Mixing
Determine whether the following statements are true or false, and then prove answers
$begingroup$
Determine whether statement is true or false, and then prove answer.
For all integers $a,b,c$:
a) if $a|bc$ and gcd$(a,b) = 1$ then $a|c$
b) if $a|c$ and $a|b$ and gcd$(a,b) = 1$ then $ab|c$.
I've answered versions of these questions without the "and gcd$(a,b) = 1$", but I'm not sure if and how that affects whether the statement is true or false.
discrete-mathematics
edited Feb 21 '16 at 0:21
Yeah..
262115
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
$endgroup$
add a comment |
$begingroup$
Determine whether statement is true or false, and then prove answer.
For all integers $a,b,c$:
a) if $a|bc$ and gcd$(a,b) = 1$ then $a|c$
b) if $a|c$ and $a|b$ and gcd$(a,b) = 1$ then $ab|c$.
I've answered versions of these questions without the "and gcd$(a,b) = 1$", but I'm not sure if and how that affects whether the statement is true or false.
discrete-mathematics
edited Feb 21 '16 at 0:21
Yeah..
262115
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
$endgroup$
1
$begingroup$
Well what does it mean for their gcd to be 1?
$endgroup$
– Brandon Thomas Van Over
Feb 21 '16 at 0:20
1
$begingroup$
Consider if $a=1$ for the second part that likely doesn't hold.
$endgroup$
– JB King
Feb 21 '16 at 0:22
$begingroup$
@JBKing: Indeed, if $a|b$ and $gcd(a,b)=1$, then $a$ has to be $1$.
$endgroup$
– TonyK
Feb 21 '16 at 0:58
$begingroup$
@TonyK, $a$ could be -1 for another possibility here.
$endgroup$
– JB King
Feb 21 '16 at 4:57
add a comment |
$begingroup$
Determine whether statement is true or false, and then prove answer.
For all integers $a,b,c$:
a) if $a|bc$ and gcd$(a,b) = 1$ then $a|c$
b) if $a|c$ and $a|b$ and gcd$(a,b) = 1$ then $ab|c$.
I've answered versions of these questions without the "and gcd$(a,b) = 1$", but I'm not sure if and how that affects whether the statement is true or false.
discrete-mathematics
edited Feb 21 '16 at 0:21
Yeah..
262115
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
$endgroup$
Determine whether statement is true or false, and then prove answer.
For all integers $a,b,c$:
a) if $a|bc$ and gcd$(a,b) = 1$ then $a|c$
b) if $a|c$ and $a|b$ and gcd$(a,b) = 1$ then $ab|c$.
I've answered versions of these questions without the "and gcd$(a,b) = 1$", but I'm not sure if and how that affects whether the statement is true or false.
discrete-mathematics
discrete-mathematics
edited Feb 21 '16 at 0:21
Yeah..
262115
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
edited Feb 21 '16 at 0:21
Yeah..
262115
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
edited Feb 21 '16 at 0:21
Yeah..
262115
edited Feb 21 '16 at 0:21
Yeah..
262115
edited Feb 21 '16 at 0:21
Yeah..
262115
262115
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
asked Feb 21 '16 at 0:08
kristof solekristof sole
1
1
1
$begingroup$
Well what does it mean for their gcd to be 1?
$endgroup$
– Brandon Thomas Van Over
Feb 21 '16 at 0:20
1
$begingroup$
Consider if $a=1$ for the second part that likely doesn't hold.
$endgroup$
– JB King
Feb 21 '16 at 0:22
$begingroup$
@JBKing: Indeed, if $a|b$ and $gcd(a,b)=1$, then $a$ has to be $1$.
$endgroup$
– TonyK
Feb 21 '16 at 0:58
$begingroup$
@TonyK, $a$ could be -1 for another possibility here.
$endgroup$
– JB King
Feb 21 '16 at 4:57
add a comment |
1
$begingroup$
Well what does it mean for their gcd to be 1?
$endgroup$
– Brandon Thomas Van Over
Feb 21 '16 at 0:20
1
$begingroup$
Consider if $a=1$ for the second part that likely doesn't hold.
