Mixing
Induction inequality check: $n!<n^n$
$begingroup$
check my proof, I feel like I made a mistake :)
so I'm looking to prove that when $p(n)$ is $n!<n^n$, $p(n)$ is true for all $n>1$.
Base Case $$ p(2) iff 2!<2^2 iff 2<4 $$
Assume p(k) is true
$$k!<k^k$$
Prove p(k+1)
$$(k+1)!<(k+1)^{(k+1)}$$
$$(k+1)(k)!<(k+1)(k+1)^k$$
$$(k+1)(k)!<(k+1)(k^k+1)$$
This part above. Can I assume that $1^k$ is always $1$ given any $k$ such that $k>1$?
$$(k+1)(k)!<(k+1)(k^k+1)$$
$$(k)!<k^k+1$$
Then above, can I factor out a k+1 from both sides?
$$(k)!<k^k+1$$
Is this a completed proof? Would my ending statement be something like "since we assumed $p(k)$ and $p(k+1)$ is still true given $p(k)$, and since $p(k+1)$ is a higher degree than $p(k)$...." (not sure what really to say here)?
trying again
prove p(k+1)
to start, im now looking to multiply the sides by (k+1)? not replace?
so
$$k!<k^k$$
$$(k+1)k!<(k+1)k^k$$
$$(k+1)!<(k^[k+1]+k^k$$
discrete-mathematics inequality induction factorial
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
asked Jul 1 '14 at 3:12
MikeMike
2125
$endgroup$
|
show 8 more comments
$begingroup$
check my proof, I feel like I made a mistake :)
so I'm looking to prove that when $p(n)$ is $n!<n^n$, $p(n)$ is true for all $n>1$.
Base Case $$ p(2) iff 2!<2^2 iff 2<4 $$
Assume p(k) is true
$$k!<k^k$$
Prove p(k+1)
$$(k+1)!<(k+1)^{(k+1)}$$
$$(k+1)(k)!<(k+1)(k+1)^k$$
$$(k+1)(k)!<(k+1)(k^k+1)$$
This part above. Can I assume that $1^k$ is always $1$ given any $k$ such that $k>1$?
$$(k+1)(k)!<(k+1)(k^k+1)$$
$$(k)!<k^k+1$$
Then above, can I factor out a k+1 from both sides?
$$(k)!<k^k+1$$
Is this a completed proof? Would my ending statement be something like "since we assumed $p(k)$ and $p(k+1)$ is still true given $p(k)$, and since $p(k+1)$ is a higher degree than $p(k)$...." (not sure what really to say here)?
trying again
prove p(k+1)
to start, im now looking to multiply the sides by (k+1)? not replace?
so
$$k!<k^k$$
$$(k+1)k!<(k+1)k^k$$
$$(k+1)!<(k^[k+1]+k^k$$
discrete-mathematics inequality induction factorial
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
asked Jul 1 '14 at 3:12
MikeMike
2125
$endgroup$
$begingroup$
You should start with $(k+1)!<k^k(k+1)$
$endgroup$
– Adam Hughes
Jul 1 '14 at 3:15
3
$begingroup$
Also, $(k+1)^kneq k^k+1^k$.
$endgroup$
– vadim123
Jul 1 '14 at 3:17
$begingroup$
$2^2 neq 2$ so this should be edited
$endgroup$
– Squirtle
Jul 1 '14 at 3:24
$begingroup$
When you are proving that $A=B$ or that $Alt B$, do not ever start from $A=B$ or $Alt B$ and manipulate.
$endgroup$
– André Nicolas
Jul 1 '14 at 3:24
$begingroup$
cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1?
$endgroup$
– Mike
Jul 1 '14 at 3:37
|
show 8 more comments
$begingroup$
check my proof, I feel like I made a mistake :)
so I'm looking to prove that when $p(n)$ is $n!<n^n$, $p(n)$ is true for all $n>1$.
Base Case $$ p(2) iff 2!<2^2 iff 2<4 $$
Assume p(k) is true
$$k!<k^k$$
Prove p(k+1)
$$(k+1)!<(k+1)^{(k+1)}$$
$$(k+1)(k)!<(k+1)(k+1)^k$$
$$(k+1)(k)!<(k+1)(k^k+1)$$
This part above. Can I assume that $1^k$ is always $1$ given any $k$ such that $k>1$?
$$(k+1)(k)!<(k+1)(k^k+1)$$
$$(k)!<k^k+1$$
Then above, can I factor out a k+1 from both sides?
$$(k)!<k^k+1$$
Is this a completed proof? Would my ending statement be something like "since we assumed $p(k)$ and $p(k+1)$ is still true given $p(k)$, and since $p(k+1)$ is a higher degree than $p(k)$...." (not sure what really to say here)?
trying again
prove p(k+1)
to start, im now looking to multiply the sides by (k+1)? not replace?
so
$$k!<k^k$$
$$(k+1)k!<(k+1)k^k$$
$$(k+1)!<(k^[k+1]+k^k$$
discrete-mathematics inequality induction factorial
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
asked Jul 1 '14 at 3:12
MikeMike
2125
$endgroup$
check my proof, I feel like I made a mistake :)
so I'm looking to prove that when $p(n)$ is $n!<n^n$, $p(n)$ is true for all $n>1$.
