Proof of $2^n > n$ by Induction
$begingroup$
I'm new to induction and trying to prove $2^n > n$ for all natural numbers.
I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.
So I show $2^1 > 1$ as the base case.
Then I assume $2^k > k$
Meaning that
$2.2^k > 2k$
i.e.
$2^{k+1} > 2k$
Or
$2^{k+1} > k + k > k + 1$
So it is considered proven.
But when $k = 1$, $k + k not> k + 1$
What am I missing please?
Do I need a special case for going from $k=1$ to $k=2$?
inequality induction
asked Dec 12 '18 at 9:42
RobinRobin
1577
$endgroup$
add a comment |
$begingroup$
I'm new to induction and trying to prove $2^n > n$ for all natural numbers.
I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.
So I show $2^1 > 1$ as the base case.
Then I assume $2^k > k$
Meaning that
$2.2^k > 2k$
i.e.
$2^{k+1} > 2k$
Or
$2^{k+1} > k + k > k + 1$
So it is considered proven.
But when $k = 1$, $k + k not> k + 1$
What am I missing please?
Do I need a special case for going from $k=1$ to $k=2$?
inequality induction
asked Dec 12 '18 at 9:42
RobinRobin
1577
$endgroup$
1
$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45
add a comment |
$begingroup$
I'm new to induction and trying to prove $2^n > n$ for all natural numbers.
I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.
So I show $2^1 > 1$ as the base case.
Then I assume $2^k > k$
Meaning that
$2.2^k > 2k$
i.e.
$2^{k+1} > 2k$
Or
$2^{k+1} > k + k > k + 1$
So it is considered proven.
But when $k = 1$, $k + k not> k + 1$
What am I missing please?
Do I need a special case for going from $k=1$ to $k=2$?
inequality induction
asked Dec 12 '18 at 9:42
RobinRobin
1577
$endgroup$
I'm new to induction and trying to prove $2^n > n$ for all natural numbers.
I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.
So I show $2^1 > 1$ as the base case.
Then I assume $2^k > k$
Meaning that
$2.2^k > 2k$
i.e.
$2^{k+1} > 2k$
Or
$2^{k+1} > k + k > k + 1$
So it is considered proven.
But when $k = 1$, $k + k not> k + 1$
What am I missing please?
Do I need a special case for going from $k=1$ to $k=2$?
inequality induction
inequality induction
asked Dec 12 '18 at 9:42
RobinRobin
1577
asked Dec 12 '18 at 9:42
RobinRobin
1577
asked Dec 12 '18 at 9:42
RobinRobin
1577
asked Dec 12 '18 at 9:42
RobinRobin
1577
asked Dec 12 '18 at 9:42
RobinRobin
1577
1577
1
$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45
add a comment |
1
$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45
1
1
$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45
$begingroup$
Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
$$2^{k+1} > k+k geq k+1$$
and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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$begingroup$
Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
$$2^{k+1} > k+k geq k+1$$
and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
$$2^{k+1} > k+k geq k+1$$
and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
$$2^{k+1} > k+k geq k+1$$
and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
$endgroup$
Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write
$$2^{k+1} > k+k geq k+1$$
and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $bgeq c$ always implies $a>c$.
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
answered Dec 12 '18 at 9:45
5xum5xum
91.8k394161
91.8k394161
add a comment |
add a comment |
$begingroup$
No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
add a comment |
$begingroup$
No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
add a comment |
$begingroup$
No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
$endgroup$
No, you don't need a special case. $2^{k+1} >2k $ and $2k geq k+1$ together impliy $2^{k+1} >k+1 $
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
answered Dec 12 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
71.7k53170
71.7k53170
add a comment |
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1
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Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+kgeq k+1$ for $kgeq1$ is probably nicer though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:45