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

walters
 one year ago
Best ResponseYou've already chosen the best response.10 is a binary operation not zero

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

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

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

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

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

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1on the first part i got \[f _{7}(x)=x\] but the second part i don't know how to do it

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

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

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1i let it to be element of T because by the definition of identity we have e*a=a*e=a

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer

walters
 one year ago
Best ResponseYou've already chosen the best response.1and we know that identity element is unique

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1if i am using f7 and f2 i have dw:1362684306098:dw how can i simplify it until i get x as my identity

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

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

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

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

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1so if is like this how wat am i gonna say about the inverse

walters
 one year ago
Best ResponseYou've already chosen the best response.1actually i mean how to write them

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

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1the inverse not always is unique but the identity must always be unique

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1the inverse is given by a*b=b*a=e

walters
 one year ago
Best ResponseYou've already chosen the best response.1where e is the identity we have found

walters
 one year ago
Best ResponseYou've already chosen the best response.1i think this is not a group

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2I am getting the same result as before have you verified associativity??

walters
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.2what a fool i've been ... f3, f5, f6 < they all are self inverse.

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

walters
 one year ago
Best ResponseYou've already chosen the best response.1ok now i see it is a group

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