Explaining the proof of Fibonacci number using inductive reasoning












0












$begingroup$


Fibonacci numbers are defined as follows.



$$F_{1}= F_{2} = 1$$
When $n geq 3$, $$F_{n} = F_{n-1} + F_{n-2}$$




Task: Prove the following statement using mathematical induction:




  • When $n geq 2$, $$F_{n-1}F_{n+1} = F_{n}^2 + (-1)^n$$





The Base Case:



enter image description here




The Inductive Step:



enter image description here




I'm really confused about the inductive step. The answer makes absolutely no sense to me.



Questions:




  1. For the inductive step, why is the yellow area equal to the green area?

  2. For the inductive step, how do we arrive at the purple and red statement?


I think the answer given to me is too simplified and doesn't demonstrate a clear logical reasoning.






induction fibonacci-numbers







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  • $begingroup$
    My apologies for anyone who is color blind. It's just easier for me to color code the statements.
    $endgroup$
    – potatoguy
    Dec 9 '18 at 22:18
















0












$begingroup$


Fibonacci numbers are defined as follows.



$$F_{1}= F_{2} = 1$$
When $n geq 3$, $$F_{n} = F_{n-1} + F_{n-2}$$




Task: Prove the following statement using mathematical induction:




  • When $n geq 2$, $$F_{n-1}F_{n+1} = F_{n}^2 + (-1)^n$$





The Base Case:



enter image description here




The Inductive Step:



enter image description here




I'm really confused about the inductive step. The answer makes absolutely no sense to me.



Questions:




  1. For the inductive step, why is the yellow area equal to the green area?

  2. For the inductive step, how do we arrive at the purple and red statement?


I think the answer given to me is too simplified and doesn't demonstrate a clear logical reasoning.






induction fibonacci-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    My apologies for anyone who is color blind. It's just easier for me to color code the statements.
    $endgroup$
    – potatoguy
    Dec 9 '18 at 22:18














0












0








0





$begingroup$


Fibonacci numbers are defined as follows.



$$F_{1}= F_{2} = 1$$
When $n geq 3$, $$F_{n} = F_{n-1} + F_{n-2}$$




Task: Prove the following statement using mathematical induction:




  • When $n geq 2$, $$F_{n-1}F_{n+1} = F_{n}^2 + (-1)^n$$





The Base Case:



enter image description here




The Inductive Step:



enter image description here




I'm really confused about the inductive step. The answer makes absolutely no sense to me.



Questions:




  1. For the inductive step, why is the yellow area equal to the green area?

  2. For the inductive step, how do we arrive at the purple and red statement?


I think the answer given to me is too simplified and doesn't demonstrate a clear logical reasoning.






induction fibonacci-numbers







share|cite|improve this question











$endgroup$




Fibonacci numbers are defined as follows.



$$F_{1}= F_{2} = 1$$
When $n geq 3$, $$F_{n} = F_{n-1} + F_{n-2}$$




Task: Prove the following statement using mathematical induction:




  • When $n geq 2$, $$F_{n-1}F_{n+1} = F_{n}^2 + (-1)^n$$





The Base Case:



enter image description here




The Inductive Step:



enter image description here




I'm really confused about the inductive step. The answer makes absolutely no sense to me.



Questions:




  1. For the inductive step, why is the yellow area equal to the green area?

  2. For the inductive step, how do we arrive at the purple and red statement?


I think the answer given to me is too simplified and doesn't demonstrate a clear logical reasoning.






induction fibonacci-numbers




induction fibonacci-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 11:44









amWhy

1




1










asked Dec 9 '18 at 22:15









potatoguypotatoguy

525




525












  • $begingroup$
    My apologies for anyone who is color blind. It's just easier for me to color code the statements.
    $endgroup$
    – potatoguy
    Dec 9 '18 at 22:18


















  • $begingroup$
    My apologies for anyone who is color blind. It's just easier for me to color code the statements.
    $endgroup$
    – potatoguy
    Dec 9 '18 at 22:18
















$begingroup$
My apologies for anyone who is color blind. It's just easier for me to color code the statements.
$endgroup$
– potatoguy
Dec 9 '18 at 22:18




$begingroup$
My apologies for anyone who is color blind. It's just easier for me to color code the statements.
$endgroup$
– potatoguy
Dec 9 '18 at 22:18










2 Answers
2






active

oldest

votes


















0












$begingroup$


  1. The green part is
    $$f^2(k)+f(k)f(k+1)=(f(k-1)+f(k))f(k+1)-(-1)^k=f^2(k+1)+(-1)^{k+1},$$
    i.e. the yellow part. Note that it's not supposed to be obvious at this stage that green = yellow; that they are is what the subsequent lines show. The important part is that red - yellow; that's why the inductive step works. The way the proof of this actually starts is by rewriting $f(k+2)$ as a sum to get the green expression.


