Correct Way of Reaching Inductive Goal
$begingroup$
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
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
asked Jan 15 at 17:44
man2006man2006
175
$endgroup$
add a comment |
$begingroup$
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
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
asked Jan 15 at 17:44
man2006man2006
175
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
asked Jan 15 at 17:44
man2006man2006
175
$endgroup$
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
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
asked Jan 15 at 17:44
man2006man2006
175
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
asked Jan 15 at 17:44
man2006man2006
175
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
edited Jan 15 at 18:18
Martin Sleziak
45.1k10123277
45.1k10123277
asked Jan 15 at 17:44
man2006man2006
175
asked Jan 15 at 17:44
man2006man2006
175
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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.
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
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add a comment |
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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.
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
$endgroup$
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.
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
answered Jan 15 at 21:48
William ElliotWilliam Elliot
9,2662820
9,2662820
add a comment |
add a comment |
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$begingroup$
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