Base Case $$ p(2) iff 2!<2^2 iff 2<4 $$
Assume p(k) is true
$$k!<k^k$$
Prove p(k+1)
$$(k+1)!<(k+1)^{(k+1)}$$
$$(k+1)(k)!<(k+1)(k+1)^k$$
$$(k+1)(k)!<(k+1)(k^k+1)$$
This part above. Can I assume that $1^k$ is always $1$ given any $k$ such that $k>1$?
$$(k+1)(k)!<(k+1)(k^k+1)$$
$$(k)!<k^k+1$$
Then above, can I factor out a k+1 from both sides?
$$(k)!<k^k+1$$
Is this a completed proof? Would my ending statement be something like "since we assumed $p(k)$ and $p(k+1)$ is still true given $p(k)$, and since $p(k+1)$ is a higher degree than $p(k)$...." (not sure what really to say here)?
trying again
prove p(k+1)
to start, im now looking to multiply the sides by (k+1)? not replace?
so
$$k!<k^k$$
$$(k+1)k!<(k+1)k^k$$
$$(k+1)!<(k^[k+1]+k^k$$
discrete-mathematics inequality induction factorial
discrete-mathematics inequality induction factorial
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
asked Jul 1 '14 at 3:12
MikeMike
2125
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
asked Jul 1 '14 at 3:12
MikeMike
2125
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
edited Jan 26 at 14:57
Martin Sleziak
44.9k10122275
44.9k10122275
asked Jul 1 '14 at 3:12
MikeMike
2125
asked Jul 1 '14 at 3:12
MikeMike
2125
asked Jul 1 '14 at 3:12
MikeMike
2125
2125
$begingroup$
You should start with $(k+1)!<k^k(k+1)$
$endgroup$
– Adam Hughes
Jul 1 '14 at 3:15
3
$begingroup$
Also, $(k+1)^kneq k^k+1^k$.
$endgroup$
– vadim123
Jul 1 '14 at 3:17
$begingroup$
$2^2 neq 2$ so this should be edited
$endgroup$
– Squirtle
Jul 1 '14 at 3:24
$begingroup$
When you are proving that $A=B$ or that $Alt B$, do not ever start from $A=B$ or $Alt B$ and manipulate.
$endgroup$
– André Nicolas
Jul 1 '14 at 3:24
$begingroup$
cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1?
$endgroup$
– Mike
Jul 1 '14 at 3:37
|
show 8 more comments
$begingroup$
You should start with $(k+1)!<k^k(k+1)$
$endgroup$
– Adam Hughes
Jul 1 '14 at 3:15
3
$begingroup$
Also, $(k+1)^kneq k^k+1^k$.
$endgroup$
– vadim123
Jul 1 '14 at 3:17
$begingroup$
$2^2 neq 2$ so this should be edited
$endgroup$
– Squirtle
Jul 1 '14 at 3:24
$begingroup$
When you are proving that $A=B$ or that $Alt B$, do not ever start from $A=B$ or $Alt B$ and manipulate.
$endgroup$
– André Nicolas
Jul 1 '14 at 3:24
$begingroup$
cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1?
$endgroup$
– Mike
Jul 1 '14 at 3:37
$begingroup$
You should start with $(k+1)!<k^k(k+1)$
$endgroup$
– Adam Hughes
Jul 1 '14 at 3:15
$begingroup$
You should start with $(k+1)!<k^k(k+1)$
$endgroup$
– Adam Hughes
Jul 1 '14 at 3:15
3
3
$begingroup$
Also, $(k+1)^kneq k^k+1^k$.
$endgroup$
– vadim123
Jul 1 '14 at 3:17
$begingroup$
Also, $(k+1)^kneq k^k+1^k$.
$endgroup$
– vadim123
Jul 1 '14 at 3:17
$begingroup$
$2^2 neq 2$ so this should be edited
$endgroup$
– Squirtle
Jul 1 '14 at 3:24
$begingroup$
$2^2 neq 2$ so this should be edited
$endgroup$
– Squirtle
Jul 1 '14 at 3:24
$begingroup$
When you are proving that $A=B$ or that $Alt B$, do not ever start from $A=B$ or $Alt B$ and manipulate.
$endgroup$
– André Nicolas
Jul 1 '14 at 3:24
$begingroup$
When you are proving that $A=B$ or that $Alt B$, do not ever start from $A=B$ or $Alt B$ and manipulate.
$endgroup$
– André Nicolas
Jul 1 '14 at 3:24
$begingroup$
cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1?
$endgroup$
– Mike
Jul 1 '14 at 3:37
$begingroup$
cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1?
