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Prove that $n! > n^{3}$ for every integer $n ge 6$ using induction [duplicate]
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This question already has an answer here:
Proof by induction: inequality $n! > n^3$ for $n > 5$
3 answers
Proof.
Base case. When $n = 6$, LHS = $6! = 720$ and RHS = $6^{3} = 216$. Thus LHS > RHS.
Inductive step. Assume that $m! > m^{3}$ for some fixed integer $m$ such that $m ge 6$. We must show that it also holds true for $(m+1)$, i.e, $(m+1)! > (m+1)^{3}$. Observe that
begin{align*}
(m+1)! &= (m+1) cdot m! \
&> (m+1) cdot m^{3} text{$quad$(by the inductive hypothesis)}\
end{align*}
(still unfinished)
I know it boils down to showing that $(m+1) cdot m^{3} > (m+1)^{3}$ but I'm unsure on how to proceed.
elementary-number-theory induction
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
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marked as duplicate by N. F. Taussig, Namaste, saz, B. Goddard, Martin Sleziak Jan 24 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Proof by induction: inequality $n! > n^3$ for $n > 5$
3 answers
Proof.
Base case. When $n = 6$, LHS = $6! = 720$ and RHS = $6^{3} = 216$. Thus LHS > RHS.
Inductive step. Assume that $m! > m^{3}$ for some fixed integer $m$ such that $m ge 6$. We must show that it also holds true for $(m+1)$, i.e, $(m+1)! > (m+1)^{3}$. Observe that
begin{align*}
(m+1)! &= (m+1) cdot m! \
&> (m+1) cdot m^{3} text{$quad$(by the inductive hypothesis)}\
end{align*}
(still unfinished)
I know it boils down to showing that $(m+1) cdot m^{3} > (m+1)^{3}$ but I'm unsure on how to proceed.
elementary-number-theory induction
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
$endgroup$
marked as duplicate by N. F. Taussig, Namaste, saz, B. Goddard, Martin Sleziak Jan 24 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
There is already a SO question related to this one but the answer to this specific part is unsatisfactory. Furthermore, there seems to be some triviality to it that I'm failing to see.
$endgroup$
– uzlxxxx
Jan 23 at 19:26
3
$begingroup$
Does this answer help?
$endgroup$
– user574848
Jan 23 at 19:29
add a comment |
$begingroup$
This question already has an answer here:
Proof by induction: inequality $n! > n^3$ for $n > 5$
3 answers
Proof.
Base case. When $n = 6$, LHS = $6! = 720$ and RHS = $6^{3} = 216$. Thus LHS > RHS.
Inductive step. Assume that $m! > m^{3}$ for some fixed integer $m$ such that $m ge 6$. We must show that it also holds true for $(m+1)$, i.e, $(m+1)! > (m+1)^{3}$. Observe that
begin{align*}
(m+1)! &= (m+1) cdot m! \
&> (m+1) cdot m^{3} text{$quad$(by the inductive hypothesis)}\
end{align*}
(still unfinished)
I know it boils down to showing that $(m+1) cdot m^{3} > (m+1)^{3}$ but I'm unsure on how to proceed.
elementary-number-theory induction
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
$endgroup$
This question already has an answer here:
Proof by induction: inequality $n! > n^3$ for $n > 5$
3 answers
Proof.
Base case. When $n = 6$, LHS = $6! = 720$ and RHS = $6^{3} = 216$. Thus LHS > RHS.
Inductive step. Assume that $m! > m^{3}$ for some fixed integer $m$ such that $m ge 6$. We must show that it also holds true for $(m+1)$, i.e, $(m+1)! > (m+1)^{3}$. Observe that
begin{align*}
(m+1)! &= (m+1) cdot m! \
&> (m+1) cdot m^{3} text{$quad$(by the inductive hypothesis)}\
end{align*}
(still unfinished)
I know it boils down to showing that $(m+1) cdot m^{3} > (m+1)^{3}$ but I'm unsure on how to proceed.
