If $g = gcd(a,b)$ prove (a,b)=(g). Furthermore, if $k = lcm(a,b)$ prove that $(a)cap(b) = (k)$
$begingroup$
$a,b in mathbb{Z}$.
$(a,b),hspace{0.4mm}(g)$ and $(k)$ are principle ideals
I'm new to this kind of problems, so I don't even know how to start it. Some help would be appreciated.
abstract-algebra ring-theory ideals greatest-common-divisor least-common-multiple
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
$endgroup$
add a comment |
$begingroup$
$a,b in mathbb{Z}$.
$(a,b),hspace{0.4mm}(g)$ and $(k)$ are principle ideals
I'm new to this kind of problems, so I don't even know how to start it. Some help would be appreciated.
abstract-algebra ring-theory ideals greatest-common-divisor least-common-multiple
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
$endgroup$
$begingroup$
What are your definitions of $(a,,b)$ and $(n)$?
$endgroup$
– J.G.
Dec 7 '18 at 11:34
$begingroup$
@J.G. (a,b), g and k are principal ideals
$endgroup$
– Neon Xd
Dec 7 '18 at 11:39
add a comment |
$begingroup$
$a,b in mathbb{Z}$.
$(a,b),hspace{0.4mm}(g)$ and $(k)$ are principle ideals
I'm new to this kind of problems, so I don't even know how to start it. Some help would be appreciated.
abstract-algebra ring-theory ideals greatest-common-divisor least-common-multiple
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
$endgroup$
$a,b in mathbb{Z}$.
$(a,b),hspace{0.4mm}(g)$ and $(k)$ are principle ideals
I'm new to this kind of problems, so I don't even know how to start it. Some help would be appreciated.
abstract-algebra ring-theory ideals greatest-common-divisor least-common-multiple
abstract-algebra ring-theory ideals greatest-common-divisor least-common-multiple
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
edited Dec 7 '18 at 11:41
Neon Xd
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
asked Dec 7 '18 at 11:19
Neon XdNeon Xd
183
183
$begingroup$
What are your definitions of $(a,,b)$ and $(n)$?
$endgroup$
– J.G.
Dec 7 '18 at 11:34
$begingroup$
@J.G. (a,b), g and k are principal ideals
$endgroup$
– Neon Xd
Dec 7 '18 at 11:39
add a comment |
$begingroup$
What are your definitions of $(a,,b)$ and $(n)$?
$endgroup$
– J.G.
Dec 7 '18 at 11:34
$begingroup$
@J.G. (a,b), g and k are principal ideals
$endgroup$
– Neon Xd
Dec 7 '18 at 11:39
$begingroup$
What are your definitions of $(a,,b)$ and $(n)$?
$endgroup$
– J.G.
Dec 7 '18 at 11:34
$begingroup$
What are your definitions of $(a,,b)$ and $(n)$?
$endgroup$
– J.G.
Dec 7 '18 at 11:34
$begingroup$
@J.G. (a,b), g and k are principal ideals
$endgroup$
– Neon Xd
Dec 7 '18 at 11:39
$begingroup$
@J.G. (a,b), g and k are principal ideals
$endgroup$
– Neon Xd
Dec 7 '18 at 11:39
add a comment |
1 Answer
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$begingroup$
You want to prove an arbitrary $ninBbb Z$ is of the form $ax+by$ iff a multiple of $g$, and divisible by both $a,,b$ iff divisible by $k$.
Let $S$ denote the set of positive integers in $(a,,b)$, which is clearly nonempty as it contains $|a|$. Clearly, its elements are all multiples of $g$. Let $d$ denote $S$'s minimal element, so $S$ is precisely the positive elements of $d$ (it includes them all, and any other element's remainder on division by $d$ would be an element of $S$ too, smaller than $d$ and thus contradicting its definition). Thus $d$ divides both $a$ and $b$, so $d|g$. Hence $d=g$.
For the second result let $T$ denote the set of positive common multiples of $a,,b$, so $k$ is the minimal element of $T$. By the same remainder-contradicts-minimality logic as above, $T$ is precisely the multiples of $k$.
