Mixing
For any integer $a$, $gcd(11a+5,2a+1)=1$.
$begingroup$
How would I go by proving this statement?
What I did was I tried using Proposition GCD Of One, so that
$(11a+5)x + (2a+1)y = 1$, and
$(11x+2y)a + (5x+y) = 1$.
But I have no idea what to do from here.
What can I do?
divisibility
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
$endgroup$
add a comment |
$begingroup$
How would I go by proving this statement?
What I did was I tried using Proposition GCD Of One, so that
$(11a+5)x + (2a+1)y = 1$, and
$(11x+2y)a + (5x+y) = 1$.
But I have no idea what to do from here.
What can I do?
divisibility
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
$endgroup$
add a comment |
$begingroup$
How would I go by proving this statement?
What I did was I tried using Proposition GCD Of One, so that
$(11a+5)x + (2a+1)y = 1$, and
$(11x+2y)a + (5x+y) = 1$.
But I have no idea what to do from here.
What can I do?
divisibility
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
$endgroup$
How would I go by proving this statement?
What I did was I tried using Proposition GCD Of One, so that
$(11a+5)x + (2a+1)y = 1$, and
$(11x+2y)a + (5x+y) = 1$.
But I have no idea what to do from here.
What can I do?
divisibility
divisibility
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
asked Apr 15 '14 at 0:29
HaxifyHaxify
150116
150116
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Can you justify the following?
begin{align*}
gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
&= gcd(a, 2a + 1)
end{align*}
Can you finish it from here?
$endgroup$
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
|
show 1 more comment
$begingroup$
Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$
Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$
$$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$
Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
$endgroup$
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
add a comment |
Your Answer
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%2f754128%2ffor-any-integer-a-gcd11a5-2a1-1%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
$begingroup$
Hint: Can you justify the following?
begin{align*}
gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
&= gcd(a, 2a + 1)
end{align*}
Can you finish it from here?
$endgroup$
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
|
show 1 more comment
$begingroup$
Hint: Can you justify the following?
begin{align*}
gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
&= gcd(a, 2a + 1)
end{align*}
Can you finish it from here?
$endgroup$
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
|
show 1 more comment
$begingroup$
Hint: Can you justify the following?
begin{align*}
gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
&= gcd(a, 2a + 1)
end{align*}
Can you finish it from here?
$endgroup$
Hint: Can you justify the following?
begin{align*}
gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
&= gcd(a, 2a + 1)
end{align*}
Can you finish it from here?
answered Apr 15 '14 at 0:30
user61527
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
|
show 1 more comment
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
Sorry, but could you explain how the first line works?
$endgroup$
– Haxify
Apr 15 '14 at 0:33
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
@Haxify $$gcd(a, c) = gcd(a - c, c)$$
$endgroup$
– user61527
Apr 15 '14 at 0:35
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Oh, okay I see how that works. Thank you!
$endgroup$
– Haxify
Apr 15 '14 at 0:37
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
$endgroup$
– Haxify
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
$begingroup$
@Haxify Exactly.
$endgroup$
– user61527
Apr 15 '14 at 0:47
|
show 1 more comment
$begingroup$
Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$
Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$
$$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$
Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
$endgroup$
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
add a comment |
$begingroup$
Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$
Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$
$$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$
Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
$endgroup$
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
add a comment |
$begingroup$
Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$
Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$
$$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$
Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
$endgroup$
Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$
Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$
$$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$
Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
edited Feb 2 at 4:08
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
answered Apr 15 '14 at 0:35
Bill DubuqueBill Dubuque
214k29197656
214k29197656
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
add a comment |
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
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%2f754128%2ffor-any-integer-a-gcd11a5-2a1-1%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
