Prove $n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0
$begingroup$
The statement I'm trying to prove is:
$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0
I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.
I don't really understand how to deal with $k + 1$, so I'm a little lost.
I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.
discrete-mathematics proof-verification induction
asked Feb 5 '16 at 17:41
123123
4551926
$endgroup$
add a comment |
$begingroup$
The statement I'm trying to prove is:
$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0
I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.
I don't really understand how to deal with $k + 1$, so I'm a little lost.
I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.
discrete-mathematics proof-verification induction
asked Feb 5 '16 at 17:41
123123
4551926
$endgroup$
$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45
$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46
2
$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03
$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18
add a comment |
$begingroup$
The statement I'm trying to prove is:
$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0
I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.
I don't really understand how to deal with $k + 1$, so I'm a little lost.
I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.
discrete-mathematics proof-verification induction
asked Feb 5 '16 at 17:41
123123
4551926
$endgroup$
The statement I'm trying to prove is:
$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0
I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.
I don't really understand how to deal with $k + 1$, so I'm a little lost.
I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.
discrete-mathematics proof-verification induction
discrete-mathematics proof-verification induction
asked Feb 5 '16 at 17:41
123123
4551926
asked Feb 5 '16 at 17:41
123123
4551926
edited Feb 5 '16 at 17:53
123
asked Feb 5 '16 at 17:41
123123
4551926
asked Feb 5 '16 at 17:41
123123
4551926
asked Feb 5 '16 at 17:41
123123
4551926
4551926
$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45
$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46
2
$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03
$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18
add a comment |
$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45
$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46
2
$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03
$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18
$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45
$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45
$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46
$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46
2
2
$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03
$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03
$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18
$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The statement is not true. Take $n=1$ as a counter example.
Since it's not true, you won't manage to prove it (by induction or otherwise).
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
$endgroup$
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
add a comment |
$begingroup$
Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
$endgroup$
add a comment |
$begingroup$
Set $n=k+1$ then your expression is equivalent modulo 3 to
$$
n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
$$
In particular, this is a multiple of $3$ iff $3$ divides $k+1$.
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
$begingroup$
The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3 is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.
I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
In that case Proof is detailed below.
P(n): n^3 - 7n + 3 is divisible by 3
P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3
Assume P(k) is true for some positive integer k, i.e
P(k): (k)^3 - 7k + 3 is divisible by 3
We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
P(k+1): (k+1)^3 - 7(k+1) + 3
= k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)
= k^3 + 1 +3k(k+1) - 7(k+1) + 3
= k^3 + 1 + 3k^2 + 3k -7k - 7 +3
= k^3 - 7k + 3 + 3k^2 +3k - 6 (by rearranging)
= 3d + 3(k^2 + k - 2 ).........{From (1)}
= 3d + 3(k^2 + 2k - k - 2 )
= 3d + 3 {k(k +2)-(k+2)}
= 3d + 3 (k-1)(k+2)
= 3 {d +(k-1)(k+2)}
= 3 q, where q = {d +(k-1)(k+2)} which is some positive integer
From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0
edited Dec 28 '18 at 21:08
amWhy
1
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement is not true. Take $n=1$ as a counter example.
Since it's not true, you won't manage to prove it (by induction or otherwise).
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
$endgroup$
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
add a comment |
$begingroup$
The statement is not true. Take $n=1$ as a counter example.
Since it's not true, you won't manage to prove it (by induction or otherwise).
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
$endgroup$
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
add a comment |
$begingroup$
The statement is not true. Take $n=1$ as a counter example.
Since it's not true, you won't manage to prove it (by induction or otherwise).
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
$endgroup$
The statement is not true. Take $n=1$ as a counter example.
Since it's not true, you won't manage to prove it (by induction or otherwise).
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
answered Feb 5 '16 at 17:42
peter.petrovpeter.petrov
5,451821
5,451821
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
add a comment |
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
$begingroup$
nice observation, +1
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:17
add a comment |
$begingroup$
Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
$endgroup$
add a comment |
$begingroup$
Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
$endgroup$
add a comment |
$begingroup$
Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
$endgroup$
Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
answered Feb 5 '16 at 17:46
ajotatxeajotatxe
53.8k23990
53.8k23990
add a comment |
add a comment |
$begingroup$
Set $n=k+1$ then your expression is equivalent modulo 3 to
$$
n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
$$
In particular, this is a multiple of $3$ iff $3$ divides $k+1$.
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
$begingroup$
Set $n=k+1$ then your expression is equivalent modulo 3 to
$$
n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
$$
In particular, this is a multiple of $3$ iff $3$ divides $k+1$.
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
$begingroup$
Set $n=k+1$ then your expression is equivalent modulo 3 to
$$
n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
$$
In particular, this is a multiple of $3$ iff $3$ divides $k+1$.
