Mixing
Prove that if ${7k:kin{mathbb Z}}subsetneq{nm:min{mathbb Z}}$, then $n=1.$
$begingroup$
Let n be a natural number. Prove that if ${7k:kin{mathbb Z}}subsetneq{nm:min{mathbb Z}}$, then $n=1$.
I know that we must show $xin{A}$ implies $xin{B}$, and that there exists $xin{B}$ such that $xnotin{A}$, which means that $xin{A}neq{xin{B}}$.
Any help is appreciated; thanks
proof-writing
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:22
macymacy
526
$endgroup$
add a comment |
$begingroup$
Let n be a natural number. Prove that if ${7k:kin{mathbb Z}}subsetneq{nm:min{mathbb Z}}$, then $n=1$.
I know that we must show $xin{A}$ implies $xin{B}$, and that there exists $xin{B}$ such that $xnotin{A}$, which means that $xin{A}neq{xin{B}}$.
Any help is appreciated; thanks
proof-writing
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:22
macymacy
526
$endgroup$
add a comment |
$begingroup$
Let n be a natural number. Prove that if ${7k:kin{mathbb Z}}subsetneq{nm:min{mathbb Z}}$, then $n=1$.
I know that we must show $xin{A}$ implies $xin{B}$, and that there exists $xin{B}$ such that $xnotin{A}$, which means that $xin{A}neq{xin{B}}$.
Any help is appreciated; thanks
proof-writing
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:22
macymacy
526
$endgroup$
Let n be a natural number. Prove that if ${7k:kin{mathbb Z}}subsetneq{nm:min{mathbb Z}}$, then $n=1$.
I know that we must show $xin{A}$ implies $xin{B}$, and that there exists $xin{B}$ such that $xnotin{A}$, which means that $xin{A}neq{xin{B}}$.
Any help is appreciated; thanks
proof-writing
proof-writing
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:22
macymacy
526
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:22
macymacy
526
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
edited Feb 3 at 1:20
Andrés E. Caicedo
66k8160252
66k8160252
asked Feb 2 at 21:22
macymacy
526
asked Feb 2 at 21:22
macymacy
526
asked Feb 2 at 21:22
macymacy
526
526
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.
answered Feb 2 at 21:33
bouncebackbounceback
446212
$endgroup$
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
add a comment |
$begingroup$
Hint $ $ Show $ aBbb Zsupseteq bBbb Ziff amid b. $ Hence $,nBbb Zsupseteq 7Bbb Ziff nmid 7$
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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votes
$begingroup$
The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.
answered Feb 2 at 21:33
bouncebackbounceback
446212
$endgroup$
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
add a comment |
$begingroup$
The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.
answered Feb 2 at 21:33
bouncebackbounceback
446212
$endgroup$
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
add a comment |
$begingroup$
The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.
answered Feb 2 at 21:33
bouncebackbounceback
446212
$endgroup$
The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.
answered Feb 2 at 21:33
bouncebackbounceback
446212
answered Feb 2 at 21:33
bouncebackbounceback
446212
answered Feb 2 at 21:33
bouncebackbounceback
446212
answered Feb 2 at 21:33
bouncebackbounceback
446212
446212
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
add a comment |
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Ah I see what you are saying! Thanks so much.
$endgroup$
– macy
Feb 2 at 21:47
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
$begingroup$
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x in 7mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process.
$endgroup$
– bounceback
Feb 2 at 22:13
add a comment |
$begingroup$
Hint $ $ Show $ aBbb Zsupseteq bBbb Ziff amid b. $ Hence $,nBbb Zsupseteq 7Bbb Ziff nmid 7$
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
add a comment |
$begingroup$
Hint $ $ Show $ aBbb Zsupseteq bBbb Ziff amid b. $ Hence $,nBbb Zsupseteq 7Bbb Ziff nmid 7$
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
add a comment |
$begingroup$
Hint $ $ Show $ aBbb Zsupseteq bBbb Ziff amid b. $ Hence $,nBbb Zsupseteq 7Bbb Ziff nmid 7$
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
Hint $ $ Show $ aBbb Zsupseteq bBbb Ziff amid b. $ Hence $,nBbb Zsupseteq 7Bbb Ziff nmid 7$
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
answered Feb 2 at 22:08
Bill DubuqueBill Dubuque
214k29197659
214k29197659
add a comment |
add a comment |
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