simple question regarding divides












0












$begingroup$


Suppose a is an integer. If 5|2a then 5|a. Prove



So i just suppose that 5|2a and so 2a=5b for some b in the integers but i dont know where to go from here.



Thanks for help






elementary-number-theory proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    $2a=5b$ implies that $5b$ is even.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:20










  • $begingroup$
    so b must be even, so b=2k and then that gives a=5k?
    $endgroup$
    – Harry
    Dec 4 '18 at 16:24










  • $begingroup$
    Yes. And in this way you are done.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:26










  • $begingroup$
    $5mid 5a-2(2a) = a $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 18:16
















0












$begingroup$


Suppose a is an integer. If 5|2a then 5|a. Prove



So i just suppose that 5|2a and so 2a=5b for some b in the integers but i dont know where to go from here.



Thanks for help






elementary-number-theory proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    $2a=5b$ implies that $5b$ is even.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:20










  • $begingroup$
    so b must be even, so b=2k and then that gives a=5k?
    $endgroup$
    – Harry
    Dec 4 '18 at 16:24










  • $begingroup$
    Yes. And in this way you are done.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:26










  • $begingroup$
    $5mid 5a-2(2a) = a $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 18:16














0












0








0





$begingroup$


Suppose a is an integer. If 5|2a then 5|a. Prove



So i just suppose that 5|2a and so 2a=5b for some b in the integers but i dont know where to go from here.



Thanks for help






elementary-number-theory proof-writing







share|cite|improve this question











$endgroup$




Suppose a is an integer. If 5|2a then 5|a. Prove



So i just suppose that 5|2a and so 2a=5b for some b in the integers but i dont know where to go from here.



Thanks for help






elementary-number-theory proof-writing




elementary-number-theory proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 16:26









Foobaz John

21.5k41351




21.5k41351










asked Dec 4 '18 at 16:18









HarryHarry

253




253












  • $begingroup$
    $2a=5b$ implies that $5b$ is even.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:20










  • $begingroup$
    so b must be even, so b=2k and then that gives a=5k?
    $endgroup$
    – Harry
    Dec 4 '18 at 16:24










  • $begingroup$
    Yes. And in this way you are done.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:26










  • $begingroup$
    $5mid 5a-2(2a) = a $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 18:16


















  • $begingroup$
    $2a=5b$ implies that $5b$ is even.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:20










  • $begingroup$
    so b must be even, so b=2k and then that gives a=5k?
    $endgroup$
    – Harry
    Dec 4 '18 at 16:24










  • $begingroup$
    Yes. And in this way you are done.
    $endgroup$
    – mfl
    Dec 4 '18 at 16:26










  • $begingroup$
    $5mid 5a-2(2a) = a $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 18:16
















$begingroup$
$2a=5b$ implies that $5b$ is even.
$endgroup$
– mfl
Dec 4 '18 at 16:20




$begingroup$
$2a=5b$ implies that $5b$ is even.
$endgroup$
– mfl
Dec 4 '18 at 16:20












$begingroup$
so b must be even, so b=2k and then that gives a=5k?
$endgroup$
– Harry
Dec 4 '18 at 16:24




$begingroup$
so b must be even, so b=2k and then that gives a=5k?
$endgroup$
– Harry
Dec 4 '18 at 16:24












$begingroup$
Yes. And in this way you are done.
$endgroup$
– mfl
Dec 4 '18 at 16:26




$begingroup$
Yes. And in this way you are done.
$endgroup$
– mfl
Dec 4 '18 at 16:26












$begingroup$
$5mid 5a-2(2a) = a $
$endgroup$
– Bill Dubuque
Dec 4 '18 at 18:16




$begingroup$
$5mid 5a-2(2a) = a $
$endgroup$
– Bill Dubuque
Dec 4 '18 at 18:16










3 Answers
3






active

oldest

votes


















0












$begingroup$

$$5vert 2a implies 2a = 5b implies a = frac{5b}{2}$$



By definition, $2a$ is even for $a in mathbb{Z}$. Therefore, $5b$ (and subsequently $b$) must be even.



$$b = 2c implies a = frac{5(2c)}{2} = 5c$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since $5$ and $2$ are relatively prime (by Bezout's lemma or a result of reversing Euclid's gcd algorithm), there exists integers $x, y$ such that
    $$
    5x+2y=1.
    $$
    Multiply through by $a$ to get that
    $$
    5ax+2ay=a
    $$
    As you noted $2a=5b$, so the LHS is divisible by $5$ and hence $5|a$ too.



    Note that the same reasoning can be used to prove Euclid's lemma which states that if $nmid ab$ and $n$ and $a$ are relatively prime, then $nmid b$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Here's another approach:



      Since $5mid 2a$, we also have $5mid 3cdot(2a)=6a$



      Also $5mid 5a$.



      So $5mid(6a-5a)=a$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
        $endgroup$
        – Harry
        Dec 4 '18 at 16:38






      • 1




        $begingroup$
        Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
        $endgroup$
        – paw88789
        Dec 4 '18 at 16:40











      Your Answer





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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

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      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      $$5vert 2a implies 2a = 5b implies a = frac{5b}{2}$$



      By definition, $2a$ is even for $a in mathbb{Z}$. Therefore, $5b$ (and subsequently $b$) must be even.



      $$b = 2c implies a = frac{5(2c)}{2} = 5c$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $$5vert 2a implies 2a = 5b implies a = frac{5b}{2}$$



        By definition, $2a$ is even for $a in mathbb{Z}$. Therefore, $5b$ (and subsequently $b$) must be even.



        $$b = 2c implies a = frac{5(2c)}{2} = 5c$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $$5vert 2a implies 2a = 5b implies a = frac{5b}{2}$$



