Question about proof of Bézout's lemma












1












$begingroup$


Picture of Proof (1)



Picture of Proof (2)



Can someone please explain me why



$r=|a|-qd=|a|-q(|a|bar{s}+bt)=|a|(1-qbar{s})-bqtin Scup {0}$



?






elementary-number-theory







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  • $begingroup$
    Th. D1 which was applied here is the remainder Theorem which states that every pair of natural numbers (a,b) has exactly one pair of natural-numbers + 0 (q,r) such that a = qb+r
    $endgroup$
    – RM777
    Dec 17 '18 at 16:55
















1












$begingroup$


Picture of Proof (1)



Picture of Proof (2)



Can someone please explain me why



$r=|a|-qd=|a|-q(|a|bar{s}+bt)=|a|(1-qbar{s})-bqtin Scup {0}$



?






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Th. D1 which was applied here is the remainder Theorem which states that every pair of natural numbers (a,b) has exactly one pair of natural-numbers + 0 (q,r) such that a = qb+r
    $endgroup$
    – RM777
    Dec 17 '18 at 16:55














1












1








1





$begingroup$


Picture of Proof (1)



Picture of Proof (2)



Can someone please explain me why



$r=|a|-qd=|a|-q(|a|bar{s}+bt)=|a|(1-qbar{s})-bqtin Scup {0}$



?






elementary-number-theory







share|cite|improve this question











$endgroup$




Picture of Proof (1)



Picture of Proof (2)



Can someone please explain me why



$r=|a|-qd=|a|-q(|a|bar{s}+bt)=|a|(1-qbar{s})-bqtin Scup {0}$



?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 16:57







RM777

















asked Dec 17 '18 at 16:47









RM777RM777

37612




37612












  • $begingroup$
    Th. D1 which was applied here is the remainder Theorem which states that every pair of natural numbers (a,b) has exactly one pair of natural-numbers + 0 (q,r) such that a = qb+r
    $endgroup$
    – RM777
    Dec 17 '18 at 16:55


















  • $begingroup$
    Th. D1 which was applied here is the remainder Theorem which states that every pair of natural numbers (a,b) has exactly one pair of natural-numbers + 0 (q,r) such that a = qb+r
    $endgroup$
    – RM777
    Dec 17 '18 at 16:55
















$begingroup$
Th. D1 which was applied here is the remainder Theorem which states that every pair of natural numbers (a,b) has exactly one pair of natural-numbers + 0 (q,r) such that a = qb+r
$endgroup$
– RM777
Dec 17 '18 at 16:55




$begingroup$
Th. D1 which was applied here is the remainder Theorem which states that every pair of natural numbers (a,b) has exactly one pair of natural-numbers + 0 (q,r) such that a = qb+r
$endgroup$
– RM777
Dec 17 '18 at 16:55










1 Answer
1






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1












$begingroup$

$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
    $endgroup$
    – RM777
    Dec 17 '18 at 17:01











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$begingroup$

$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
    $endgroup$
    – RM777
    Dec 17 '18 at 17:01
















1












$begingroup$

$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
    $endgroup$
    – RM777
    Dec 17 '18 at 17:01














1












1








1





$begingroup$

$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.






share|cite|improve this answer









$endgroup$



$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 16:56









Thomas ShelbyThomas Shelby

3,1971524




3,1971524








  • 1




    $begingroup$
    Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
    $endgroup$
    – RM777
    Dec 17 '18 at 17:01














  • 1




    $begingroup$
    Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
    $endgroup$
    – RM777
    Dec 17 '18 at 17:01








1




1




$begingroup$
Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
$endgroup$
– RM777
Dec 17 '18 at 17:01




$begingroup$
Okay I got it thanks a lot. r is positive or equal to zero by default therefore the linear combination must also be positive.
$endgroup$
– RM777
Dec 17 '18 at 17:01


















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