Bezout Identity: Is r∈S U {0} Necessary to Prove?

[Suyogya]

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.

 

[fresh_42]

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.

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]

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!

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).

 

[fresh_42]

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).

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.

If in our example, $d \in S$ is minimal, and $r if \in S \cup \{\,0\,\}$, then either $r=0$ or $d \leq r$. As $r < d$, the second is impossible, leaving $r=0$ as only possibility. This entire argument needs $r if \in S \cup \{\,0\,\}$ and that $d\in S$ is the smallest integer there.