Induction on a list of integers? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Mathematical InductionProve by induction that sum of an odd number of odd numbers is oddStrong induction on property of integers involving setsProof By Induction for arbitrary integersInduction to prove $a_n=2^n+1$Proving a bijection with contradiction and inductionRecursive sequence induction problemUse complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$Prove through induction that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|a in A_i| $ is oddProve by strong induction that $a_n < 2^n$ for all integers $n ≥ 1$, given a list of $a_n$ values.
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Induction on a list of integers?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Mathematical InductionProve by induction that sum of an odd number of odd numbers is oddStrong induction on property of integers involving setsProof By Induction for arbitrary integersInduction to prove $a_n=2^n+1$Proving a bijection with contradiction and inductionRecursive sequence induction problemUse complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$Prove through induction that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|i| $ is oddProve by strong induction that $a_n < 2^n$ for all integers $n ≥ 1$, given a list of $a_n$ values.
$begingroup$
let $(a_1,cdots,a_n)$ be a list of integers prove that if $a_1$ is even and $a_n$ is odd then there is an index $i$, $1leq i < n$ such that $a_i$ is even and $a_i+1$ is odd. I’m extremely confused on how to even preform induction on this. What would the base case even be?
induction
edited Apr 9 at 3:51
Bruno Andrades
24611
asked Apr 9 at 3:04
Dani JoDani Jo
43
$endgroup$
add a comment |
$begingroup$
let $(a_1,cdots,a_n)$ be a list of integers prove that if $a_1$ is even and $a_n$ is odd then there is an index $i$, $1leq i < n$ such that $a_i$ is even and $a_i+1$ is odd. I’m extremely confused on how to even preform induction on this. What would the base case even be?
induction
edited Apr 9 at 3:51
Bruno Andrades
24611
asked Apr 9 at 3:04
Dani JoDani Jo
43
$endgroup$
add a comment |
$begingroup$
let $(a_1,cdots,a_n)$ be a list of integers prove that if $a_1$ is even and $a_n$ is odd then there is an index $i$, $1leq i < n$ such that $a_i$ is even and $a_i+1$ is odd. I’m extremely confused on how to even preform induction on this. What would the base case even be?
induction
edited Apr 9 at 3:51
Bruno Andrades
24611
asked Apr 9 at 3:04
Dani JoDani Jo
43
$endgroup$
let $(a_1,cdots,a_n)$ be a list of integers prove that if $a_1$ is even and $a_n$ is odd then there is an index $i$, $1leq i < n$ such that $a_i$ is even and $a_i+1$ is odd. I’m extremely confused on how to even preform induction on this. What would the base case even be?
induction
induction
edited Apr 9 at 3:51
Bruno Andrades
24611
asked Apr 9 at 3:04
Dani JoDani Jo
43
edited Apr 9 at 3:51
Bruno Andrades
24611
asked Apr 9 at 3:04
Dani JoDani Jo
43
edited Apr 9 at 3:51
Bruno Andrades
24611
edited Apr 9 at 3:51
Bruno Andrades
24611
edited Apr 9 at 3:51
Bruno Andrades
24611
24611
asked Apr 9 at 3:04
Dani JoDani Jo
43
asked Apr 9 at 3:04
Dani JoDani Jo
43
asked Apr 9 at 3:04
Dani JoDani Jo
43
43
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Induction on the value of $n$. For $n=2$ is trivial. If it’s true for $n$, take $(a_1,cdots,a_n+1)$ and choose some intermediate element $a_k$ with $1 < k < n+1$. If $a_k$ is odd, apply induction on $(a_k,cdots,a_n+1)$. If it’s even, apply induction on $(a_1,cdots,a_k)$.
edited Apr 9 at 3:51
Bruno Andrades
24611
answered Apr 9 at 3:21
BorelianBorelian
733
$endgroup$
add a comment |
$begingroup$
If $a_2$ is odd then we are done. Suppose $a_2$ is even. Now if $a_3$ is odd we are similarly done. So assume that $a_3$ is even. Continuing this argument $n-2$ times we find that for all $1 leq i <n$ we have $a_i$ is even. So in particular $a_n-1$ is even.Take $j=n-1.$ Then $a_j$ and $a_j+1$ have the required property.
