RE: Inductive proof
- To: mathgroup at smc.vnet.net
- Subject: [mg37900] RE: [mg37897] Inductive proof
- From: "David Park" <djmp at earthlink.net>
- Date: Mon, 18 Nov 2002 00:53:16 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
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