gcd(2a+1, 9a+4) = 1












1












$begingroup$


The question is from Burton's Elementary Number Theory. I want to know if my proof is legible.
Proof : Let $$d=gcd(2a+1, 9a+4)$$
Then $$d|2a+1$$ and $$d|9a+4$$
$$2a+1=db$$ and $$9a+4=dc$$
$$ a = frac{db-1}{2}$$and$$a= frac{dc-4}{9}$$
Equating both equations :
$$ 9db-9=2dc-8 $$
$$ d(9b-2c) = 1 $$
$$ 9b-2c = frac{1}{d}$$
Now, since b and c are integers, therefore $frac{1}{d}$ is an integer, i.e d divides 1 and therefore $$gcd(a, b)= d = 1$$.






proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows: $$bmod d!:, 2a!+!1equiv 0equiv 9a!+!4 Rightarrow color{#c00}0equiv 9(2a!+!1)-2(9a!+!4)equivcolor{#c00} 1 Rightarrow dmid color{#c00}{1!-!0}qquad$$
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:02












  • $begingroup$
    @BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it.
    $endgroup$
    – P.Jo
    Jan 4 at 17:22










  • $begingroup$
    We can eliminate congruence notation too! Namely the above means $$dmid 2a!+!1,9a!+!4 Rightarrow dmid 1!=! 9(2a!+!1)-2(9a!+!4)qquad $$ since multiples of $d$ are closed under integral linear combinations.
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:27










  • $begingroup$
    @BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'?
    $endgroup$
    – P.Jo
    Jan 5 at 4:19










  • $begingroup$
    @BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b?
    $endgroup$
    – P.Jo
    Jan 5 at 4:27
















1












$begingroup$


The question is from Burton's Elementary Number Theory. I want to know if my proof is legible.
Proof : Let $$d=gcd(2a+1, 9a+4)$$
Then $$d|2a+1$$ and $$d|9a+4$$
$$2a+1=db$$ and $$9a+4=dc$$
$$ a = frac{db-1}{2}$$and$$a= frac{dc-4}{9}$$
Equating both equations :
$$ 9db-9=2dc-8 $$
$$ d(9b-2c) = 1 $$
$$ 9b-2c = frac{1}{d}$$
Now, since b and c are integers, therefore $frac{1}{d}$ is an integer, i.e d divides 1 and therefore $$gcd(a, b)= d = 1$$.






proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows: $$bmod d!:, 2a!+!1equiv 0equiv 9a!+!4 Rightarrow color{#c00}0equiv 9(2a!+!1)-2(9a!+!4)equivcolor{#c00} 1 Rightarrow dmid color{#c00}{1!-!0}qquad$$
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:02












  • $begingroup$
    @BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it.
    $endgroup$
    – P.Jo
    Jan 4 at 17:22










  • $begingroup$
    We can eliminate congruence notation too! Namely the above means $$dmid 2a!+!1,9a!+!4 Rightarrow dmid 1!=! 9(2a!+!1)-2(9a!+!4)qquad $$ since multiples of $d$ are closed under integral linear combinations.
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:27










  • $begingroup$
    @BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'?
    $endgroup$
    – P.Jo
    Jan 5 at 4:19










  • $begingroup$
    @BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b?
    $endgroup$
    – P.Jo
    Jan 5 at 4:27














1












1








1





$begingroup$


The question is from Burton's Elementary Number Theory. I want to know if my proof is legible.
Proof : Let $$d=gcd(2a+1, 9a+4)$$
Then $$d|2a+1$$ and $$d|9a+4$$
$$2a+1=db$$ and $$9a+4=dc$$
$$ a = frac{db-1}{2}$$and$$a= frac{dc-4}{9}$$
Equating both equations :
$$ 9db-9=2dc-8 $$
$$ d(9b-2c) = 1 $$
$$ 9b-2c = frac{1}{d}$$
Now, since b and c are integers, therefore $frac{1}{d}$ is an integer, i.e d divides 1 and therefore $$gcd(a, b)= d = 1$$.