  2. The purple part is equal to the line above it, by summing the coefficients of $f(k+1)$ and taking care with powers of $-1$. The red part rewrites the summed coefficient from the recursion relation, and then notes a square is present.







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    1. Because, by definition, $f(k+2)=f(k+1)+f(k)$.

    2. First of all, $-(-1)^k=(-1)times(-1)^k=(-1)^{k+1}$. Then,$$f(k)f(k+1)+f(k-1)f(k+1)=bigl(f(k)+f(k-1)bigr)f(k+1).$$So,$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k)+f(k-1)bigr)f(k+1)+(-1)^{k+1}.tag1$$But, by definition, $f(k+1)=f(k)+f(k-1)$. So, $(1)$ becomes$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k+1)bigr)^2+(-1)^{k+1}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      1. asked about the green expression, not the blue one.
      $endgroup$
      – J.G.
      Dec 9 '18 at 22:25










    • $begingroup$
      I've edited my answer. I hope that everything is clear now.
      $endgroup$
      – José Carlos Santos
      Dec 9 '18 at 22:29











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$


    1. The green part is
      $$f^2(k)+f(k)f(k+1)=(f(k-1)+f(k))f(k+1)-(-1)^k=f^2(k+1)+(-1)^{k+1},$$
      i.e. the yellow part. Note that it's not supposed to be obvious at this stage that green = yellow; that they are is what the subsequent lines show. The important part is that red - yellow; that's why the inductive step works. The way the proof of this actually starts is by rewriting $f(k+2)$ as a sum to get the green expression.


    2. The purple part is equal to the line above it, by summing the coefficients of $f(k+1)$ and taking care with powers of $-1$. The red part rewrites the summed coefficient from the recursion relation, and then notes a square is present.







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$


      1. The green part is
        $$f^2(k)+f(k)f(k+1)=(f(k-1)+f(k))f(k+1)-(-1)^k=f^2(k+1)+(-1)^{k+1},$$
        i.e. the yellow part. Note that it's not supposed to be obvious at this stage that green = yellow; that they are is what the subsequent lines show. The important part is that red - yellow; that's why the inductive step works. The way the proof of this actually starts is by rewriting $f(k+2)$ as a sum to get the green expression.


      2. The purple part is equal to the line above it, by summing the coefficients of $f(k+1)$ and taking care with powers of $-1$. The red part rewrites the summed coefficient from the recursion relation, and then notes a square is present.







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$


        1. The green part is
          $$f^2(k)+f(k)f(k+1)=(f(k-1)+f(k))f(k+1)-(-1)^k=f^2(k+1)+(-1)^{k+1},$$
          i.e. the yellow part. Note that it's not supposed to be obvious at this stage that green = yellow; that they are is what the subsequent lines show. The important part is that red - yellow; that's why the inductive step works. The way the proof of this actually starts is by rewriting $f(k+2)$ as a sum to get the green expression.


        2. The purple part is equal to the line above it, by summing the coefficients of $f(k+1)$ and taking care with powers of $-1$. The red part rewrites the summed coefficient from the recursion relation, and then notes a square is present.







        share|cite|improve this answer









        $endgroup$




        1. The green part is
          $$f^2(k)+f(k)f(k+1)=(f(k-1)+f(k))f(k+1)-(-1)^k=f^2(k+1)+(-1)^{k+1},$$
          i.e. the yellow part. Note that it's not supposed to be obvious at this stage that green = yellow; that they are is what the subsequent lines show. The important part is that red - yellow; that's why the inductive step works. The way the proof of this actually starts is by rewriting $f(k+2)$ as a sum to get the green expression.


        2. The purple part is equal to the line above it, by summing the coefficients of $f(k+1)$ and taking care with powers of $-1$. The red part rewrites the summed coefficient from the recursion relation, and then notes a square is present.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 22:25









        J.G.J.G.

