Prove ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$is a covering system
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
add a comment |
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
add a comment |
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
modular-arithmetic
edited Nov 28 '18 at 4:53
Alex Vong
1,286819
1,286819
asked Nov 28 '18 at 2:59
Jingting931015
828
828
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016641%2fprove-0-bmod-2-0-bmod-3-1-bmod-4-1-bmod-6-11-bmod-12-is-a-covering%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
add a comment |
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
add a comment |
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
edited Nov 28 '18 at 3:26
answered Nov 28 '18 at 3:07
Valborg
542
542
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
add a comment |
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 '18 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 '18 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 '18 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 '18 at 3:39
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
answered Nov 28 '18 at 3:33
fleablood
68.4k22685
68.4k22685
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016641%2fprove-0-bmod-2-0-bmod-3-1-bmod-4-1-bmod-6-11-bmod-12-is-a-covering%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown