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
$endgroup$
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
$endgroup$
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
$endgroup$
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
150116
add a comment |
add a comment |
2 Answers
2
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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 $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
$endgroup$
$begingroup$
Oh, I see. Thank you!!
$endgroup$
– Haxify
Apr 15 '14 at 0:38
add a comment |
Your Answer
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2 Answers
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oldest
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2 Answers
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active
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active
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active
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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 $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
$endgroup$
$begingroup$
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 $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
$endgroup$
$begingroup$
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 $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
$endgroup$
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 $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.
edited Feb 2 at 4:08
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 |
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