The discussion centers on the necessity of demonstrating that r belongs to the set S U {0} in the proof of Bézout's Identity. Participants assert that if r is not in S U {0}, the minimality of d does not apply, preventing the conclusion that r equals zero. The argument emphasizes that for comparisons to be valid, both elements must belong to the same set, specifically S U {0} in this case. Thus, establishing r's membership in S U {0} is crucial for the proof's integrity.
PREREQUISITESMathematicians, students of number theory, and anyone interested in understanding the foundational proofs related to Bézout's Identity and set theory comparisons.
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).