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anonymous
 3 years ago
Abstract Algebra help
anonymous
 3 years ago
Abstract Algebra help

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1x },f _{3}(x)\frac{ x }{ x1 },f _{4}(x)=\frac{ x1 }{ x },f _{5}(x)=1x,f _{6}(x)=\frac{ 1 }{ x }\right\}\] show that T is a group or not <T,0> i've shown that 0 is associative on T can u pls help on showing the identity and the inverse

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.00 is a binary operation not zero

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2use \\ in latex for new line

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f _{6}(x)=\frac{ 1 }{ x }\]} this is the lst element

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2\[ T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1x },f _{3}(x)\frac{ x }{ x1 }, \\ f _{4}(x)=\frac{ x1 }{ x },f _{5}(x)=1x,f _{6}(x)=\frac{ 1 }{ x }\right\} \]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2hmm ... f1 is the identity element.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let \[f _{7}(x) \in T\] be the identity with respect to 0,then \[(f _{7}0f _{1})(x)=(f _{1}0f _{7})(x)=f _{1}(x)\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2unfortunately, the last element ... has inverse as itself.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from there i did this \[(f _{7}0f _{1})(x)=f _{1}(x)\] and \[(f _{1} 0f _{7})(x)=f _{1}(x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0on the first part i got \[f _{7}(x)=x\] but the second part i don't know how to do it

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2huh?? how did you do that??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(f _{7}0f _{1})(x)=f _{1}(x)\] then \[f _{7}(f _{1}(x))=f _{1}(x)\] \[f _{7}(x)=x\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2hmm .. what is f7, where is f7 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i let it to be element of T because by the definition of identity we have e*a=a*e=a

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2isn't dw:1362683941681:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and we know that identity element is unique

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2there isn't any other identity element on the group.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if i am using f7 and f2 i have dw:1362684306098:dw how can i simplify it until i get x as my identity

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2I think you are not getting the meaningdw:1362684677725:dw of identity

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1362684807109:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2hell ... the page refreshed and my work is gone.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1362685396384:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2looks like i made some mistake.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if is like this how wat am i gonna say about the inverse

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually i mean how to write them

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2could you check those stuffs?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2if inverse exist, then it must be unique.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the inverse not always is unique but the identity must always be unique

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2the identity is unique on this case ... f(x) = x is the only identity element.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the inverse is given by a*b=b*a=e

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where e is the identity we have found

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think this is not a group

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2why do you think it's not a group?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.2here, 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2I am getting the same result as before have you verified associativity??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.2what a fool i've been ... f3, f5, f6 < they all are self inverse.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2assuming 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
 3 years ago
Best ResponseYou've already chosen the best response.2for every operation this is closed. earliser I showed, f2 and f4 are inverses of each other.dw:1362690024527:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1362690149777:dw these element are inverses of themselves.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok now i see it is a group

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2most likely yes ... as always I am not sure.
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