$endgroup$
– Mike
Jul 1 '14 at 3:37
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Finishing up the induction proof,
begin{align}
(k+1)!&=(k+1)k! \
&<(k+1)k^k & text{by hypothesis, } k! < k^k\
&<(k+1)(k+1)^k & k < k+1 text{ for all }k>1\
&=(k+1)^{k+1}
end{align}
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
$endgroup$
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
add a comment |
$begingroup$
As an alternative solution, we could use the AM-GM inequality:
$$frac{1 + 2 + 3 + dots + n}{n} ge sqrt[n]{n!}$$
$$frac{n(n+1)}{2n}ge sqrt[n]{n!}$$
$$frac{n+1}{2}gesqrt[n]{n!}$$
Since $n> frac{n+1}{2}$ for $n>1$, we have
$$n>sqrt[n]{n!}$$
And finally,
$$n^n > n!$$
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
Finishing up the induction proof,
begin{align}
(k+1)!&=(k+1)k! \
&<(k+1)k^k & text{by hypothesis, } k! < k^k\
&<(k+1)(k+1)^k & k < k+1 text{ for all }k>1\
&=(k+1)^{k+1}
end{align}
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
$endgroup$
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
add a comment |
$begingroup$
Finishing up the induction proof,
begin{align}
(k+1)!&=(k+1)k! \
&<(k+1)k^k & text{by hypothesis, } k! < k^k\
&<(k+1)(k+1)^k & k < k+1 text{ for all }k>1\
&=(k+1)^{k+1}
end{align}
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
$endgroup$
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
add a comment |
$begingroup$
Finishing up the induction proof,
begin{align}
(k+1)!&=(k+1)k! \
&<(k+1)k^k & text{by hypothesis, } k! < k^k\
&<(k+1)(k+1)^k & k < k+1 text{ for all }k>1\
&=(k+1)^{k+1}
end{align}
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
$endgroup$
Finishing up the induction proof,
begin{align}
(k+1)!&=(k+1)k! \
&<(k+1)k^k & text{by hypothesis, } k! < k^k\
&<(k+1)(k+1)^k & k < k+1 text{ for all }k>1\
&=(k+1)^{k+1}
end{align}
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
answered Jul 1 '14 at 3:36
CookieCookie
8,790123885
8,790123885
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
add a comment |
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1
$endgroup$
– Mike
Jul 1 '14 at 3:58
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something?
$endgroup$
– Mike
Jul 1 '14 at 4:06
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on.
$endgroup$
– Cookie
Jul 1 '14 at 4:27
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
$begingroup$
cool, thanks for clarifying that
$endgroup$
– Mike
Jul 1 '14 at 4:35
add a comment |
$begingroup$
As an alternative solution, we could use the AM-GM inequality:
$$frac{1 + 2 + 3 + dots + n}{n} ge sqrt[n]{n!}$$
$$frac{n(n+1)}{2n}ge sqrt[n]{n!}$$
$$frac{n+1}{2}gesqrt[n]{n!}$$
Since $n> frac{n+1}{2}$ for $n>1$, we have
$$n>sqrt[n]{n!}$$
And finally,
$$n^n > n!$$
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
add a comment |
$begingroup$
As an alternative solution, we could use the AM-GM inequality:
$$frac{1 + 2 + 3 + dots + n}{n} ge sqrt[n]{n!}$$
$$frac{n(n+1)}{2n}ge sqrt[n]{n!}$$
$$frac{n+1}{2}gesqrt[n]{n!}$$
Since $n> frac{n+1}{2}$ for $n>1$, we have
$$n>sqrt[n]{n!}$$
And finally,
$$n^n > n!$$
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
add a comment |
$begingroup$
As an alternative solution, we could use the AM-GM inequality:
$$frac{1 + 2 + 3 + dots + n}{n} ge sqrt[n]{n!}$$
$$frac{n(n+1)}{2n}ge sqrt[n]{n!}$$
$$frac{n+1}{2}gesqrt[n]{n!}$$
Since $n> frac{n+1}{2}$ for $n>1$, we have
$$n>sqrt[n]{n!}$$
And finally,
$$n^n > n!$$
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
As an alternative solution, we could use the AM-GM inequality:
$$frac{1 + 2 + 3 + dots + n}{n} ge sqrt[n]{n!}$$
$$frac{n(n+1)}{2n}ge sqrt[n]{n!}$$
$$frac{n+1}{2}gesqrt[n]{n!}$$
Since $n> frac{n+1}{2}$ for $n>1$, we have
$$n>sqrt[n]{n!}$$
And finally,
$$n^n > n!$$
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
answered Jul 1 '14 at 3:39
Yiyuan LeeYiyuan Lee
12.9k42960
12.9k42960
add a comment |
add a comment |
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$begingroup$
You should start with $(k+1)!<k^k(k+1)$
$endgroup$
– Adam Hughes
Jul 1 '14 at 3:15
3
$begingroup$
Also, $(k+1)^kneq k^k+1^k$.
$endgroup$
– vadim123
Jul 1 '14 at 3:17
$begingroup$
$2^2 neq 2$ so this should be edited
$endgroup$
– Squirtle
Jul 1 '14 at 3:24
$begingroup$
When you are proving that $A=B$ or that $Alt B$, do not ever start from $A=B$ or $Alt B$ and manipulate.
$endgroup$
– André Nicolas
Jul 1 '14 at 3:24
$begingroup$
cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1?
$endgroup$
– Mike
Jul 1 '14 at 3:37