Let us consider the common combinatorics problem centered around a group of six people . Now , among six people we must have at least one triad which consists of three friends or a triad which consists of three strangers.
I was trying to find a proof to it and proceeded as follows :
Let $\ k $ denote the friendship value between two people ,which assumes value $\ f $ if they are friends and value $\ s $ if they are strangers.
Now for a tuple (a,b,c) let ( $\ K_(a,b) ,K_(b,c) ,K_(a,c) $) be used to denote their relation.
Let us assume that we cannot have a triad with all its pair having same friendship value.
Now let a,b,c,d,e,f be the six people.
let us assign the pair (a,b) the friendship value : $\ f$ . Now let us form the triads with (a,b) being one of the pairs .
(a,b,c) -> ( $\ K_(a,b) ,K_(b,c) ,K_(a,c) $) (a,b,d)-> ( $\ K_(a,b) ,K_(b,d) ,K_(a,d) $) (a,b,e) -> ( $\ K_(a,b) ,K_(b,e) ,K_(a,e) $) (a,b,f)-> ( $\ K_(a,b) ,K_(b,f) ,K_(a,f) $)
Suppose $\ K_(a,b) $ = $\ f $ . Now $\ K_(a,b) $ =$\ K_(c,b) $ =$\ K_(a,c) $ is impossible according to our assumption.
Now Let us consider the common combinatorics problem centered around a group of six people . Now , among six people we must have at least one triad which consists of three friends or a triad which consists of three strangers.
I was trying to find a proof to it and proceeded as follows :
Let $\ k $ denote the friendship value between two people ,which assumes value $\ f $ if they are friends and value $\ s $ if they are strangers.
Now for a tuple (a,b,c) let ( $\ K_(a,b) ,K_(b,c) ,K_(a,c) $) be used to denote their relation.
Let us assume that we cannot have a triad with all its pair having same friendship value.
Now let a,b,c,d,e,f be the six people.
let us assign the pair (a,b) the friendship value : $\ f$ . Now let us form the triads with (a,b) being one of the pairs .
(a,b,c) -> ( $\ K_(a,b) ,K_(b,c) ,K_(a,c) $) (a,b,d)-> ( $\ K_(a,b) ,K_(b,d) ,K_(a,d) $) (a,b,e) -> ( $\ K_(a,b) ,K_(b,e) ,K_(a,e) $) (a,b,f)-> ( $\ K_(a,b) ,K_(b,f) ,K_(a,f) $)
Suppose$\ K_(a,b) $ = $\ f $ . Now $\ K_(a,b) $ =$\ K_(c,b) $ =$\ K_(a,c) $ is impossible according to our assumption. Let's suppose $\ K_(c,a) $ =$\ f $ and hence $\ K_(b,c) $ = $\ s $
Now suppose $\ K_(a,b) $ = $\ K_(b,c) $ =$\ K_(b,d) $ .
Now let us take the triad (b,c,d) as $\ K_(b,c) $ = $\ K_(b,d) $ ,hence to keep up with our assumption $\ K_(c,d) $ can not be equal to the friendship value of other two pairs in the triad.
hence it assumes the other friendship value which in this case is $\ s $.
Without loss of generality we can say that we can't also have $\ K_(b,c) $ =$\ K_(b,d) $ = $\ K_(b,e) $ cause from the aforementioned argument it would make the friendship value of the pairs in the triad (c,d,e) same which is not permissible.
Now let us consider the pair (c,d). As $\ K_(c,b) $ = $\ K_(d,b) $ = $\ s $, hence by taking the triad (b,c,d) , we can say the friendship value of (c,d) to be $\ f $ but now $\ K_(a,c) $ = $\ K_(a,d) $ = $\ f $ and hence the friendship value of (c,d) turns out to be $\ s $, which is a contradiction .
Hence our assumption is wrong .
However , I am unable to find the specialty of number 6 here and as per this argument even a group of four people can also have a triad with all pairs having the same friendship value ,which can be checked to be false .
I must have made a mistake somewhere , I would be really grateful on being pointed out where I have I erred.
Where you write "but now $K(a,c)=K(a,d)=f$", I don't see where you get $K(a,d)$ from – you hadn't mentioned it before as far as I can tell.
More generally speaking, your presentation is rather confusing. The first few paragraphs seem to be repeated verbatim, and all the opening parentheses are written as subscripts without apparent reason. Also you might want to add some justifications for the many statements of the form "let us assume ..." / "let's suppose ...", not all of which are self-evidently admissible.