n! > n^2 for every integer n >= 4
Infinity
n^2 for every integer n >= 4.
However, I am unable to do the induction step correctly. That is, I am
unable to show that (n + 1)! > (n + 1)^2 when n! > n^2. Could someone
please help me?
jonnie
"Infinity" <countabl...@gmail.com> wrote in message
news:46c43b9f-b101-429f...@s31g2000yqs.googlegroups.com...
is a mistake, not induction
try again
A N Niel
<46c43b9f-b101-429f...@s31g2000yqs.googlegroups.com>,
Infinity <countabl...@gmail.com> wrote:
The key thing to use is: (n+1)! is the product of (n+1) and n!
Infinity
> In article
> <46c43b9f-b101-429f-b746-436499af3...@s31g2000yqs.googlegroups.com>,
>
> Infinity <countableinfin...@gmail.com> wrote:
> > I want to show by the principle of mathematical induction that n! >
> > n^2 for every integer n >= 4.
>
> > However, I am unable to do the induction step correctly. That is, I am
> > unable to show that (n + 1)! > (n + 1)^2 when n! > n^2. Could someone
> > please help me?
>
> The key thing to use is: (n+1)! is the product of (n+1) and n!
yes, I tried it. I got
(n + 1)! > (n + 1)n^2
Now, I need to show that (n + 1)n^2 > (n + 1)^2. This would involve
showing that n^2 > (n + 1). So I have to solve one more problem of
mathematical induction to show that n^2 > (n + 1) for all n >= 4. Is
there an easier way to prove this where I can avoid solving one more
problem of mathematical induction?
Rick Decker
Yup. You need to show n^2 - n - 1 > 0 for n >= 4. Write
n^2 - n - 1 = (n - 1/2)^2 - 5/4
and then for n >= 4 you'll see that the right hand side is > 0.
Regards,
Rick
jonnie
yes, is trivial.
Bill Dubuque
> Infinity wrote:
>> On Jan 6, 12:24 am, A N Niel <ann...@nym.alias.net.invalid> wrote:
>>> Infinity <countableinfin...@gmail.com> wrote:
>>>>
>>>> I want to show by the principle of mathematical induction that n! >
>>>> n^2 for every integer n >= 4.
>>>> However, I am unable to do the induction step correctly. That is,
>>>> I am> unable to show that (n + 1)! > (n + 1)^2 when n! > n^2.
>>>
>>
>> yes, I tried it. I got (n + 1)! > (n + 1)n^2
>> Now, I need to show that (n + 1)n^2 > (n + 1)^2. This would involve
>> showing that n^2 > (n + 1). So I have to solve one more problem of
>> mathematical induction to show that n^2 > (n + 1) for all n >= 4. Is
>> there an easier way to prove this where I can avoid solving one more
>> problem of mathematical induction?
>
> Yup. You need to show n^2 - n - 1 > 0 for n >= 4. Write
>
> n^2 - n - 1 = (n - 1/2)^2 - 5/4
>
> and then for n >= 4 you'll see that the right hand side is > 0.
SIMPLER n>2 => n, n-1 > 1 => n(n-1) > 1
Rick Decker
Cute, Bill.
Sheepishly,
Rick
Susam Pal
> Infinity wrote:
> > On Jan 6, 12:24 am, A N Niel <ann...@nym.alias.net.invalid> wrote:
> >> In article
> >> <46c43b9f-b101-429f-b746-436499af3...@s31g2000yqs.googlegroups.com>,
>
> >> Infinity <countableinfin...@gmail.com> wrote:
> >>> I want to show by the principle of mathematical induction thatn! >
> >>> However, I am unable to do the induction step correctly. That is, I am
> >>> please help me?
> >> The key thing to use is: (n+1)! is the product of (n+1) andn!
>
> > yes, I tried it. I got
>
> > (n+1)! > (n+1)n^2
>
> > Now, I need to show that (n+1)n^2 > (n+1)^2. This would involve
> > mathematical induction to show thatn^2 > (n+1) for alln>= 4. Is
> > there an easier way to prove this where I can avoid solving one more
> > problem of mathematical induction?
>
> Yup. You need to shown^2 -n-1> 0 forn>= 4. Write
>
> n^2 -n-1= (n-1/2)^2 - 5/4
>
>
> Regards,
>
> Rick
n >= 2 => n^2 >= 2n = n + n > n + 1
Susam Pal
> >> Infinity wrote:
> >>> On Jan 6, 12:24 am, A N Niel <ann...@nym.alias.net.invalid> wrote:
> >>>> Infinity <countableinfin...@gmail.com> wrote:
> >>>>> However, I am unable to do the induction step correctly. That is,
> >>>> The key thing to use is: (n+1)! is the product of (n+1) andn!
> >>> yes, I tried it. I got (n+1)! > (n+1)n^2
> >>> mathematical induction to show thatn^2 > (n+1) for alln>= 4. Is
> >>> there an easier way to prove this where I can avoid solving one more
> >>> problem of mathematical induction?
>
> >> n^2 -n-1= (n-1/2)^2 - 5/4
>
>
> > SIMPLER n>2 => n,n-1>1 => n(n-1) >1
>
> Cute, Bill.
>
> Sheepishly,
>
> Rick
n >= 2 => n^2 >= 2n = n + n > n + 1
Zdislav V. Kovarik
On Thu, 7 Jan 2010, Susam Pal wrote:
> On Jan 6, 2:27 am, Rick Decker <rdec...@hamilton.edu> wrote:
> > Infinity wrote:
> > > On Jan 6, 12:24 am, A N Niel <ann...@nym.alias.net.invalid> wrote:
> > >> In article
> > >> <46c43b9f-b101-429f-b746-436499af3...@s31g2000yqs.googlegroups.com>,
> >
> > >> Infinity <countableinfin...@gmail.com> wrote:
> > >>> I want to show by the principle of mathematical induction thatn! >
> > >>>n^2 for every integern>= 4.
> > >>> However, I am unable to do the induction step correctly. That is, I am
> > >>> unable to show that (n+1)! > (n+1)^2 whenn! >n^2. Could someone
> > >>> please help me?
> > >> The key thing to use is: (n+1)! is the product of (n+1) andn!
> >
> > > yes, I tried it. I got
> >
> > > (n+1)! > (n+1)n^2
> >
> > > Now, I need to show that (n+1)n^2 > (n+1)^2. This would involve
> > > showing that
n^2 > (n+1).
Write down an identity
n^2 = n + 1 + ((n-4)^2 + 7*(n-4) + 11)
which proves what you need, for n>=4.
How to discover that? In the polynomial n^2-n-1, change the variable by
n=m+4, expand and substitute back: m=n-4.
Cheers, ZVK(Slavek).
[Nothing added by me from here down.]