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I have to prove $n for all $n>2$ by mathematical induction.

I did it as follows. I proved the base case.

Then let it be true for $K>2$:

$$ K

I have to prove,

$$ K+1<(K+1)! $$

$$ K

Adding $1$ on both sides

$$ K+1

Hence

$$ K+1<(K+1)! $$

Is the last step I did ($K!+1<(K+1)!$) Right?

Please help me out. Thanks!

  • 0
    I think you should prove $K!+1<(K+1)!$ which is not so hard.2011-11-05
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    No. For the n=3 case, you just check if 3<3! which is true. Then assume for n=k case and prove for the n=k+1 case.2011-11-05
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    @KaratugOzanBircan: Thanks2011-11-05
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    What do you mean by n2?2011-11-05
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    @Rankeya: I guess, it should be $n>2.$2011-11-05
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    Then Akito's question reads "I have to prove $n > 2$ by mathematical induction", and this does not make sense either does it?2011-11-05
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    Akito what is the exact statement that you need to prove using induction?2011-11-05

3 Answers 3

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Why use induction at all when you can see plainly that $n! = n\cdot (n-1) \cdots 1 \ge n\cdot (n-1) > n$ if $n>2$ ?

If this is an exercise, it's a silly one. It's exercises like this that give a bad name to induction...

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    Amen. Though to be fair, exercises that specify an unreasonable method are hardly limited to induction, never mind exercises that specify a reasonable method when others would do as well.2011-11-05
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    or similarly $n! = n\cdots 2\cdot 1 \ge n\cdot 2 \gt n$ if $n \gt 2$, making it clearer why $2$ matters.2011-11-06
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    @Henry, sure. It proves something stronger than required but so does my argument, and mine provides a quadratic lower bound :-)2011-11-06
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Starting with $K \lt K!$,

multiplying both sides by $K+1 \gt 0$ you have $K(K+1) \lt (K+1)!$,

but since $K\gt 2$ you have $K+1 \lt K(K+1) $ and so $K+1 \lt (K+1)!$

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Since in your case $K > 2$, the statement $K!+1 < (K+1)!$ is true. This is because $(K+1)! = K!(K+1) = K!K + K!$, and $K!K > 1$ for $K > 1$.