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
Set $n=k+1$ then your expression is equivalent modulo 3 to
$$
n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
$$
In particular, this is a multiple of $3$ iff $3$ divides $k+1$.
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
answered Feb 5 '16 at 17:47
Paolo LeonettiPaolo Leonetti
11.5k21550
11.5k21550
add a comment |
add a comment |
$begingroup$
The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3 is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.
I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
In that case Proof is detailed below.
P(n): n^3 - 7n + 3 is divisible by 3
P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3
Assume P(k) is true for some positive integer k, i.e
P(k): (k)^3 - 7k + 3 is divisible by 3
We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
P(k+1): (k+1)^3 - 7(k+1) + 3
= k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)
= k^3 + 1 +3k(k+1) - 7(k+1) + 3
= k^3 + 1 + 3k^2 + 3k -7k - 7 +3
= k^3 - 7k + 3 + 3k^2 +3k - 6 (by rearranging)
= 3d + 3(k^2 + k - 2 ).........{From (1)}
= 3d + 3(k^2 + 2k - k - 2 )
= 3d + 3 {k(k +2)-(k+2)}
= 3d + 3 (k-1)(k+2)
= 3 {d +(k-1)(k+2)}
= 3 q, where q = {d +(k-1)(k+2)} which is some positive integer
From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0
edited Dec 28 '18 at 21:08
amWhy
1
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
$endgroup$
add a comment |
$begingroup$
The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3 is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.
I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
In that case Proof is detailed below.
P(n): n^3 - 7n + 3 is divisible by 3
P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3
Assume P(k) is true for some positive integer k, i.e
P(k): (k)^3 - 7k + 3 is divisible by 3
We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
P(k+1): (k+1)^3 - 7(k+1) + 3
= k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)
= k^3 + 1 +3k(k+1) - 7(k+1) + 3
= k^3 + 1 + 3k^2 + 3k -7k - 7 +3
= k^3 - 7k + 3 + 3k^2 +3k - 6 (by rearranging)
= 3d + 3(k^2 + k - 2 ).........{From (1)}
= 3d + 3(k^2 + 2k - k - 2 )
= 3d + 3 {k(k +2)-(k+2)}
= 3d + 3 (k-1)(k+2)
= 3 {d +(k-1)(k+2)}
= 3 q, where q = {d +(k-1)(k+2)} which is some positive integer
From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0
edited Dec 28 '18 at 21:08
amWhy
1
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
$endgroup$
add a comment |
$begingroup$
The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3 is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.
I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
In that case Proof is detailed below.
P(n): n^3 - 7n + 3 is divisible by 3
P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3
Assume P(k) is true for some positive integer k, i.e
P(k): (k)^3 - 7k + 3 is divisible by 3
We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
P(k+1): (k+1)^3 - 7(k+1) + 3
= k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)
= k^3 + 1 +3k(k+1) - 7(k+1) + 3
= k^3 + 1 + 3k^2 + 3k -7k - 7 +3
= k^3 - 7k + 3 + 3k^2 +3k - 6 (by rearranging)
= 3d + 3(k^2 + k - 2 ).........{From (1)}
= 3d + 3(k^2 + 2k - k - 2 )
= 3d + 3 {k(k +2)-(k+2)}
= 3d + 3 (k-1)(k+2)
= 3 {d +(k-1)(k+2)}
= 3 q, where q = {d +(k-1)(k+2)} which is some positive integer
From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0
edited Dec 28 '18 at 21:08
amWhy
1
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
$endgroup$
The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3 is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.
I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
In that case Proof is detailed below.
P(n): n^3 - 7n + 3 is divisible by 3
P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3
Assume P(k) is true for some positive integer k, i.e
P(k): (k)^3 - 7k + 3 is divisible by 3
We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
P(k+1): (k+1)^3 - 7(k+1) + 3
= k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)
= k^3 + 1 +3k(k+1) - 7(k+1) + 3
= k^3 + 1 + 3k^2 + 3k -7k - 7 +3
= k^3 - 7k + 3 + 3k^2 +3k - 6 (by rearranging)
= 3d + 3(k^2 + k - 2 ).........{From (1)}
= 3d + 3(k^2 + 2k - k - 2 )
= 3d + 3 {k(k +2)-(k+2)}
= 3d + 3 (k-1)(k+2)
= 3 {d +(k-1)(k+2)}
= 3 q, where q = {d +(k-1)(k+2)} which is some positive integer
From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0
edited Dec 28 '18 at 21:08
amWhy
1
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
edited Dec 28 '18 at 21:08
amWhy
1
edited Dec 28 '18 at 21:08
amWhy
1
edited Dec 28 '18 at 21:08
amWhy
1
1
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
answered Dec 28 '18 at 3:09
Nammalwar ThiruvengadamNammalwar Thiruvengadam
11
11
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$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45
$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46
2
$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03
$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18