Number Theory: Divisibilty
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Q: Prove that if a and b are positive integers satisfying (a,b)=[a,b}, then a=b.
My approach (with ans):
(a,b) means gcd, and I let g=(a,b)
since g is the gcd,
then I get these two: g|a and g|b, while g
Next is [a,b], the lcm of a and b, and I let it be h.
then I get another two of these: a|h and b|h while a
Combine the inequalities together will get:
g
Since the question provides that g=h,
so h
Is the a correct explanation? As I hope to get full credits, I hope I can present my answers as detail as possible (without any faults). Thank you.
number-theory divisibility greatest-common-divisor least-common-multiple
asked Sep 7 at 7:12
Jason Ng
517
add a comment |Â
up vote
0
down vote
favorite
Q: Prove that if a and b are positive integers satisfying (a,b)=[a,b}, then a=b.
My approach (with ans):
(a,b) means gcd, and I let g=(a,b)
since g is the gcd,
then I get these two: g|a and g|b, while g
Next is [a,b], the lcm of a and b, and I let it be h.
then I get another two of these: a|h and b|h while a
Combine the inequalities together will get:
g
Since the question provides that g=h,
so h
Is the a correct explanation? As I hope to get full credits, I hope I can present my answers as detail as possible (without any faults). Thank you.
number-theory divisibility greatest-common-divisor least-common-multiple
asked Sep 7 at 7:12
Jason Ng
517
1
Combine the inequalitiesWhich inequalities? There is no inequality in the posted question. However, it would be enough to argue that $,(a,b) le a le [a,b],$.
– dxiv
Sep 7 at 7:16
Oh, i may miss typing (a,b)≤a≤[a,b]. Thank you very much.
– Jason Ng
Sep 7 at 7:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Q: Prove that if a and b are positive integers satisfying (a,b)=[a,b}, then a=b.
My approach (with ans):
(a,b) means gcd, and I let g=(a,b)
since g is the gcd,
then I get these two: g|a and g|b, while g
Next is [a,b], the lcm of a and b, and I let it be h.
then I get another two of these: a|h and b|h while a
Combine the inequalities together will get:
g
Since the question provides that g=h,
so h
Is the a correct explanation? As I hope to get full credits, I hope I can present my answers as detail as possible (without any faults). Thank you.
number-theory divisibility greatest-common-divisor least-common-multiple
asked Sep 7 at 7:12
Jason Ng
517
Q: Prove that if a and b are positive integers satisfying (a,b)=[a,b}, then a=b.
My approach (with ans):
(a,b) means gcd, and I let g=(a,b)
since g is the gcd,
then I get these two: g|a and g|b, while g
Next is [a,b], the lcm of a and b, and I let it be h.
then I get another two of these: a|h and b|h while a
Combine the inequalities together will get:
g
Since the question provides that g=h,
so h
Is the a correct explanation? As I hope to get full credits, I hope I can present my answers as detail as possible (without any faults). Thank you.
number-theory divisibility greatest-common-divisor least-common-multiple
number-theory divisibility greatest-common-divisor least-common-multiple
asked Sep 7 at 7:12
Jason Ng
517
asked Sep 7 at 7:12
Jason Ng
517
asked Sep 7 at 7:12
Jason Ng
517
asked Sep 7 at 7:12
Jason Ng
517
asked Sep 7 at 7:12
Jason Ng
517
517
1
Combine the inequalitiesWhich inequalities? There is no inequality in the posted question. However, it would be enough to argue that $,(a,b) le a le [a,b],$.
– dxiv
Sep 7 at 7:16
Oh, i may miss typing (a,b)≤a≤[a,b]. Thank you very much.
– Jason Ng
Sep 7 at 7:26
add a comment |Â
1
Combine the inequalitiesWhich inequalities? There is no inequality in the posted question. However, it would be enough to argue that $,(a,b) le a le [a,b],$.
– dxiv
Sep 7 at 7:16
Oh, i may miss typing (a,b)≤a≤[a,b]. Thank you very much.
– Jason Ng
Sep 7 at 7:26
1
1
Combine the inequalities Which inequalities? There is no inequality in the posted question. However, it would be enough to argue that $,(a,b) le a le [a,b],$.– dxiv
Sep 7 at 7:16
Combine the inequalities Which inequalities? There is no inequality in the posted question. However, it would be enough to argue that $,(a,b) le a le [a,b],$.– dxiv
Sep 7 at 7:16
Oh, i may miss typing (a,b)≤a≤[a,b]. Thank you very much.
– Jason Ng
Sep 7 at 7:26
Oh, i may miss typing (a,b)≤a≤[a,b]. Thank you very much.
– Jason Ng
Sep 7 at 7:26
add a comment |Â
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1
Combine the inequalitiesWhich inequalities? There is no inequality in the posted question. However, it would be enough to argue that $,(a,b) le a le [a,b],$.– dxiv
Sep 7 at 7:16
Oh, i may miss typing (a,b)≤a≤[a,b]. Thank you very much.
– Jason Ng
Sep 7 at 7:26