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I'm doing discrete maths as a subject at my uni and I've been asked to solve the following equation, yet I'm having trouble understanding both what it's asking me to do and how I need to go about getting the answer.

I need to find the smallest natural number $a$ such that $a! > 3^{a}$. Now for $n$ a natural number let CLAIM($n$) be the statement:

$$n! > 3^{n}$$

Prove that CLAIM($n$) is true for all $n \ge a$.

5xum
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SteppingHat
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  • First, plug in various natural numbers to find the lowest one that works. Is it true for 1? 2? 3? Then from that point on, use induction – Alan Aug 31 '15 at 06:36
  • I think that $n=7$ is the smallest number such that $n! > 3^n$. – GAVD Aug 31 '15 at 06:53

1 Answers1

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You actually have two things to do:

  1. Find the smallest natural number $a$ such that $a!>3^a$. To do this, first try to plug in some values for $a$ and see what happens. $$1!=1, 3^1=3\\2!=2, 3^2=9\\3!=6, 3^3=27\\4!=24, 3^4=81$$ Can you continue?
  2. Prove that the claim is true for all $n\geq a$. Once you have found $a$, you can prove this claim by induction. The claim is obviously true for $n=a$ (since that's how $a$ is defined), so you have to just prove that if it is true for $n$, it is also true for $n+1$. So, you assume that $$n!>3^n$$ and you try to prove that $$(n+1)!>3^{n+1}.$$ Using the fact that $(n+1)!=(n+1)\cdot n!$ and $3^{n+1}=3\cdot 3^n$ should make the task easy.
5xum
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