Correct Way of Reaching Inductive Goal












0












$begingroup$


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

To satisfy the above, set a8 = 1 and b8 = 1, therefore 3(1) + 5(1) = 3 + 5 = 8



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

Set a9 = 3 and b9 = 0, therefore 3(3) + 5(0) = 9 + 0 = 9



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



Set a10 = 0 and b10 = 2, therefore 3(0) + 5(2) = 0 + 10 = 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



Since for every integer n, 8 ≤ n ≤ k, there are ak-2 and bk-2, then it follows that:



3ak-2 + 5bk-2 = k - 2 (Starting Point)



..



..



..



3ak+1 + 5bk+1 = k+1 (Goal)




Now I know where to begin and the goal I need to reach for my inductive step but I am unsure as to how to reach my goal.



Any guidance would be much appreciated!






discrete-mathematics proof-verification induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    Am I missing something? Is it not sufficient to set $a_n = a_{n-3}+1$ and $b_n = b_{n-3}$?
    $endgroup$
    – Brian Tung
    Jan 15 at 18:21






  • 2




    $begingroup$
    You can have a look at For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$. Found using Approach0. More tips on searching: How to search on this site?
    $endgroup$
    – Martin Sleziak
    Jan 15 at 18:23










  • $begingroup$
    Simply 3 + 5 = 8, 3×3 = 9, 5x2 = 10 is sufficient to establish the base step. Toss in a 5×0 or a 3×0 if you'd like.
    $endgroup$
    – William Elliot
    Jan 15 at 21:33
















0












$begingroup$


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

To satisfy the above, set a8 = 1 and b8 = 1, therefore 3(1) + 5(1) = 3 + 5 = 8



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

Set a9 = 3 and b9 = 0, therefore 3(3) + 5(0) = 9 + 0 = 9



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



Set a10 = 0 and b10 = 2, therefore 3(0) + 5(2) = 0 + 10 = 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



Since for every integer n, 8 ≤ n ≤ k, there are ak-2 and bk-2, then it follows that:



3ak-2 + 5bk-2 = k - 2 (Starting Point)



..



..



..



3ak+1 + 5bk+1 = k+1 (Goal)




Now I know where to begin and the goal I need to reach for my inductive step but I am unsure as to how to reach my goal.



Any guidance would be much appreciated!






discrete-mathematics proof-verification induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    Am I missing something? Is it not sufficient to set $a_n = a_{n-3}+1$ and $b_n = b_{n-3}$?
    $endgroup$
    – Brian Tung
    Jan 15 at 18:21






  • 2




    $begingroup$
    You can have a look at For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$. Found using Approach0. More tips on searching: How to search on this site?
    $endgroup$
    – Martin Sleziak
    Jan 15 at 18:23










  • $begingroup$
    Simply 3 + 5 = 8, 3×3 = 9, 5x2 = 10 is sufficient to establish the base step. Toss in a 5×0 or a 3×0 if you'd like.
    $endgroup$
    – William Elliot
    Jan 15 at 21:33














0












0








0





$begingroup$


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

To satisfy the above, set a8 = 1 and b8 = 1, therefore 3(1) + 5(1) = 3 + 5 = 8



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

Set a9 = 3 and b9 = 0, therefore 3(3) + 5(0) = 9 + 0 = 9



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



Set a10 = 0 and b10 = 2, therefore 3(0) + 5(2) = 0 + 10 = 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



Since for every integer n, 8 ≤ n ≤ k, there are ak-2 and bk-2, then it follows that:



3ak-2 + 5bk-2 = k - 2 (Starting Point)



..



..



..



3ak+1 + 5bk+1 = k+1 (Goal)




Now I know where to begin and the goal I need to reach for my inductive step but I am unsure as to how to reach my goal.



Any guidance would be much appreciated!






discrete-mathematics proof-verification 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

To satisfy the above, set a8 = 1 and b8 = 1, therefore 3(1) + 5(1) = 3 + 5 = 8



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

Set a9 = 3 and b9 = 0, therefore 3(3) + 5(0) = 9 + 0 = 9



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



Set a10 = 0 and b10 = 2, therefore 3(0) + 5(2) = 0 + 10 = 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



Since for every integer n, 8 ≤ n ≤ k, there are ak-2 and bk-2, then it follows that:



3ak-2 + 5bk-2 = k - 2 (Starting Point)



..



..



..



3ak+1 + 5bk+1 = k+1 (Goal)




Now I know where to begin and the goal I need to reach for my inductive step but I am unsure as to how to reach my goal.