This question already has an answer here:
Proof by induction: inequality $n! > n^3$ for $n > 5$
3 answers
elementary-number-theory induction
elementary-number-theory induction
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
asked Jan 23 at 19:26
uzlxxxxuzlxxxx
354
354
marked as duplicate by N. F. Taussig, Namaste, saz, B. Goddard, Martin Sleziak Jan 24 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by N. F. Taussig, Namaste, saz, B. Goddard, Martin Sleziak Jan 24 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
There is already a SO question related to this one but the answer to this specific part is unsatisfactory. Furthermore, there seems to be some triviality to it that I'm failing to see.
$endgroup$
– uzlxxxx
Jan 23 at 19:26
3
$begingroup$
Does this answer help?
$endgroup$
– user574848
Jan 23 at 19:29
add a comment |
$begingroup$
There is already a SO question related to this one but the answer to this specific part is unsatisfactory. Furthermore, there seems to be some triviality to it that I'm failing to see.
$endgroup$
– uzlxxxx
Jan 23 at 19:26
3
$begingroup$
Does this answer help?
$endgroup$
– user574848
Jan 23 at 19:29
$begingroup$
There is already a SO question related to this one but the answer to this specific part is unsatisfactory. Furthermore, there seems to be some triviality to it that I'm failing to see.
$endgroup$
– uzlxxxx
Jan 23 at 19:26
$begingroup$
There is already a SO question related to this one but the answer to this specific part is unsatisfactory. Furthermore, there seems to be some triviality to it that I'm failing to see.
$endgroup$
– uzlxxxx
Jan 23 at 19:26
3
3
$begingroup$
Does this answer help?
$endgroup$
– user574848
Jan 23 at 19:29
$begingroup$
Does this answer help?
$endgroup$
– user574848
Jan 23 at 19:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you want to show $(m+1)cdot m^3>(m+1)^3$, this is equivalent to
$$m^3>(m+1)^2.$$
As $m>2$, this is implied by
$$2m^2>(m+1)^2,$$
which is equivalent to
$$m^2>2m+1$$
or
$$(m-1)^2>2,$$
which is true if $mgeq 3$.
(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want to show $(m+1)cdot m^3>(m+1)^3$, this is equivalent to
$$m^3>(m+1)^2.$$
As $m>2$, this is implied by
$$2m^2>(m+1)^2,$$
which is equivalent to
$$m^2>2m+1$$
or
$$(m-1)^2>2,$$
which is true if $mgeq 3$.
(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
add a comment |
$begingroup$
If you want to show $(m+1)cdot m^3>(m+1)^3$, this is equivalent to
$$m^3>(m+1)^2.$$
As $m>2$, this is implied by
$$2m^2>(m+1)^2,$$
which is equivalent to
$$m^2>2m+1$$
or
$$(m-1)^2>2,$$
which is true if $mgeq 3$.
(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
add a comment |
$begingroup$
If you want to show $(m+1)cdot m^3>(m+1)^3$, this is equivalent to
$$m^3>(m+1)^2.$$
As $m>2$, this is implied by
$$2m^2>(m+1)^2,$$
which is equivalent to
$$m^2>2m+1$$
or
$$(m-1)^2>2,$$
which is true if $mgeq 3$.
(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
If you want to show $(m+1)cdot m^3>(m+1)^3$, this is equivalent to
$$m^3>(m+1)^2.$$
As $m>2$, this is implied by
$$2m^2>(m+1)^2,$$
which is equivalent to
$$m^2>2m+1$$
or
$$(m-1)^2>2,$$
which is true if $mgeq 3$.
(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Jan 23 at 19:36
Carl SchildkrautCarl Schildkraut
11.7k11443
11.7k11443
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
add a comment |
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
$begingroup$
Thanks. This makes things a lot clearer.
$endgroup$
– uzlxxxx
Jan 24 at 16:47
add a comment |
$begingroup$
There is already a SO question related to this one but the answer to this specific part is unsatisfactory. Furthermore, there seems to be some triviality to it that I'm failing to see.
$endgroup$
– uzlxxxx
Jan 23 at 19:26
3
$begingroup$
Does this answer help?
$endgroup$
– user574848
Jan 23 at 19:29