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0 is a binary operation not zero

use \\ in latex for new line

\[f _{6}(x)=\frac{ 1 }{ x }\]} this is the lst element

hmm ... f1 is the identity element.

unfortunately, the last element ... has inverse as itself.

from there i did this
\[(f _{7}0f _{1})(x)=f _{1}(x)\] and \[(f _{1} 0f _{7})(x)=f _{1}(x)\]

on the first part i got \[f _{7}(x)=x\]
but the second part i don't know how to do it

huh?? how did you do that??

\[(f _{7}0f _{1})(x)=f _{1}(x)\]
then \[f _{7}(f _{1}(x))=f _{1}(x)\]
\[f _{7}(x)=x\]

hmm .. what is f7, where is f7 ??

i let it to be element of T because by the definition of identity we have
e*a=a*e=a

isn't |dw:1362683941681:dw|

i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer

and we know that identity element is unique

there isn't any other identity element on the group.

I think you are not getting the meaning|dw:1362684677725:dw| of identity

|dw:1362684807109:dw|

hell ... the page refreshed and my work is gone.

|dw:1362685396384:dw|

looks like i made some mistake.

so if is like this how wat am i gonna say about the inverse

actually i mean how to write them

could you check those stuffs?

if inverse exist, then it must be unique.

the inverse not always is unique but the identity must always be unique

the identity is unique on this case ... f(x) = x is the only identity element.

the inverse is given by
a*b=b*a=e

where e is the identity we have found

i think this is not a group

why do you think it's not a group?

ok

I am getting the same result as before
have you verified associativity??

yes it holds

what a fool i've been ...
f3, f5, f6 <-- they all are self inverse.

this is a group.

can u show them pls

|dw:1362690149777:dw|
these element are inverses of themselves.

ok now i see it is a group

most likely yes ... as always I am not sure.