          By definition, $2a$ is even for $a in mathbb{Z}$. Therefore, $5b$ (and subsequently $b$) must be even.



          $$b = 2c implies a = frac{5(2c)}{2} = 5c$$






          share|cite|improve this answer









          $endgroup$



          $$5vert 2a implies 2a = 5b implies a = frac{5b}{2}$$



          By definition, $2a$ is even for $a in mathbb{Z}$. Therefore, $5b$ (and subsequently $b$) must be even.



          $$b = 2c implies a = frac{5(2c)}{2} = 5c$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 16:26









          KM101KM101

          5,8711423




          5,8711423























              0












              $begingroup$

              Since $5$ and $2$ are relatively prime (by Bezout's lemma or a result of reversing Euclid's gcd algorithm), there exists integers $x, y$ such that
              $$
              5x+2y=1.
              $$
              Multiply through by $a$ to get that
              $$
              5ax+2ay=a
              $$
              As you noted $2a=5b$, so the LHS is divisible by $5$ and hence $5|a$ too.



              Note that the same reasoning can be used to prove Euclid's lemma which states that if $nmid ab$ and $n$ and $a$ are relatively prime, then $nmid b$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $5$ and $2$ are relatively prime (by Bezout's lemma or a result of reversing Euclid's gcd algorithm), there exists integers $x, y$ such that
                $$
                5x+2y=1.
                $$
                Multiply through by $a$ to get that
                $$
                5ax+2ay=a
                $$
                As you noted $2a=5b$, so the LHS is divisible by $5$ and hence $5|a$ too.



                Note that the same reasoning can be used to prove Euclid's lemma which states that if $nmid ab$ and $n$ and $a$ are relatively prime, then $nmid b$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $5$ and $2$ are relatively prime (by Bezout's lemma or a result of reversing Euclid's gcd algorithm), there exists integers $x, y$ such that
                  $$
                  5x+2y=1.
                  $$
                  Multiply through by $a$ to get that
                  $$
                  5ax+2ay=a
                  $$
                  As you noted $2a=5b$, so the LHS is divisible by $5$ and hence $5|a$ too.



                  Note that the same reasoning can be used to prove Euclid's lemma which states that if $nmid ab$ and $n$ and $a$ are relatively prime, then $nmid b$.






                  share|cite|improve this answer









                  $endgroup$



                  Since $5$ and $2$ are relatively prime (by Bezout's lemma or a result of reversing Euclid's gcd algorithm), there exists integers $x, y$ such that
                  $$
                  5x+2y=1.
                  $$
                  Multiply through by $a$ to get that
                  $$
                  5ax+2ay=a
                  $$
                  As you noted $2a=5b$, so the LHS is divisible by $5$ and hence $5|a$ too.



                  Note that the same reasoning can be used to prove Euclid's lemma which states that if $nmid ab$ and $n$ and $a$ are relatively prime, then $nmid b$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 16:25









                  Foobaz JohnFoobaz John

                  21.5k41351




                  21.5k41351























                      0












                      $begingroup$

                      Here's another approach:



                      Since $5mid 2a$, we also have $5mid 3cdot(2a)=6a$



                      Also $5mid 5a$.



                      So $5mid(6a-5a)=a$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
                        $endgroup$
                        – Harry
                        Dec 4 '18 at 16:38






                      • 1




                        $begingroup$
                        Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
                        $endgroup$
                        – paw88789
                        Dec 4 '18 at 16:40
















                      0












                      $begingroup$

                      Here's another approach:



                      Since $5mid 2a$, we also have $5mid 3cdot(2a)=6a$



                      Also $5mid 5a$.



                      So $5mid(6a-5a)=a$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
                        $endgroup$
                        – Harry
                        Dec 4 '18 at 16:38






                      • 1




                        $begingroup$
                        Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
                        $endgroup$
                        – paw88789
                        Dec 4 '18 at 16:40














                      0












                      0








                      0





                      $begingroup$

                      Here's another approach:



                      Since $5mid 2a$, we also have $5mid 3cdot(2a)=6a$



                      Also $5mid 5a$.



                      So $5mid(6a-5a)=a$






                      share|cite|improve this answer









                      $endgroup$



                      Here's another approach:



                      Since $5mid 2a$, we also have $5mid 3cdot(2a)=6a$



                      Also $5mid 5a$.



                      So $5mid(6a-5a)=a$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 4 '18 at 16:29









                      paw88789paw88789

                      29k12349




                      29k12349












                      • $begingroup$
                        if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
                        $endgroup$
                        – Harry
                        Dec 4 '18 at 16:38






                      • 1




                        $begingroup$
                        Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
                        $endgroup$
                        – paw88789
                        Dec 4 '18 at 16:40


















                      • $begingroup$
                        if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
                        $endgroup$
                        – Harry
                        Dec 4 '18 at 16:38






                      • 1




                        $begingroup$
                        Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
                        $endgroup$
                        – paw88789
                        Dec 4 '18 at 16:40
















                      $begingroup$
                      if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
                      $endgroup$
                      – Harry
                      Dec 4 '18 at 16:38




                      $begingroup$
                      if you go from 5|2a to 5|6a why cant you go from 5|2a to 5|1/2*2*a?
                      $endgroup$
                      – Harry
                      Dec 4 '18 at 16:38




                      1




                      1




                      $begingroup$
                      Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
                      $endgroup$
                      – paw88789
                      Dec 4 '18 at 16:40




                      $begingroup$
                      Because for any integers $a,b,c$, $amid b$ implies $amid bc$; but not necessarily $amid b/c$.
                      $endgroup$
                      – paw88789
                      Dec 4 '18 at 16:40


















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