Show that $forall_{n,m,kin mathbb{N}}(k|nm$ and gcd$(n,k)=1 Rightarrow k|m)$
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Definitons
gcd := greatest common divisor
Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$
For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$
algebra-precalculus prime-numbers
asked Nov 21 at 11:15
RM777
1808
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up vote
1
down vote
favorite
Please help me
Definitons
gcd := greatest common divisor
Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$
For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$
algebra-precalculus prime-numbers
asked Nov 21 at 11:15
RM777
1808
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Please help me
Definitons
gcd := greatest common divisor
Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$
For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$
algebra-precalculus prime-numbers
asked Nov 21 at 11:15
RM777
1808
Please help me
Definitons
gcd := greatest common divisor
Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$
For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$
algebra-precalculus prime-numbers
algebra-precalculus prime-numbers
asked Nov 21 at 11:15
RM777
1808
asked Nov 21 at 11:15
RM777
1808
asked Nov 21 at 11:15
RM777
1808
asked Nov 21 at 11:15
RM777
1808
asked Nov 21 at 11:15
RM777
1808
1808
add a comment |
add a comment |
1 Answer
1
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up vote
2
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accepted
If $m = 0$ the implication is obvious
If $m ne 0$
Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$
We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$
And $gcd(mn,mk) = m$
$$implies k|m$$
answered Nov 21 at 12:39
TheD0ubleT
37718
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $m = 0$ the implication is obvious
If $m ne 0$
Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$
We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$
And $gcd(mn,mk) = m$
$$implies k|m$$
answered Nov 21 at 12:39
TheD0ubleT
37718
add a comment |
up vote
2
down vote
accepted
If $m = 0$ the implication is obvious
If $m ne 0$
Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$
We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$
And $gcd(mn,mk) = m$
$$implies k|m$$
answered Nov 21 at 12:39
TheD0ubleT
37718
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $m = 0$ the implication is obvious
If $m ne 0$
Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$
We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$
And $gcd(mn,mk) = m$
$$implies k|m$$
answered Nov 21 at 12:39
TheD0ubleT
37718
If $m = 0$ the implication is obvious
If $m ne 0$
Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$
We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$
And $gcd(mn,mk) = m$
$$implies k|m$$
answered Nov 21 at 12:39
TheD0ubleT
37718
answered Nov 21 at 12:39
TheD0ubleT
37718
answered Nov 21 at 12:39
TheD0ubleT
37718
answered Nov 21 at 12:39
TheD0ubleT
37718
37718
add a comment |
add a comment |
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