proof-verification







share|cite|improve this question









$endgroup$




The question is from Burton's Elementary Number Theory. I want to know if my proof is legible.
Proof : Let $$d=gcd(2a+1, 9a+4)$$
Then $$d|2a+1$$ and $$d|9a+4$$
$$2a+1=db$$ and $$9a+4=dc$$
$$ a = frac{db-1}{2}$$and$$a= frac{dc-4}{9}$$
Equating both equations :
$$ 9db-9=2dc-8 $$
$$ d(9b-2c) = 1 $$
$$ 9b-2c = frac{1}{d}$$
Now, since b and c are integers, therefore $frac{1}{d}$ is an integer, i.e d divides 1 and therefore $$gcd(a, b)= d = 1$$.






proof-verification




proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 15:31









P.JoP.Jo

85




85












  • $begingroup$
    It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows: $$bmod d!:, 2a!+!1equiv 0equiv 9a!+!4 Rightarrow color{#c00}0equiv 9(2a!+!1)-2(9a!+!4)equivcolor{#c00} 1 Rightarrow dmid color{#c00}{1!-!0}qquad$$
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:02












  • $begingroup$
    @BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it.
    $endgroup$
    – P.Jo
    Jan 4 at 17:22










  • $begingroup$
    We can eliminate congruence notation too! Namely the above means $$dmid 2a!+!1,9a!+!4 Rightarrow dmid 1!=! 9(2a!+!1)-2(9a!+!4)qquad $$ since multiples of $d$ are closed under integral linear combinations.
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:27










  • $begingroup$
    @BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'?
    $endgroup$
    – P.Jo
    Jan 5 at 4:19










  • $begingroup$
    @BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b?
    $endgroup$
    – P.Jo
    Jan 5 at 4:27


















  • $begingroup$
    It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows: $$bmod d!:, 2a!+!1equiv 0equiv 9a!+!4 Rightarrow color{#c00}0equiv 9(2a!+!1)-2(9a!+!4)equivcolor{#c00} 1 Rightarrow dmid color{#c00}{1!-!0}qquad$$
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:02












  • $begingroup$
    @BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it.
    $endgroup$
    – P.Jo
    Jan 4 at 17:22










  • $begingroup$
    We can eliminate congruence notation too! Namely the above means $$dmid 2a!+!1,9a!+!4 Rightarrow dmid 1!=! 9(2a!+!1)-2(9a!+!4)qquad $$ since multiples of $d$ are closed under integral linear combinations.
    $endgroup$
    – Bill Dubuque
    Jan 4 at 17:27










  • $begingroup$
    @BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'?
    $endgroup$
    – P.Jo
    Jan 5 at 4:19










  • $begingroup$
    @BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b?
    $endgroup$
    – P.Jo
    Jan 5 at 4:27
















$begingroup$
It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows: $$bmod d!:, 2a!+!1equiv 0equiv 9a!+!4 Rightarrow color{#c00}0equiv 9(2a!+!1)-2(9a!+!4)equivcolor{#c00} 1 Rightarrow dmid color{#c00}{1!-!0}qquad$$
$endgroup$
– Bill Dubuque
Jan 4 at 17:02






$begingroup$
It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows: $$bmod d!:, 2a!+!1equiv 0equiv 9a!+!4 Rightarrow color{#c00}0equiv 9(2a!+!1)-2(9a!+!4)equivcolor{#c00} 1 Rightarrow dmid color{#c00}{1!-!0}qquad$$
$endgroup$
– Bill Dubuque
Jan 4 at 17:02














$begingroup$
@BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it.
$endgroup$
– P.Jo
Jan 4 at 17:22




$begingroup$
@BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it.
$endgroup$
– P.Jo
Jan 4 at 17:22












$begingroup$
We can eliminate congruence notation too! Namely the above means $$dmid 2a!+!1,9a!+!4 Rightarrow dmid 1!=! 9(2a!+!1)-2(9a!+!4)qquad $$ since multiples of $d$ are closed under integral linear combinations.
$endgroup$
– Bill Dubuque
Jan 4 at 17:27




$begingroup$
We can eliminate congruence notation too! Namely the above means $$dmid 2a!+!1,9a!+!4 Rightarrow dmid 1!=! 9(2a!+!1)-2(9a!+!4)qquad $$ since multiples of $d$ are closed under integral linear combinations.
$endgroup$
– Bill Dubuque
Jan 4 at 17:27












$begingroup$
@BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'?
$endgroup$
– P.Jo
Jan 5 at 4:19




$begingroup$
@BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'?
$endgroup$
– P.Jo
Jan 5 at 4:19












$begingroup$
@BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b?
$endgroup$
– P.Jo
Jan 5 at 4:27




$begingroup$
@BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b?
$endgroup$
– P.Jo
Jan 5 at 4:27










4 Answers
4






active

oldest

votes


















0












$begingroup$

Seems fine.



