Correct Application of Inductive Hypothesis












1












$begingroup$


To provide full context of the practice question I'm attempting, it is as follows:




For every integer n such that n ≥ 8, there exist nonnegative integers an and bn such that 3an + 5bn = n.



Write a proof of the claim using the strong form of mathematical induction with the integer 10 as breakpoint — such that that n = 8, n = 9 and n = 10 would all be considered in the basis.




This is my attempt at the proof, but I can't seem to get past applying the I.H. in order to reach my goal in the inductive step.




Basis

If n = 8, then 3a8 + 5b8 = 8

If n = 9, then 3a9 + 5b9 = 9

If n = 10, then 3a10 + 5b10 = 10



Inductive Hypothesis

Let k be an integer such that k ≥ 10. It is necessary and sufficient to use the following:




  • Inductive Hypothesis: 3an + 5bn = n for every integer n such that 8 ≤ n ≤ k.


Inductive Claim
3ak+1 + 5bk+1 = k+1




Now I have to complete the remainder of the proof but I'm not even sure if I was proving it correctly so far or how to complete it from where I'm at.



Any guidance would be much appreciated!






discrete-mathematics induction







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    To provide full context of the practice question I'm attempting, it is as follows:




    For every integer n such that n ≥ 8, there exist nonnegative integers an and bn such that 3an + 5bn = n.



    Write a proof of the claim using the strong form of mathematical induction with the integer 10 as breakpoint — such that that n = 8, n = 9 and n = 10 would all be considered in the basis.




    This is my attempt at the proof, but I can't seem to get past applying the I.H. in order to reach my goal in the inductive step.




    Basis

    If n = 8, then 3a8 + 5b8 = 8

    If n = 9, then 3a9 + 5b9 = 9

    If n = 10, then 3a10 + 5b10 = 10



    Inductive Hypothesis

    Let k be an integer such that k ≥ 10. It is necessary and sufficient to use the following:




    • Inductive Hypothesis: 3an + 5bn = n for every integer n such that 8 ≤ n ≤ k.


    Inductive Claim
    3ak+1 + 5bk+1 = k+1




    Now I have to complete the remainder of the proof but I'm not even sure if I was proving it correctly so far or how to complete it from where I'm at.



    Any guidance would be much appreciated!






    discrete-mathematics induction







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      To provide full context of the practice question I'm attempting, it is as follows:




      For every integer n such that n ≥ 8, there exist nonnegative integers an and bn such that 3an + 5bn = n.



      Write a proof of the claim using the strong form of mathematical induction with the integer 10 as breakpoint — such that that n = 8, n = 9 and n = 10 would all be considered in the basis.




      This is my attempt at the proof, but I can't seem to get past applying the I.H. in order to reach my goal in the inductive step.




      Basis

      If n = 8, then 3a8 + 5b8 = 8

      If n = 9, then 3a9 + 5b9 = 9

      If n = 10, then 3a10 + 5b10 = 10



      Inductive Hypothesis

      Let k be an integer such that k ≥ 10. It is necessary and sufficient to use the following:




      • Inductive Hypothesis: 3an + 5bn = n for every integer n such that 8 ≤ n ≤ k.


      Inductive Claim
      3ak+1 + 5bk+1 = k+1




      Now I have to complete the remainder of the proof but I'm not even sure if I was proving it correctly so far or how to complete it from where I'm at.



      Any guidance would be much appreciated!






      discrete-mathematics induction







      share|cite|improve this question











      $endgroup$




      To provide full context of the practice question I'm attempting, it is as follows:




      For every integer n such that n ≥ 8, there exist nonnegative integers an and bn such that 3an + 5bn = n.



      Write a proof of the claim using the strong form of mathematical induction with the integer 10 as breakpoint — such that that n = 8, n = 9 and n = 10 would all be considered in the basis.




      This is my attempt at the proof, but I can't seem to get past applying the I.H. in order to reach my goal in the inductive step.




      Basis

      If n = 8, then 3a8 + 5b8 = 8

      If n = 9, then 3a9 + 5b9 = 9

      If n = 10, then 3a10 + 5b10 = 10



      Inductive Hypothesis

      Let k be an integer such that k ≥ 10. It is necessary and sufficient to use the following:




      • Inductive Hypothesis: 3an + 5bn = n for every integer n such that 8 ≤ n ≤ k.


      Inductive Claim
      3ak+1 + 5bk+1 = k+1




      Now I have to complete the remainder of the proof but I'm not even sure if I was proving it correctly so far or how to complete it from where I'm at.



      Any guidance would be much appreciated!






      discrete-mathematics induction




      discrete-mathematics induction






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 0:11







      man2006

















      asked Jan 10 at 23:05









      man2006man2006

      175




      175






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Note that, in particular, there are $a_{k-2}$ and $b_{k-2}$ such that $3a_{k-2} + 5b_{k-2} = k-2$. Now add $3$ to both sides.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
            $endgroup$
            – man2006
            Jan 10 at 23:32










          • $begingroup$
            @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
            $endgroup$
            – Bram28
            Jan 10 at 23:42










          • $begingroup$
            In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
            $endgroup$
            – man2006
            Jan 10 at 23:49










          • $begingroup$
            @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
            $endgroup$
            – user3482749
            Jan 11 at 9:27



















          0












          $begingroup$

          For the step, see answer by @user3482749



          But other than the inductive step, you also need to show that you can find actual values for $a_n$ and $b_n$ (and the others) in your base cases to make sure these work!



