When was the generalization easier to prove than the specific case? [duplicate]












1












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This question already has an answer here:




  • Generalizing a problem to make it easier

    45 answers




I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.



But I don't remember any such situation and would like to know of one or more.



(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)










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marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    mathoverflow.net/questions/40005/…
    $endgroup$
    – Wojowu
    Feb 14 at 15:41






  • 2




    $begingroup$
    @HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment.
    $endgroup$
    – LSpice
    Feb 14 at 16:06






  • 5




    $begingroup$
    @LSpice It gets added automatically when you vote to close with the reason being duplicate.
    $endgroup$
    – Harry Gindi
    Feb 14 at 16:11










  • $begingroup$
    Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
    $endgroup$
    – Wojowu
    Feb 14 at 16:35










  • $begingroup$
    Ah, thanks for the pointers to the duplicate, I didn't find that.
    $endgroup$
    – davidbak
    Feb 14 at 18:09
















1












$begingroup$



This question already has an answer here:




  • Generalizing a problem to make it easier

    45 answers




I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.



But I don't remember any such situation and would like to know of one or more.



(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)










share|cite|improve this question











$endgroup$



marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    mathoverflow.net/questions/40005/…
    $endgroup$
    – Wojowu
    Feb 14 at 15:41






  • 2




    $begingroup$
    @HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment.
    $endgroup$
    – LSpice
    Feb 14 at 16:06






  • 5




    $begingroup$
    @LSpice It gets added automatically when you vote to close with the reason being duplicate.
    $endgroup$
    – Harry Gindi
    Feb 14 at 16:11










  • $begingroup$
    Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
    $endgroup$
    – Wojowu
    Feb 14 at 16:35










  • $begingroup$
    Ah, thanks for the pointers to the duplicate, I didn't find that.
    $endgroup$
    – davidbak
    Feb 14 at 18:09














1












1








1





$begingroup$



This question already has an answer here:




  • Generalizing a problem to make it easier

    45 answers




I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.



But I don't remember any such situation and would like to know of one or more.



(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Generalizing a problem to make it easier

    45 answers




I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.



But I don't remember any such situation and would like to know of one or more.



(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)





This question already has an answer here:




  • Generalizing a problem to make it easier

    45 answers








soft-question big-list examples






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share|cite|improve this question













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share|cite|improve this question








edited Feb 14 at 15:40


























community wiki





davidbak





marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    $begingroup$
    mathoverflow.net/questions/40005/…
    $endgroup$
    – Wojowu
    Feb 14 at 15:41






  • 2




    $begingroup$
    @HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment.
    $endgroup$
    – LSpice
    Feb 14 at 16:06






  • 5




    $begingroup$
    @LSpice It gets added automatically when you vote to close with the reason being duplicate.
    $endgroup$
    – Harry Gindi
    Feb 14 at 16:11










  • $begingroup$
    Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
    $endgroup$
    – Wojowu
    Feb 14 at 16:35










  • $begingroup$
    Ah, thanks for the pointers to the duplicate, I didn't find that.
    $endgroup$
    – davidbak
    Feb 14 at 18:09














  • 2




    $begingroup$
    mathoverflow.net/questions/40005/…
    $endgroup$
    – Wojowu
    Feb 14 at 15:41






  • 2




    $begingroup$
    @HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment.
    $endgroup$
    – LSpice
    Feb 14 at 16:06






  • 5




    $begingroup$
    @LSpice It gets added automatically when you vote to close with the reason being duplicate.
    $endgroup$
    – Harry Gindi
    Feb 14 at 16:11










  • $begingroup$
    Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
    $endgroup$
    – Wojowu
    Feb 14 at 16:35










  • $begingroup$
    Ah, thanks for the pointers to the duplicate, I didn't find that.
    $endgroup$
    – davidbak
    Feb 14 at 18:09








2




2




$begingroup$
mathoverflow.net/questions/40005/…
$endgroup$
– Wojowu
Feb 14 at 15:41




$begingroup$
mathoverflow.net/questions/40005/…
$endgroup$
– Wojowu
Feb 14 at 15:41




2




2




$begingroup$
@HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment.
$endgroup$
– LSpice
Feb 14 at 16:06




$begingroup$
@HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment.
$endgroup$
– LSpice
Feb 14 at 16:06




5




5




$begingroup$
@LSpice It gets added automatically when you vote to close with the reason being duplicate.
$endgroup$
– Harry Gindi
Feb 14 at 16:11




$begingroup$
@LSpice It gets added automatically when you vote to close with the reason being duplicate.
$endgroup$
– Harry Gindi
Feb 14 at 16:11












$begingroup$
Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
$endgroup$
– Wojowu
Feb 14 at 16:35




$begingroup$
Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
$endgroup$
– Wojowu
Feb 14 at 16:35












$begingroup$
Ah, thanks for the pointers to the duplicate, I didn't find that.
$endgroup$
– davidbak
Feb 14 at 18:09




$begingroup$
Ah, thanks for the pointers to the duplicate, I didn't find that.
$endgroup$
– davidbak
Feb 14 at 18:09










2 Answers
2






active

oldest

votes


















8












$begingroup$

Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
and then specialize your result to $z=1$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$


    1. There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.


    Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.



    Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$




    1. A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.


    Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The first example is terrific, thank you!
      $endgroup$
      – davidbak
      Feb 14 at 18:10


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
    and then specialize your result to $z=1$.






    share|cite|improve this answer











    $endgroup$


















      8












      $begingroup$

      Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
      and then specialize your result to $z=1$.






      share|cite|improve this answer











      $endgroup$
















        8












        8








        8





        $begingroup$

        Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
        and then specialize your result to $z=1$.






        share|cite|improve this answer











        $endgroup$



        Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
        and then specialize your result to $z=1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Feb 14 at 15:23


























        community wiki





        Robert Israel
























            1












            $begingroup$


            1. There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.


            Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.



            Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$




            1. A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.


            Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The first example is terrific, thank you!
              $endgroup$
              – davidbak
              Feb 14 at 18:10
















            1












            $begingroup$


            1. There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.


            Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.



            Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$




            1. A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.


            Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The first example is terrific, thank you!
              $endgroup$
              – davidbak
              Feb 14 at 18:10














            1












            1








            1





            $begingroup$


            1. There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.


            Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.



            Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$




            1. A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.


            Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)






            share|cite|improve this answer











            $endgroup$




            1. There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.


            Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.



            Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$




            1. A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.


            Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            answered Feb 14 at 17:24


























            community wiki





            Max













            • $begingroup$
              The first example is terrific, thank you!
              $endgroup$
              – davidbak
              Feb 14 at 18:10


















            • $begingroup$
              The first example is terrific, thank you!
              $endgroup$
              – davidbak
              Feb 14 at 18:10
















            $begingroup$
            The first example is terrific, thank you!
            $endgroup$
            – davidbak
            Feb 14 at 18:10




            $begingroup$
            The first example is terrific, thank you!
            $endgroup$
            – davidbak
            Feb 14 at 18:10



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