        27.1k22843




        27.1k22843























            0












            $begingroup$


            1. Because, by definition, $f(k+2)=f(k+1)+f(k)$.

            2. First of all, $-(-1)^k=(-1)times(-1)^k=(-1)^{k+1}$. Then,$$f(k)f(k+1)+f(k-1)f(k+1)=bigl(f(k)+f(k-1)bigr)f(k+1).$$So,$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k)+f(k-1)bigr)f(k+1)+(-1)^{k+1}.tag1$$But, by definition, $f(k+1)=f(k)+f(k-1)$. So, $(1)$ becomes$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k+1)bigr)^2+(-1)^{k+1}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              1. asked about the green expression, not the blue one.
              $endgroup$
              – J.G.
              Dec 9 '18 at 22:25










            • $begingroup$
              I've edited my answer. I hope that everything is clear now.
              $endgroup$
              – José Carlos Santos
              Dec 9 '18 at 22:29
















            0












            $begingroup$


            1. Because, by definition, $f(k+2)=f(k+1)+f(k)$.

            2. First of all, $-(-1)^k=(-1)times(-1)^k=(-1)^{k+1}$. Then,$$f(k)f(k+1)+f(k-1)f(k+1)=bigl(f(k)+f(k-1)bigr)f(k+1).$$So,$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k)+f(k-1)bigr)f(k+1)+(-1)^{k+1}.tag1$$But, by definition, $f(k+1)=f(k)+f(k-1)$. So, $(1)$ becomes$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k+1)bigr)^2+(-1)^{k+1}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              1. asked about the green expression, not the blue one.
              $endgroup$
              – J.G.
              Dec 9 '18 at 22:25










            • $begingroup$
              I've edited my answer. I hope that everything is clear now.
              $endgroup$
              – José Carlos Santos
              Dec 9 '18 at 22:29














            0












            0








            0





            $begingroup$


            1. Because, by definition, $f(k+2)=f(k+1)+f(k)$.

            2. First of all, $-(-1)^k=(-1)times(-1)^k=(-1)^{k+1}$. Then,$$f(k)f(k+1)+f(k-1)f(k+1)=bigl(f(k)+f(k-1)bigr)f(k+1).$$So,$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k)+f(k-1)bigr)f(k+1)+(-1)^{k+1}.tag1$$But, by definition, $f(k+1)=f(k)+f(k-1)$. So, $(1)$ becomes$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k+1)bigr)^2+(-1)^{k+1}.$$






            share|cite|improve this answer











            $endgroup$




            1. Because, by definition, $f(k+2)=f(k+1)+f(k)$.

            2. First of all, $-(-1)^k=(-1)times(-1)^k=(-1)^{k+1}$. Then,$$f(k)f(k+1)+f(k-1)f(k+1)=bigl(f(k)+f(k-1)bigr)f(k+1).$$So,$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k)+f(k-1)bigr)f(k+1)+(-1)^{k+1}.tag1$$But, by definition, $f(k+1)=f(k)+f(k-1)$. So, $(1)$ becomes$$f(k)f(k+1)+f(k-1)f(k+1)-(-1)^k=bigl(f(k+1)bigr)^2+(-1)^{k+1}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 22:29

























            answered Dec 9 '18 at 22:23









            José Carlos SantosJosé Carlos Santos

            162k22129233




            162k22129233












            • $begingroup$
              1. asked about the green expression, not the blue one.
              $endgroup$
              – J.G.
              Dec 9 '18 at 22:25










            • $begingroup$
              I've edited my answer. I hope that everything is clear now.
              $endgroup$
              – José Carlos Santos
              Dec 9 '18 at 22:29


















            • $begingroup$
              1. asked about the green expression, not the blue one.
              $endgroup$
              – J.G.
              Dec 9 '18 at 22:25










            • $begingroup$
              I've edited my answer. I hope that everything is clear now.
              $endgroup$
              – José Carlos Santos
              Dec 9 '18 at 22:29
















            $begingroup$
            1. asked about the green expression, not the blue one.
            $endgroup$
            – J.G.
            Dec 9 '18 at 22:25




            $begingroup$
            1. asked about the green expression, not the blue one.
            $endgroup$
            – J.G.
            Dec 9 '18 at 22:25












            $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Dec 9 '18 at 22:29




            $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Dec 9 '18 at 22:29


















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