$endgroup$
– JB King
Feb 21 '16 at 0:22
$begingroup$
@JBKing: Indeed, if $a|b$ and $gcd(a,b)=1$, then $a$ has to be $1$.
$endgroup$
– TonyK
Feb 21 '16 at 0:58
$begingroup$
@TonyK, $a$ could be -1 for another possibility here.
$endgroup$
– JB King
Feb 21 '16 at 4:57
1
1
$begingroup$
Well what does it mean for their gcd to be 1?
$endgroup$
– Brandon Thomas Van Over
Feb 21 '16 at 0:20
$begingroup$
Well what does it mean for their gcd to be 1?
$endgroup$
– Brandon Thomas Van Over
Feb 21 '16 at 0:20
1
1
$begingroup$
Consider if $a=1$ for the second part that likely doesn't hold.
$endgroup$
– JB King
Feb 21 '16 at 0:22
$begingroup$
Consider if $a=1$ for the second part that likely doesn't hold.
$endgroup$
– JB King
Feb 21 '16 at 0:22
$begingroup$
@JBKing: Indeed, if $a|b$ and $gcd(a,b)=1$, then $a$ has to be $1$.
$endgroup$
– TonyK
Feb 21 '16 at 0:58
$begingroup$
@JBKing: Indeed, if $a|b$ and $gcd(a,b)=1$, then $a$ has to be $1$.
$endgroup$
– TonyK
Feb 21 '16 at 0:58
$begingroup$
@TonyK, $a$ could be -1 for another possibility here.
$endgroup$
– JB King
Feb 21 '16 at 4:57
$begingroup$
@TonyK, $a$ could be -1 for another possibility here.
$endgroup$
– JB King
Feb 21 '16 at 4:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Bezout's identity. If $a$ and $b$ are relatively prime then there exist integers $x,y$ such that $ax+by=1$
Multiply both sides by $c$ for the first part and see what you can deduce.
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
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%2f1664879%2fdetermine-whether-the-following-statements-are-true-or-false-and-then-prove-ans%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Bezout's identity. If $a$ and $b$ are relatively prime then there exist integers $x,y$ such that $ax+by=1$
Multiply both sides by $c$ for the first part and see what you can deduce.
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
add a comment |
$begingroup$
Bezout's identity. If $a$ and $b$ are relatively prime then there exist integers $x,y$ such that $ax+by=1$
Multiply both sides by $c$ for the first part and see what you can deduce.
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
add a comment |
$begingroup$
Bezout's identity. If $a$ and $b$ are relatively prime then there exist integers $x,y$ such that $ax+by=1$
Multiply both sides by $c$ for the first part and see what you can deduce.
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
Bezout's identity. If $a$ and $b$ are relatively prime then there exist integers $x,y$ such that $ax+by=1$
Multiply both sides by $c$ for the first part and see what you can deduce.
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
edited Feb 21 '16 at 0:32
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
answered Feb 21 '16 at 0:26
Foobaz JohnFoobaz John
22.1k41452
22.1k41452
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
add a comment |
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
how does having c = c(ax + by) help?
$endgroup$
– kristof sole
Feb 21 '16 at 23:04
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
$begingroup$
$a$ divides the RHS by assumption and hence divides the LHS
$endgroup$
– Foobaz John
Feb 22 '16 at 22:09
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.
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%2f1664879%2fdetermine-whether-the-following-statements-are-true-or-false-and-then-prove-ans%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
1
$begingroup$
Well what does it mean for their gcd to be 1?
$endgroup$
– Brandon Thomas Van Over
Feb 21 '16 at 0:20
1
$begingroup$
Consider if $a=1$ for the second part that likely doesn't hold.
$endgroup$
– JB King
Feb 21 '16 at 0:22
$begingroup$
@JBKing: Indeed, if $a|b$ and $gcd(a,b)=1$, then $a$ has to be $1$.
$endgroup$
– TonyK
Feb 21 '16 at 0:58
$begingroup$
@TonyK, $a$ could be -1 for another possibility here.
$endgroup$
– JB King
Feb 21 '16 at 4:57