Any guidance would be much appreciated!






discrete-mathematics proof-verification induction




discrete-mathematics proof-verification induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 18:18









Martin Sleziak

45.1k10123277




45.1k10123277










asked Jan 15 at 17:44









man2006man2006

175




175












  • $begingroup$
    Am I missing something? Is it not sufficient to set $a_n = a_{n-3}+1$ and $b_n = b_{n-3}$?
    $endgroup$
    – Brian Tung
    Jan 15 at 18:21






  • 2




    $begingroup$
    You can have a look at For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$. Found using Approach0. More tips on searching: How to search on this site?
    $endgroup$
    – Martin Sleziak
    Jan 15 at 18:23










  • $begingroup$
    Simply 3 + 5 = 8, 3×3 = 9, 5x2 = 10 is sufficient to establish the base step. Toss in a 5×0 or a 3×0 if you'd like.
    $endgroup$
    – William Elliot
    Jan 15 at 21:33


















  • $begingroup$
    Am I missing something? Is it not sufficient to set $a_n = a_{n-3}+1$ and $b_n = b_{n-3}$?
    $endgroup$
    – Brian Tung
    Jan 15 at 18:21






  • 2




    $begingroup$
    You can have a look at For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$. Found using Approach0. More tips on searching: How to search on this site?
    $endgroup$
    – Martin Sleziak
    Jan 15 at 18:23










  • $begingroup$
    Simply 3 + 5 = 8, 3×3 = 9, 5x2 = 10 is sufficient to establish the base step. Toss in a 5×0 or a 3×0 if you'd like.
    $endgroup$
    – William Elliot
    Jan 15 at 21:33
















$begingroup$
Am I missing something? Is it not sufficient to set $a_n = a_{n-3}+1$ and $b_n = b_{n-3}$?
$endgroup$
– Brian Tung
Jan 15 at 18:21




$begingroup$
Am I missing something? Is it not sufficient to set $a_n = a_{n-3}+1$ and $b_n = b_{n-3}$?
$endgroup$
– Brian Tung
Jan 15 at 18:21




2




2




$begingroup$
You can have a look at For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$. Found using Approach0. More tips on searching: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 15 at 18:23




$begingroup$
You can have a look at For all non-negative integers $n$ such that $n ≥ 8$, there exist non-negative integers $a_n$ and $b_n$ such that $3a_n+5b_n = n$. Found using Approach0. More tips on searching: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 15 at 18:23












$begingroup$
Simply 3 + 5 = 8, 3×3 = 9, 5x2 = 10 is sufficient to establish the base step. Toss in a 5×0 or a 3×0 if you'd like.
$endgroup$
– William Elliot
Jan 15 at 21:33




$begingroup$
Simply 3 + 5 = 8, 3×3 = 9, 5x2 = 10 is sufficient to establish the base step. Toss in a 5×0 or a 3×0 if you'd like.
$endgroup$
– William Elliot
Jan 15 at 21:33










1 Answer
1






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3 + 5 = 8, 3×3 = 9, 5x2 = 10



The next step is 11 = 8 + 3.

Since 8 has been answered, an answer for 11 is apparent.

Likewise for n > 10, by induction hypothesis,

there is a solution for n - 3, so a solution for n.



Exercise. Show every integer >= 8 is a multiple of 3

or 5 + a multiple of 3 or 10 + a multiple of 3

as an immediate result of the above proof.






share|cite|improve this answer









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    $begingroup$

    3 + 5 = 8, 3×3 = 9, 5x2 = 10



    The next step is 11 = 8 + 3.

    Since 8 has been answered, an answer for 11 is apparent.

    Likewise for n > 10, by induction hypothesis,

    there is a solution for n - 3, so a solution for n.



    Exercise. Show every integer >= 8 is a multiple of 3

    or 5 + a multiple of 3 or 10 + a multiple of 3

    as an immediate result of the above proof.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      3 + 5 = 8, 3×3 = 9, 5x2 = 10



      The next step is 11 = 8 + 3.

      Since 8 has been answered, an answer for 11 is apparent.

      Likewise for n > 10, by induction hypothesis,

      there is a solution for n - 3, so a solution for n.



      Exercise. Show every integer >= 8 is a multiple of 3

      or 5 + a multiple of 3 or 10 + a multiple of 3

      as an immediate result of the above proof.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        3 + 5 = 8, 3×3 = 9, 5x2 = 10



        The next step is 11 = 8 + 3.

        Since 8 has been answered, an answer for 11 is apparent.

        Likewise for n > 10, by induction hypothesis,

        there is a solution for n - 3, so a solution for n.



        Exercise. Show every integer >= 8 is a multiple of 3

        or 5 + a multiple of 3 or 10 + a multiple of 3

        as an immediate result of the above proof.






        share|cite|improve this answer









        $endgroup$



        3 + 5 = 8, 3×3 = 9, 5x2 = 10



        The next step is 11 = 8 + 3.

        Since 8 has been answered, an answer for 11 is apparent.

        Likewise for n > 10, by induction hypothesis,

        there is a solution for n - 3, so a solution for n.



        Exercise. Show every integer >= 8 is a multiple of 3

        or 5 + a multiple of 3 or 10 + a multiple of 3

        as an immediate result of the above proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 21:48









        William ElliotWilliam Elliot

        9,2662820




        9,2662820






























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