An error in a proof due to variable creep?












2














I'm working through some exam practice questions and I came across this one:




Identify the error in the following “proof.”



Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.




Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.



Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.






discrete-mathematics







share|cite|improve this question



























    2














    I'm working through some exam practice questions and I came across this one:




    Identify the error in the following “proof.”



    Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
    If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
    us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
    implies that m|1, and therefore m = ±1.




    Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
    And now the substitution doesn't work and we cannot proceed.



    Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.






    discrete-mathematics







    share|cite|improve this question

























      2












      2








      2







      I'm working through some exam practice questions and I came across this one:




      Identify the error in the following “proof.”



      Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
      If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
      us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
      implies that m|1, and therefore m = ±1.




      Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
      And now the substitution doesn't work and we cannot proceed.



      Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.






      discrete-mathematics







      share|cite|improve this question













      I'm working through some exam practice questions and I came across this one:




      Identify the error in the following “proof.”



      Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
      If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
      us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
      implies that m|1, and therefore m = ±1.




      Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
      And now the substitution doesn't work and we cannot proceed.



      Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.






      discrete-mathematics




      discrete-mathematics






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 26 '18 at 19:08









      metisMusings

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          You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:




          We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.







          share|cite|improve this answer





















          • Thanks, this helps a lot :)
            – metisMusings
            Nov 26 '18 at 19:20











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          1 Answer
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          active

          oldest

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          active

          oldest

          votes









          1














          You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:




          We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.







          share|cite|improve this answer





















          • Thanks, this helps a lot :)
            – metisMusings
            Nov 26 '18 at 19:20
















          1














          You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:




          We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.







          share|cite|improve this answer





















          • Thanks, this helps a lot :)
            – metisMusings
            Nov 26 '18 at 19:20














          1












          1








          1






          You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:




          We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.







          share|cite|improve this answer












          You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:




          We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.








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          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 19:16









          Patrick Stevens

          28.5k52874




          28.5k52874












          • Thanks, this helps a lot :)
            – metisMusings
            Nov 26 '18 at 19:20


















          • Thanks, this helps a lot :)
            – metisMusings
            Nov 26 '18 at 19:20
















          Thanks, this helps a lot :)
          – metisMusings
          Nov 26 '18 at 19:20




          Thanks, this helps a lot :)
          – metisMusings
          Nov 26 '18 at 19:20


















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