Alternatively,



$$9a+4=4(2a+1)+a$$



$$2a+1=2(a)+1$$



Hence,



$$1=(2a+1)-2a=(2a+1)-2(9a+4)+8=9(2a+1)-2(9a+4)$$



That is the greatest common divisor of $2a+1$ and $9a+4$ must divide $1$. Hence the greatest common divisor is $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What did you do here really?
    $endgroup$
    – Michael Munta
    Jan 6 at 10:57










  • $begingroup$
    I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 11:08










  • $begingroup$
    Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
    $endgroup$
    – Michael Munta
    Jan 6 at 11:16










  • $begingroup$
    For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 11:21










  • $begingroup$
    This is the extended algorithm?
    $endgroup$
    – Michael Munta
    Jan 6 at 12:36



















1












$begingroup$

Looks good to me. Some nit picks follow:



A bit of formatting may help readability. For much of your answer you have pairs of equations which belong together, but the pairs which belong together aren't the pairs which appear close to one another.



After "therefore $frac{1}{d}$ is an integer", you can just say "so $d=1$ and we're done". Remember that $d=1$ is the very thing we want to prove, so it's unnecessary to continue with more calculations once you've reached that conclusion.



And personally I would probably have used the Euclidean algorithm instead.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
    $endgroup$
    – P.Jo
    Jan 4 at 16:21










  • $begingroup$
    @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
    $endgroup$
    – Arthur
    Jan 4 at 16:30










  • $begingroup$
    Nice preview before starting. Sounds like it's gonna make this easier.
    $endgroup$
    – P.Jo
    Jan 4 at 17:11



















1












$begingroup$

Divide $2a+1$ into $9a+4$ giving $9a+4 = 9/2cdot (2a+1) - 1/2$ or $2(9a+4) = 9(2a+1)-1$, or $1= -2(9a+4) + 9(2a+1)$. So each divisor of $9a+4$ and $2a+1$ divides $1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
    $endgroup$
    – Michael Munta
    Jan 7 at 18:27










  • $begingroup$
    Indeed, you're right. Thanks.
    $endgroup$
    – Wuestenfux
    Jan 8 at 8:37



















0












$begingroup$

As Siong Thye Goh said, we can use the Euclidean algorithm to guide us to a proof. The Euclidean algorithm is based on the fact that $gcd(x,y)=gcd(x,y-tx)$. If we take $x=2a+1$, $y=9a+4$, and $t=4$, then we get $gcd(2a+1,9a+4)=gcd(2a+1,a)$. Applying it again with $x=a$, $y=2a+1$, $t=2$, we get $gcd(2a+1,a)=gcd(1,a)$. From there, it's easy to see that the gcd is $1$.






share|cite|improve this answer









$endgroup$














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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Seems fine.



    Alternatively,



    $$9a+4=4(2a+1)+a$$



    $$2a+1=2(a)+1$$



    Hence,



    $$1=(2a+1)-2a=(2a+1)-2(9a+4)+8=9(2a+1)-2(9a+4)$$



    That is the greatest common divisor of $2a+1$ and $9a+4$ must divide $1$. Hence the greatest common divisor is $1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What did you do here really?
      $endgroup$
      – Michael Munta
      Jan 6 at 10:57










    • $begingroup$
      I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:08










    • $begingroup$
      Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
      $endgroup$
      – Michael Munta
      Jan 6 at 11:16










    • $begingroup$
      For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:21










    • $begingroup$
      This is the extended algorithm?
      $endgroup$
      – Michael Munta
      Jan 6 at 12:36
















    0












    $begingroup$

    Seems fine.



