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 Post subject: Proof By InductionPosted: Tue, 21 Jun 2011 13:59:10 UTC
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Joined: Sat, 23 Oct 2004 05:23:15 UTC
Posts: 435
Hi all..

I'm trying to prove the following...

3^n > 3n + 1 for all positive integers n > 1... (*)

I'm using induction on n to prove the above..

Basis Step:

n = 2: 3^2 = 9 > 3(2) + 1 = 7

Inductive Step:

Assume (*) is true for some value of n > 1 (Inductive Hypothesis)

Need to show it is also true for n + 1

That is need to show:

3^(n+1) > 3(n+1) + 1

3^(n+1) > 3n + 4

Now

3^(n+1) =

3 * 3^n > 3 * (3n+1) (By I.H.) = 9n + 3

My question is, do I just say that 9n + 3 > 3n + 4 at this point and conclude that (*) is true? Or is there some other justification I need to show.

Thank you

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 Post subject: Re: Proof By InductionPosted: Tue, 21 Jun 2011 20:09:17 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12075
Location: Austin, TX
MathWhiz wrote:
Hi all..

I'm trying to prove the following...

3^n > 3n + 1 for all positive integers n > 1... (*)

I'm using induction on n to prove the above..

Basis Step:

n = 2: 3^2 = 9 > 3(2) + 1 = 7

Inductive Step:

Assume (*) is true for some value of n > 1 (Inductive Hypothesis)

Need to show it is also true for n + 1

That is need to show:

3^(n+1) > 3(n+1) + 1

3^(n+1) > 3n + 4

Now

3^(n+1) =

3 * 3^n > 3 * (3n+1) (By I.H.) = 9n + 3 this is not by the IH, this is by the distributive law

My question is, do I just say that 9n + 3 > 3n + 4 at this point and conclude that (*) is true? Or is there some other justification I need to show.

Thank you

NO! Almost, but not quite. What you need to show is that, for that , this will complete your proof.

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 Post subject: Re: Proof By InductionPosted: Tue, 21 Jun 2011 20:38:06 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sat, 23 Oct 2004 05:23:15 UTC
Posts: 435
MathWhiz wrote:
Hi all..

I'm trying to prove the following...

3^n > 3n + 1 for all positive integers n > 1... (*)

I'm using induction on n to prove the above..

Basis Step:

n = 2: 3^2 = 9 > 3(2) + 1 = 7

Inductive Step:

Assume (*) is true for some value of n > 1 (Inductive Hypothesis)

Need to show it is also true for n + 1

That is need to show:

3^(n+1) > 3(n+1) + 1

3^(n+1) > 3n + 4

Now

3^(n+1) =

3 * 3^n > 3 * (3n+1) (By I.H.) = 9n + 3 this is not by the IH, this is by the distributive law

My question is, do I just say that 9n + 3 > 3n + 4 at this point and conclude that (*) is true? Or is there some other justification I need to show.

Thank you

NO! Almost, but not quite. What you need to show is that, for that , this will complete your proof.

Thanks..

You mean for n > 1

and I meant 3 * 3^n > 3 * (3n+1) (By I.H.)

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Mathematics is one of the essential emanations of the human spirit, a thing to be valued in and for itself, like art or poetry. -- Oswald Veblen, 1924.

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 Post subject: Posted: Tue, 21 Jun 2011 20:52:59 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12075
Location: Austin, TX
It doesn't matter for the second part, but yes, if you show it for n>0, then in particular it's true for n>1. And you are correct with your application of the I.H. if that is where you intended it.

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