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RE: RE: Inductive proof

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37910] RE: [mg37900] RE: [mg37897] Inductive proof
  • From: "David Park" <djmp at earthlink.net>
  • Date: Tue, 19 Nov 2002 03:51:25 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Whoops. I made an error in my proof in using n instead of n + 1. This
corrects it.

Print["We wish to show the following for n > 6"]
step1 = 3^n < n!
Print["It is true for n == 7"]
step1 /. n -> 7
Print["Assuming it is true for n, then for n+1..."]
step2 = step1 /. n -> n + 1
step3 = step2 /. {3^(n + 1) -> 3*HoldForm[3^n],
    (1 + n)! -> (n + 1)*HoldForm[n!]}
Print["Using 3 < n + 1 and \!\(3\^n\)< n!.."]
Simplify[step3, 3 < n + 1 && HoldForm[3^n] < HoldForm[n!] &&
   HoldForm[3^n] > 0]


From: David Park [mailto:djmp at earthlink.net]
To: mathgroup at smc.vnet.net

Sam,

Using Mathematica we might do it this way. First check that it is true for
some low numbers greater than 6.

Table[{n, 3^n, n!}, {n, 6, 15}] // TableForm

It looks like it is so maybe it's worthwhile attempting a proof. Paste in
and evaluate the following for the proof.

Print["We wish to show the following for n > 6"]
step1 = 3^n < n!
Print["It is true for n == 7"]
step1 /. n -> 7
Print["Assuming it is true for n, then for n+1..."]
step2 = step1 /. n -> n + 1
step3 = step2 /. {3^(n + 1) -> 3*HoldForm[3^n],
    (1 + n)! -> n*HoldForm[n!]}
Print["Using 3 < n and \!\(3\^n\)< n!.."]
Simplify[step3, 3 < n && HoldForm[3^n] < HoldForm[n!] &&
   HoldForm[3^n] > 0]

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/


From: Sam [mailto:skapoor2 at hotmail.com]
To: mathgroup at smc.vnet.net

Hi:

Can anyone help me solve this by induction

3 to the power n < n!, when n is > 6

Thanks:

-Sam






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