    Alternatively,



    $$9a+4=4(2a+1)+a$$



    $$2a+1=2(a)+1$$



    Hence,



    $$1=(2a+1)-2a=(2a+1)-2(9a+4)+8=9(2a+1)-2(9a+4)$$



    That is the greatest common divisor of $2a+1$ and $9a+4$ must divide $1$. Hence the greatest common divisor is $1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What did you do here really?
      $endgroup$
      – Michael Munta
      Jan 6 at 10:57










    • $begingroup$
      I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:08










    • $begingroup$
      Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
      $endgroup$
      – Michael Munta
      Jan 6 at 11:16










    • $begingroup$
      For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:21










    • $begingroup$
      This is the extended algorithm?
      $endgroup$
      – Michael Munta
      Jan 6 at 12:36














    0












    0








    0





    $begingroup$

    Seems fine.



    Alternatively,



    $$9a+4=4(2a+1)+a$$



    $$2a+1=2(a)+1$$



    Hence,



    $$1=(2a+1)-2a=(2a+1)-2(9a+4)+8=9(2a+1)-2(9a+4)$$



    That is the greatest common divisor of $2a+1$ and $9a+4$ must divide $1$. Hence the greatest common divisor is $1$.






    share|cite|improve this answer









    $endgroup$



    Seems fine.



    Alternatively,



    $$9a+4=4(2a+1)+a$$



    $$2a+1=2(a)+1$$



    Hence,



    $$1=(2a+1)-2a=(2a+1)-2(9a+4)+8=9(2a+1)-2(9a+4)$$



    That is the greatest common divisor of $2a+1$ and $9a+4$ must divide $1$. Hence the greatest common divisor is $1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 15:42









    Siong Thye GohSiong Thye Goh

    103k1468120




    103k1468120












    • $begingroup$
      What did you do here really?
      $endgroup$
      – Michael Munta
      Jan 6 at 10:57










    • $begingroup$
      I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:08










    • $begingroup$
      Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
      $endgroup$
      – Michael Munta
      Jan 6 at 11:16










    • $begingroup$
      For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:21










    • $begingroup$
      This is the extended algorithm?
      $endgroup$
      – Michael Munta
      Jan 6 at 12:36


















    • $begingroup$
      What did you do here really?
      $endgroup$
      – Michael Munta
      Jan 6 at 10:57










    • $begingroup$
      I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:08










    • $begingroup$
      Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
      $endgroup$
      – Michael Munta
      Jan 6 at 11:16










    • $begingroup$
      For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
      $endgroup$
      – Siong Thye Goh
      Jan 6 at 11:21










    • $begingroup$
      This is the extended algorithm?
      $endgroup$
      – Michael Munta
      Jan 6 at 12:36
















    $begingroup$
    What did you do here really?
    $endgroup$
    – Michael Munta
    Jan 6 at 10:57




    $begingroup$
    What did you do here really?
    $endgroup$
    – Michael Munta
    Jan 6 at 10:57












    $begingroup$
    I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 11:08




    $begingroup$
    I am using Euclidean algorithm to express the gcd as a combination of the two numbers.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 11:08












    $begingroup$
    Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
    $endgroup$
    – Michael Munta
    Jan 6 at 11:16




    $begingroup$
    Can you please explain how this happens? I know how it works with integers, but not with expressions like these.
    $endgroup$
    – Michael Munta
    Jan 6 at 11:16












    $begingroup$
    For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 11:21




    $begingroup$
    For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 11:21












    $begingroup$
    This is the extended algorithm?
    $endgroup$
    – Michael Munta
    Jan 6 at 12:36




    $begingroup$
    This is the extended algorithm?
    $endgroup$
    – Michael Munta
    Jan 6 at 12:36











    1












    $begingroup$

    Looks good to me. Some nit picks follow:



    A bit of formatting may help readability. For much of your answer you have pairs of equations which belong together, but the pairs which belong together aren't the pairs which appear close to one another.



    After "therefore $frac{1}{d}$ is an integer", you can just say "so $d=1$ and we're done". Remember that $d=1$ is the very thing we want to prove, so it's unnecessary to continue with more calculations once you've reached that conclusion.



    And personally I would probably have used the Euclidean algorithm instead.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
      $endgroup$
      – P.Jo
      Jan 4 at 16:21










    • $begingroup$
      @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
      $endgroup$
      – Arthur
      Jan 4 at 16:30










    • $begingroup$
      Nice preview before starting. Sounds like it's gonna make this easier.
      $endgroup$
      – P.Jo
      Jan 4 at 17:11
















    1












    $begingroup$

    Looks good to me. Some nit picks follow:



    A bit of formatting may help readability. For much of your answer you have pairs of equations which belong together, but the pairs which belong together aren't the pairs which appear close to one another.