          For example, to satisfy that $3a_8+5b_8=8$, you can set $a_8=1$ and $b_8=1$



          Now do this for the other base cases as well






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
            $endgroup$
            – man2006
            Jan 10 at 23:18












          • $begingroup$
            @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
            $endgroup$
            – Bram28
            Jan 10 at 23:39










          • $begingroup$
            Got it, I just have to keep trying different numbers that work. Thank you for your help!
            $endgroup$
            – man2006
            Jan 10 at 23:45










          • $begingroup$
            @man2006 you're welcome :)
            $endgroup$
            – Bram28
            Jan 11 at 3:03











          Your Answer





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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Note that, in particular, there are $a_{k-2}$ and $b_{k-2}$ such that $3a_{k-2} + 5b_{k-2} = k-2$. Now add $3$ to both sides.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
            $endgroup$
            – man2006
            Jan 10 at 23:32










          • $begingroup$
            @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
            $endgroup$
            – Bram28
            Jan 10 at 23:42










          • $begingroup$
            In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
            $endgroup$
            – man2006
            Jan 10 at 23:49










          • $begingroup$
            @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
            $endgroup$
            – user3482749
            Jan 11 at 9:27
















          1












          $begingroup$

          Note that, in particular, there are $a_{k-2}$ and $b_{k-2}$ such that $3a_{k-2} + 5b_{k-2} = k-2$. Now add $3$ to both sides.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
            $endgroup$
            – man2006
            Jan 10 at 23:32










          • $begingroup$
            @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
            $endgroup$
            – Bram28
            Jan 10 at 23:42










          • $begingroup$
            In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
            $endgroup$
            – man2006
            Jan 10 at 23:49










          • $begingroup$
            @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
            $endgroup$
            – user3482749
            Jan 11 at 9:27














          1












          1








          1





          $begingroup$

          Note that, in particular, there are $a_{k-2}$ and $b_{k-2}$ such that $3a_{k-2} + 5b_{k-2} = k-2$. Now add $3$ to both sides.






          share|cite|improve this answer











          $endgroup$



          Note that, in particular, there are $a_{k-2}$ and $b_{k-2}$ such that $3a_{k-2} + 5b_{k-2} = k-2$. Now add $3$ to both sides.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 23:11









          Bram28

          63.2k44793




          63.2k44793










          answered Jan 10 at 23:09









          user3482749user3482749

          4,296919




          4,296919












          • $begingroup$
            Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
            $endgroup$
            – man2006
            Jan 10 at 23:32










          • $begingroup$
            @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
            $endgroup$
            – Bram28
            Jan 10 at 23:42










          • $begingroup$
            In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
            $endgroup$
            – man2006
            Jan 10 at 23:49










          • $begingroup$
            @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
            $endgroup$
            – user3482749
            Jan 11 at 9:27


















          • $begingroup$
            Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
            $endgroup$
            – man2006
            Jan 10 at 23:32










          • $begingroup$
            @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
            $endgroup$
            – Bram28
            Jan 10 at 23:42










          • $begingroup$
            In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
            $endgroup$
            – man2006
            Jan 10 at 23:49










          • $begingroup$
            @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
            $endgroup$
            – user3482749
            Jan 11 at 9:27
















          $begingroup$
          Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
          $endgroup$
          – man2006
          Jan 10 at 23:32




          $begingroup$
          Am I allowed to add 3 with no restrictions arbitrarily? Also, how would I be able to deduce a_k-2 and b_k-2 from my previous steps?
          $endgroup$
          – man2006
          Jan 10 at 23:32












          $begingroup$
          @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
          $endgroup$
          – Bram28
          Jan 10 at 23:42




          $begingroup$
          @man2006 You don;t deduce $a_{k-2}$ and $b_{k-2}$ You just assume that you have those numbers such that $3a_{k-2} + 5b_{k-2} = k-2$. Now, how can you change those two numbers such that that $3a_{k+1} + 5b_{k+1} = k+1$? What is the relation between $a_{k+1}$ and $a_{k-2}$, and what is the relation between $a_{k+1}$ and $a_{k-2}$?
          $endgroup$
          – Bram28
          Jan 10 at 23:42












          $begingroup$
          In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
          $endgroup$
          – man2006
          Jan 10 at 23:49




          $begingroup$
          In this case, the best way to change those two numbers would be through the application of the Inductive Hypothesis. The relation between both a_k+1 and a_k-2 would be through the base which is a if I'm not mistaken.
          $endgroup$
          – man2006
          Jan 10 at 23:49












          $begingroup$
          @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
          $endgroup$
          – user3482749
          Jan 11 at 9:27




          $begingroup$
          @man2006 Because $3$ is a number, you can add it to everything. Why wouldn't you be able to add $3$ to something? More generally, because you can do anything you like to an equality, providing you apply the same function to the value of each side. The existence of $a_{k-2}$ and $b_{k-2}$ follows immediately from the inductive hypothesis. You don't need to know anything about them other than that it exists.
          $endgroup$
          – user3482749
          Jan 11 at 9:27











          0












          $begingroup$

          For the step, see answer by @user3482749



          But other than the inductive step, you also need to show that you can find actual values for $a_n$ and $b_n$ (and the others) in your base cases to make sure these work!