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
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$begingroup$
Induction on the value of $n$. For $n=2$ is trivial. If it’s true for $n$, take $(a_1,cdots,a_n+1)$ and choose some intermediate element $a_k$ with $1 < k < n+1$. If $a_k$ is odd, apply induction on $(a_k,cdots,a_n+1)$. If it’s even, apply induction on $(a_1,cdots,a_k)$.
edited Apr 9 at 3:51
Bruno Andrades
24611
answered Apr 9 at 3:21
BorelianBorelian
733
$endgroup$
add a comment |
$begingroup$
Induction on the value of $n$. For $n=2$ is trivial. If it’s true for $n$, take $(a_1,cdots,a_n+1)$ and choose some intermediate element $a_k$ with $1 < k < n+1$. If $a_k$ is odd, apply induction on $(a_k,cdots,a_n+1)$. If it’s even, apply induction on $(a_1,cdots,a_k)$.
edited Apr 9 at 3:51
Bruno Andrades
24611
answered Apr 9 at 3:21
BorelianBorelian
733
$endgroup$
add a comment |
$begingroup$
Induction on the value of $n$. For $n=2$ is trivial. If it’s true for $n$, take $(a_1,cdots,a_n+1)$ and choose some intermediate element $a_k$ with $1 < k < n+1$. If $a_k$ is odd, apply induction on $(a_k,cdots,a_n+1)$. If it’s even, apply induction on $(a_1,cdots,a_k)$.
edited Apr 9 at 3:51
Bruno Andrades
24611
answered Apr 9 at 3:21
BorelianBorelian
733
$endgroup$
Induction on the value of $n$. For $n=2$ is trivial. If it’s true for $n$, take $(a_1,cdots,a_n+1)$ and choose some intermediate element $a_k$ with $1 < k < n+1$. If $a_k$ is odd, apply induction on $(a_k,cdots,a_n+1)$. If it’s even, apply induction on $(a_1,cdots,a_k)$.
edited Apr 9 at 3:51
Bruno Andrades
24611
answered Apr 9 at 3:21
BorelianBorelian
733
edited Apr 9 at 3:51
Bruno Andrades
24611
edited Apr 9 at 3:51
Bruno Andrades
24611
edited Apr 9 at 3:51
Bruno Andrades
24611
24611
answered Apr 9 at 3:21
BorelianBorelian
733
answered Apr 9 at 3:21
BorelianBorelian
733
answered Apr 9 at 3:21
BorelianBorelian
733
733
add a comment |
add a comment |
$begingroup$
If $a_2$ is odd then we are done. Suppose $a_2$ is even. Now if $a_3$ is odd we are similarly done. So assume that $a_3$ is even. Continuing this argument $n-2$ times we find that for all $1 leq i <n$ we have $a_i$ is even. So in particular $a_n-1$ is even.Take $j=n-1.$ Then $a_j$ and $a_j+1$ have the required property.
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
$endgroup$
add a comment |
$begingroup$
If $a_2$ is odd then we are done. Suppose $a_2$ is even. Now if $a_3$ is odd we are similarly done. So assume that $a_3$ is even. Continuing this argument $n-2$ times we find that for all $1 leq i <n$ we have $a_i$ is even. So in particular $a_n-1$ is even.Take $j=n-1.$ Then $a_j$ and $a_j+1$ have the required property.
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
$endgroup$
add a comment |
$begingroup$
If $a_2$ is odd then we are done. Suppose $a_2$ is even. Now if $a_3$ is odd we are similarly done. So assume that $a_3$ is even. Continuing this argument $n-2$ times we find that for all $1 leq i <n$ we have $a_i$ is even. So in particular $a_n-1$ is even.Take $j=n-1.$ Then $a_j$ and $a_j+1$ have the required property.
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
$endgroup$
If $a_2$ is odd then we are done. Suppose $a_2$ is even. Now if $a_3$ is odd we are similarly done. So assume that $a_3$ is even. Continuing this argument $n-2$ times we find that for all $1 leq i <n$ we have $a_i$ is even. So in particular $a_n-1$ is even.Take $j=n-1.$ Then $a_j$ and $a_j+1$ have the required property.
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
answered Apr 9 at 3:20
Dbchatto67Dbchatto67
3,185625
3,185625
add a comment |
add a comment |
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