    After "therefore $frac{1}{d}$ is an integer", you can just say "so $d=1$ and we're done". Remember that $d=1$ is the very thing we want to prove, so it's unnecessary to continue with more calculations once you've reached that conclusion.



    And personally I would probably have used the Euclidean algorithm instead.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
      $endgroup$
      – P.Jo
      Jan 4 at 16:21










    • $begingroup$
      @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
      $endgroup$
      – Arthur
      Jan 4 at 16:30










    • $begingroup$
      Nice preview before starting. Sounds like it's gonna make this easier.
      $endgroup$
      – P.Jo
      Jan 4 at 17:11














    1












    1








    1





    $begingroup$

    Looks good to me. Some nit picks follow:



    A bit of formatting may help readability. For much of your answer you have pairs of equations which belong together, but the pairs which belong together aren't the pairs which appear close to one another.



    After "therefore $frac{1}{d}$ is an integer", you can just say "so $d=1$ and we're done". Remember that $d=1$ is the very thing we want to prove, so it's unnecessary to continue with more calculations once you've reached that conclusion.



    And personally I would probably have used the Euclidean algorithm instead.






    share|cite|improve this answer









    $endgroup$



    Looks good to me. Some nit picks follow:



    A bit of formatting may help readability. For much of your answer you have pairs of equations which belong together, but the pairs which belong together aren't the pairs which appear close to one another.



    After "therefore $frac{1}{d}$ is an integer", you can just say "so $d=1$ and we're done". Remember that $d=1$ is the very thing we want to prove, so it's unnecessary to continue with more calculations once you've reached that conclusion.



    And personally I would probably have used the Euclidean algorithm instead.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 15:39









    ArthurArthur

    122k7122210




    122k7122210












    • $begingroup$
      The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
      $endgroup$
      – P.Jo
      Jan 4 at 16:21










    • $begingroup$
      @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
      $endgroup$
      – Arthur
      Jan 4 at 16:30










    • $begingroup$
      Nice preview before starting. Sounds like it's gonna make this easier.
      $endgroup$
      – P.Jo
      Jan 4 at 17:11


















    • $begingroup$
      The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
      $endgroup$
      – P.Jo
      Jan 4 at 16:21










    • $begingroup$
      @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
      $endgroup$
      – Arthur
      Jan 4 at 16:30










    • $begingroup$
      Nice preview before starting. Sounds like it's gonna make this easier.
      $endgroup$
      – P.Jo
      Jan 4 at 17:11
















    $begingroup$
    The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
    $endgroup$
    – P.Jo
    Jan 4 at 16:21




    $begingroup$
    The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book.
    $endgroup$
    – P.Jo
    Jan 4 at 16:21












    $begingroup$
    @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
    $endgroup$
    – Arthur
    Jan 4 at 16:30




    $begingroup$
    @P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours.
    $endgroup$
    – Arthur
    Jan 4 at 16:30












    $begingroup$
    Nice preview before starting. Sounds like it's gonna make this easier.
    $endgroup$
    – P.Jo
    Jan 4 at 17:11




    $begingroup$
    Nice preview before starting. Sounds like it's gonna make this easier.
    $endgroup$
    – P.Jo
    Jan 4 at 17:11











    1












    $begingroup$

    Divide $2a+1$ into $9a+4$ giving $9a+4 = 9/2cdot (2a+1) - 1/2$ or $2(9a+4) = 9(2a+1)-1$, or $1= -2(9a+4) + 9(2a+1)$. So each divisor of $9a+4$ and $2a+1$ divides $1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
      $endgroup$
      – Michael Munta
      Jan 7 at 18:27










    • $begingroup$
      Indeed, you're right. Thanks.
      $endgroup$
      – Wuestenfux
      Jan 8 at 8:37
















    1












    $begingroup$

    Divide $2a+1$ into $9a+4$ giving $9a+4 = 9/2cdot (2a+1) - 1/2$ or $2(9a+4) = 9(2a+1)-1$, or $1= -2(9a+4) + 9(2a+1)$. So each divisor of $9a+4$ and $2a+1$ divides $1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
      $endgroup$
      – Michael Munta
      Jan 7 at 18:27










