Examples where Strong Induction is more useful than “Regular” Induction?

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The Principles of "Regular" and Strong Induction are equivalent (see for example here).



But are there any examples where something is more easily/elegantly proven with Strong rather than "Regular" Induction?



(And if not, what use is Strong Induction -- why do we even bother mentioning it?)






proof-writing induction







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    up vote
    1
    down vote

    favorite












    The Principles of "Regular" and Strong Induction are equivalent (see for example here).



    But are there any examples where something is more easily/elegantly proven with Strong rather than "Regular" Induction?



    (And if not, what use is Strong Induction -- why do we even bother mentioning it?)






    proof-writing induction







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      The Principles of "Regular" and Strong Induction are equivalent (see for example here).



      But are there any examples where something is more easily/elegantly proven with Strong rather than "Regular" Induction?



      (And if not, what use is Strong Induction -- why do we even bother mentioning it?)






      proof-writing induction







      share|cite|improve this question













      The Principles of "Regular" and Strong Induction are equivalent (see for example here).



      But are there any examples where something is more easily/elegantly proven with Strong rather than "Regular" Induction?



      (And if not, what use is Strong Induction -- why do we even bother mentioning it?)






      proof-writing induction




      proof-writing induction






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      share|cite|improve this question











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      asked 1 hour ago









      dtcm840

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          One common proof of the existence component of the fundamental theorem of arithmetic uses strong induction.



          Suppose that every natural number less than n factors as a product of primes.



          If $n$ is prime then it factors uniquely as a product of primes. If $n$ is not prime then it can be written as the product of two numbers less than $n$ which by the hypothesis of strong induction, can both be written as the product of primes. Hence $n$ can be written as a product of primes.



          Notice that the argument wouldn't work with 'weak' induction.






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            up vote
            4
            down vote



            accepted










            One common proof of the existence component of the fundamental theorem of arithmetic uses strong induction.



            Suppose that every natural number less than n factors as a product of primes.



            If $n$ is prime then it factors uniquely as a product of primes. If $n$ is not prime then it can be written as the product of two numbers less than $n$ which by the hypothesis of strong induction, can both be written as the product of primes. Hence $n$ can be written as a product of primes.



            Notice that the argument wouldn't work with 'weak' induction.






            share|cite|improve this answer
























              up vote
              4
              down vote



              accepted










              One common proof of the existence component of the fundamental theorem of arithmetic uses strong induction.



              Suppose that every natural number less than n factors as a product of primes.



              If $n$ is prime then it factors uniquely as a product of primes. If $n$ is not prime then it can be written as the product of two numbers less than $n$ which by the hypothesis of strong induction, can both be written as the product of primes. Hence $n$ can be written as a product of primes.



              Notice that the argument wouldn't work with 'weak' induction.






              share|cite|improve this answer






















                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                One common proof of the existence component of the fundamental theorem of arithmetic uses strong induction.



                Suppose that every natural number less than n factors as a product of primes.



                If $n$ is prime then it factors uniquely as a product of primes. If $n$ is not prime then it can be written as the product of two numbers less than $n$ which by the hypothesis of strong induction, can both be written as the product of primes. Hence $n$ can be written as a product of primes.



                Notice that the argument wouldn't work with 'weak' induction.






                share|cite|improve this answer












                One common proof of the existence component of the fundamental theorem of arithmetic uses strong induction.



                Suppose that every natural number less than n factors as a product of primes.



                If $n$ is prime then it factors uniquely as a product of primes. If $n$ is not prime then it can be written as the product of two numbers less than $n$ which by the hypothesis of strong induction, can both be written as the product of primes. Hence $n$ can be written as a product of primes.



                Notice that the argument wouldn't work with 'weak' induction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Bernard W

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