Prove that $n! > n^{3}$ for every integer $n ge 6$ using induction [duplicate]












5












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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







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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.


















  • $begingroup$
    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
















5












$begingroup$



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







share|cite|improve this question









$endgroup$



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














5












5








5





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. This makes things a lot clearer.
    $endgroup$
    – uzlxxxx
    Jan 24 at 16:47


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. This makes things a lot clearer.
    $endgroup$
    – uzlxxxx
    Jan 24 at 16:47
















3












$begingroup$

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.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. This makes things a lot clearer.
    $endgroup$
    – uzlxxxx
    Jan 24 at 16:47














3












3








3





$begingroup$

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.)






share|cite|improve this answer









$endgroup$



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.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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



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