    • $begingroup$
      Indeed, you're right. Thanks.
      $endgroup$
      – Wuestenfux
      Jan 8 at 8:37














    1












    1








    1





    $begingroup$

    Divide $2a+1$ into $9a+4$ giving $9a+4 = 9/2cdot (2a+1) - 1/2$ or $2(9a+4) = 9(2a+1)-1$, or $1= -2(9a+4) + 9(2a+1)$. So each divisor of $9a+4$ and $2a+1$ divides $1$.






    share|cite|improve this answer











    $endgroup$



    Divide $2a+1$ into $9a+4$ giving $9a+4 = 9/2cdot (2a+1) - 1/2$ or $2(9a+4) = 9(2a+1)-1$, or $1= -2(9a+4) + 9(2a+1)$. So each divisor of $9a+4$ and $2a+1$ divides $1$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 8 at 8:36

























    answered Jan 4 at 15:39









    WuestenfuxWuestenfux

    5,3631513




    5,3631513












    • $begingroup$
      You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
      $endgroup$
      – Michael Munta
      Jan 7 at 18:27










    • $begingroup$
      Indeed, you're right. Thanks.
      $endgroup$
      – Wuestenfux
      Jan 8 at 8:37


















    • $begingroup$
      You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
      $endgroup$
      – Michael Munta
      Jan 7 at 18:27










    • $begingroup$
      Indeed, you're right. Thanks.
      $endgroup$
      – Wuestenfux
      Jan 8 at 8:37
















    $begingroup$
    You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
    $endgroup$
    – Michael Munta
    Jan 7 at 18:27




    $begingroup$
    You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed.
    $endgroup$
    – Michael Munta
    Jan 7 at 18:27












    $begingroup$
    Indeed, you're right. Thanks.
    $endgroup$
    – Wuestenfux
    Jan 8 at 8:37




    $begingroup$
    Indeed, you're right. Thanks.
    $endgroup$
    – Wuestenfux
    Jan 8 at 8:37











    0












    $begingroup$

    As Siong Thye Goh said, we can use the Euclidean algorithm to guide us to a proof. The Euclidean algorithm is based on the fact that $gcd(x,y)=gcd(x,y-tx)$. If we take $x=2a+1$, $y=9a+4$, and $t=4$, then we get $gcd(2a+1,9a+4)=gcd(2a+1,a)$. Applying it again with $x=a$, $y=2a+1$, $t=2$, we get $gcd(2a+1,a)=gcd(1,a)$. From there, it's easy to see that the gcd is $1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      As Siong Thye Goh said, we can use the Euclidean algorithm to guide us to a proof. The Euclidean algorithm is based on the fact that $gcd(x,y)=gcd(x,y-tx)$. If we take $x=2a+1$, $y=9a+4$, and $t=4$, then we get $gcd(2a+1,9a+4)=gcd(2a+1,a)$. Applying it again with $x=a$, $y=2a+1$, $t=2$, we get $gcd(2a+1,a)=gcd(1,a)$. From there, it's easy to see that the gcd is $1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As Siong Thye Goh said, we can use the Euclidean algorithm to guide us to a proof. The Euclidean algorithm is based on the fact that $gcd(x,y)=gcd(x,y-tx)$. If we take $x=2a+1$, $y=9a+4$, and $t=4$, then we get $gcd(2a+1,9a+4)=gcd(2a+1,a)$. Applying it again with $x=a$, $y=2a+1$, $t=2$, we get $gcd(2a+1,a)=gcd(1,a)$. From there, it's easy to see that the gcd is $1$.






        share|cite|improve this answer









        $endgroup$



        As Siong Thye Goh said, we can use the Euclidean algorithm to guide us to a proof. The Euclidean algorithm is based on the fact that $gcd(x,y)=gcd(x,y-tx)$. If we take $x=2a+1$, $y=9a+4$, and $t=4$, then we get $gcd(2a+1,9a+4)=gcd(2a+1,a)$. Applying it again with $x=a$, $y=2a+1$, $t=2$, we get $gcd(2a+1,a)=gcd(1,a)$. From there, it's easy to see that the gcd is $1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 19:25









        AcccumulationAcccumulation

        7,2152619




        7,2152619






























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