The discussion focuses on the necessity of showing that a value \( r \) belongs to the set \( S \cup \{0\} \) in the context of proving Bézout's identity. Participants explore the implications of this condition on the proof's validity, particularly regarding the minimality of certain elements.
Participants express differing views on the necessity of \( r \) belonging to \( S \cup \{0\} \) for the proof of Bézout's identity. No consensus is reached, as multiple competing perspectives remain regarding the implications of this condition.
The discussion highlights the dependence on definitions and the implications of minimality in set comparisons, which remain unresolved within the context of the proof.
If ##r \notin S\cup \{\,0\,\}##, then the minimality of ##d## has nothing to do with ##r## and we cannot conclude ##r=0##. It is zero, because it is part of this set!Suyogya said:what is the need to show that r belongs to S U {0} in proof (https://en.wikipedia.org/wiki/Bézout's_identity#Proof)
r is zero afterall, whether it lies in S U {0} or not doesn't affect.
please tell the following:fresh_42 said:If ##r \notin S\cup \{\,0\,\}##, then the minimality of ##d## has nothing to do with ##r## and we cannot conclude ##r=0##. It is zero, because it is part of this set!
We start with a set ##M##. Then we choose a minimal element of ##M##, called ##m##. In order to compare any other element ##n## to ##m##, we can only do this, if ##n \in M##, because then we know, that ##m \leq n## as ##m## was chosen minimal. Otherwise we can't say anything.Suyogya said:please tell the following:
is the need to show that r lies in S U {0} just because to compare r is less than or equal to d (as d also belongs to S U {0})?
So are only the same set elements can be compared? If yes, then consider a counter example, set a={10}, set b={5} couldn't we say 5<10 (as both belongs to different sets).