Question about proof of Bézout's lemma
1
$begingroup$
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
asked Dec 17 '18 at 16:47
RM777RM777
37612
$endgroup$
add a comment |
1
$begingroup$
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
asked Dec 17 '18 at 16:47
RM777RM777
37612
$endgroup$
$begingroup$
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
add a comment |
1
1
1
$begingroup$
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
asked Dec 17 '18 at 16:47
RM777RM777
37612
$endgroup$
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
asked Dec 17 '18 at 16:47
RM777RM777
37612
asked Dec 17 '18 at 16:47
RM777RM777
37612
edited Dec 17 '18 at 16:57
RM777
asked Dec 17 '18 at 16:47
RM777RM777
37612
asked Dec 17 '18 at 16:47
RM777RM777
37612
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
$endgroup$
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
add a comment |
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1 Answer
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1
$begingroup$
$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
$endgroup$
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
add a comment |
1
$begingroup$
$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
$endgroup$
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
add a comment |
1
1
1
$begingroup$
$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
$endgroup$
$r $ is of the form $xa+yb $ with $x=pm (1-qbar{s})$ and $y=-qt $. Hence $rin Scup {0}$.
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
answered Dec 17 '18 at 16:56
Thomas ShelbyThomas Shelby
3,1971524
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
add a comment |
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
add a comment |
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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