          For example, to satisfy that $3a_8+5b_8=8$, you can set $a_8=1$ and $b_8=1$



          Now do this for the other base cases as well






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
            $endgroup$
            – man2006
            Jan 10 at 23:18












          • $begingroup$
            @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
            $endgroup$
            – Bram28
            Jan 10 at 23:39










          • $begingroup$
            Got it, I just have to keep trying different numbers that work. Thank you for your help!
            $endgroup$
            – man2006
            Jan 10 at 23:45










          • $begingroup$
            @man2006 you're welcome :)
            $endgroup$
            – Bram28
            Jan 11 at 3:03
















          0












          $begingroup$

          For the step, see answer by @user3482749



          But other than the inductive step, you also need to show that you can find actual values for $a_n$ and $b_n$ (and the others) in your base cases to make sure these work!



          For example, to satisfy that $3a_8+5b_8=8$, you can set $a_8=1$ and $b_8=1$



          Now do this for the other base cases as well






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
            $endgroup$
            – man2006
            Jan 10 at 23:18












          • $begingroup$
            @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
            $endgroup$
            – Bram28
            Jan 10 at 23:39










          • $begingroup$
            Got it, I just have to keep trying different numbers that work. Thank you for your help!
            $endgroup$
            – man2006
            Jan 10 at 23:45










          • $begingroup$
            @man2006 you're welcome :)
            $endgroup$
            – Bram28
            Jan 11 at 3:03














          0












          0








          0





          $begingroup$

          For the step, see answer by @user3482749



          But other than the inductive step, you also need to show that you can find actual values for $a_n$ and $b_n$ (and the others) in your base cases to make sure these work!



          For example, to satisfy that $3a_8+5b_8=8$, you can set $a_8=1$ and $b_8=1$



          Now do this for the other base cases as well






          share|cite|improve this answer









          $endgroup$



          For the step, see answer by @user3482749



          But other than the inductive step, you also need to show that you can find actual values for $a_n$ and $b_n$ (and the others) in your base cases to make sure these work!



          For example, to satisfy that $3a_8+5b_8=8$, you can set $a_8=1$ and $b_8=1$



          Now do this for the other base cases as well







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 23:10









          Bram28Bram28

          63.2k44793




          63.2k44793












          • $begingroup$
            Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
            $endgroup$
            – man2006
            Jan 10 at 23:18












          • $begingroup$
            @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
            $endgroup$
            – Bram28
            Jan 10 at 23:39










          • $begingroup$
            Got it, I just have to keep trying different numbers that work. Thank you for your help!
            $endgroup$
            – man2006
            Jan 10 at 23:45










          • $begingroup$
            @man2006 you're welcome :)
            $endgroup$
            – Bram28
            Jan 11 at 3:03


















          • $begingroup$
            Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
            $endgroup$
            – man2006
            Jan 10 at 23:18












          • $begingroup$
            @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
            $endgroup$
            – Bram28
            Jan 10 at 23:39










          • $begingroup$
            Got it, I just have to keep trying different numbers that work. Thank you for your help!
            $endgroup$
            – man2006
            Jan 10 at 23:45










          • $begingroup$
            @man2006 you're welcome :)
            $endgroup$
            – Bram28
            Jan 11 at 3:03
















          $begingroup$
          Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
          $endgroup$
          – man2006
          Jan 10 at 23:18






          $begingroup$
          Along the same lines, would I be able to set a_9 = 2 or any other number such that it satisfies the right hand side is equal to 9 and the same for b_9?
          $endgroup$
          – man2006
          Jan 10 at 23:18














          $begingroup$
          @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
          $endgroup$
          – Bram28
          Jan 10 at 23:39




          $begingroup$
          @man2006 Yes, you need to find non-negative integers that work. Though $a_9=2$ is not going to work ...
          $endgroup$
          – Bram28
          Jan 10 at 23:39












          $begingroup$
          Got it, I just have to keep trying different numbers that work. Thank you for your help!
          $endgroup$
          – man2006
          Jan 10 at 23:45




          $begingroup$
          Got it, I just have to keep trying different numbers that work. Thank you for your help!
          $endgroup$
          – man2006
          Jan 10 at 23:45












          $begingroup$
          @man2006 you're welcome :)
          $endgroup$
          – Bram28
          Jan 11 at 3:03




          $begingroup$
          @man2006 you're welcome :)
          $endgroup$
          – Bram28
          Jan 11 at 3:03


















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