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You want to prove an arbitrary $ninBbb Z$ is of the form $ax+by$ iff a multiple of $g$, and divisible by both $a,,b$ iff divisible by $k$.
Let $S$ denote the set of positive integers in $(a,,b)$, which is clearly nonempty as it contains $|a|$. Clearly, its elements are all multiples of $g$. Let $d$ denote $S$'s minimal element, so $S$ is precisely the positive elements of $d$ (it includes them all, and any other element's remainder on division by $d$ would be an element of $S$ too, smaller than $d$ and thus contradicting its definition). Thus $d$ divides both $a$ and $b$, so $d|g$. Hence $d=g$.
For the second result let $T$ denote the set of positive common multiples of $a,,b$, so $k$ is the minimal element of $T$. By the same remainder-contradicts-minimality logic as above, $T$ is precisely the multiples of $k$.
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
$endgroup$
add a comment |
$begingroup$
You want to prove an arbitrary $ninBbb Z$ is of the form $ax+by$ iff a multiple of $g$, and divisible by both $a,,b$ iff divisible by $k$.
Let $S$ denote the set of positive integers in $(a,,b)$, which is clearly nonempty as it contains $|a|$. Clearly, its elements are all multiples of $g$. Let $d$ denote $S$'s minimal element, so $S$ is precisely the positive elements of $d$ (it includes them all, and any other element's remainder on division by $d$ would be an element of $S$ too, smaller than $d$ and thus contradicting its definition). Thus $d$ divides both $a$ and $b$, so $d|g$. Hence $d=g$.
For the second result let $T$ denote the set of positive common multiples of $a,,b$, so $k$ is the minimal element of $T$. By the same remainder-contradicts-minimality logic as above, $T$ is precisely the multiples of $k$.
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
$endgroup$
add a comment |
$begingroup$
You want to prove an arbitrary $ninBbb Z$ is of the form $ax+by$ iff a multiple of $g$, and divisible by both $a,,b$ iff divisible by $k$.
Let $S$ denote the set of positive integers in $(a,,b)$, which is clearly nonempty as it contains $|a|$. Clearly, its elements are all multiples of $g$. Let $d$ denote $S$'s minimal element, so $S$ is precisely the positive elements of $d$ (it includes them all, and any other element's remainder on division by $d$ would be an element of $S$ too, smaller than $d$ and thus contradicting its definition). Thus $d$ divides both $a$ and $b$, so $d|g$. Hence $d=g$.
For the second result let $T$ denote the set of positive common multiples of $a,,b$, so $k$ is the minimal element of $T$. By the same remainder-contradicts-minimality logic as above, $T$ is precisely the multiples of $k$.
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
$endgroup$
You want to prove an arbitrary $ninBbb Z$ is of the form $ax+by$ iff a multiple of $g$, and divisible by both $a,,b$ iff divisible by $k$.
Let $S$ denote the set of positive integers in $(a,,b)$, which is clearly nonempty as it contains $|a|$. Clearly, its elements are all multiples of $g$. Let $d$ denote $S$'s minimal element, so $S$ is precisely the positive elements of $d$ (it includes them all, and any other element's remainder on division by $d$ would be an element of $S$ too, smaller than $d$ and thus contradicting its definition). Thus $d$ divides both $a$ and $b$, so $d|g$. Hence $d=g$.
For the second result let $T$ denote the set of positive common multiples of $a,,b$, so $k$ is the minimal element of $T$. By the same remainder-contradicts-minimality logic as above, $T$ is precisely the multiples of $k$.
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
answered Dec 7 '18 at 11:47
J.G.J.G.
29.6k22946
29.6k22946
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add a comment |
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$begingroup$
What are your definitions of $(a,,b)$ and $(n)$?
$endgroup$
– J.G.
Dec 7 '18 at 11:34
$begingroup$
@J.G. (a,b), g and k are principal ideals
$endgroup$
– Neon Xd
Dec 7 '18 at 11:39