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walters Group TitleBest 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
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.1
0 is a binary operation not zero
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.1
@experimentX
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

walters Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

walters Group TitleBest ResponseYou've already chosen the best response.1
yes it holds
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.2
this is a group.
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.1
can u show them pls
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
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 \( \neq 1\)
 one year ago

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

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

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

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