024527:dw|
experimentX
  • experimentX
|dw:1362690149777:dw| these element are inverses of themselves.
walters
  • walters
ok now i see it is a group
experimentX
  • experimentX
most likely yes ... as always I am not sure.

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