2

I just want to know if my proof to this question is correct. First, I proved it was true for $n = 4$. $$4!>2^4$$ $$24>16$$ Then, I assumed that it was true for $n=k$. $$k!>2^k$$ Afterwards, I tried to prove it for $n=k+1$. $$(k+1)!>2^{k+1}$$ $$(k+1)k!>(2)(2^k)$$ Since $k!>2^k$, am I allowed to simply do $$(k+1)(2^k)>(2)(2^k)$$ $$(k+1)>2$$ and this is true for all $k\geq3$. Would the last two steps be illegal because I only assumed that $k!>2^k$ is true?

joejacobz
  • 1,199
  • 1
    The basic idea is fine. The writeup should be changed somewhat. For example at an early stage you have the line $(k+1)!\gt 2^{k+1}$. That is allowed in secondary school, but not later. You should say that you want to show that $\dots$. Writing $\dots$ without explanation is the assertion that $\dots$ is true. – André Nicolas Feb 07 '13 at 01:41
  • About your question at the end, there is no problem. We have already that $k+1\gt 2$ if $k\ge 2$. It so happens you only need it for $k\ge 4$. – André Nicolas Feb 07 '13 at 01:46
  • Ok, so is that the only thing I would need to change, or are the some other things? – joejacobz Feb 07 '13 at 02:04
  • 1
    You wouldn't need to change the stuff at the end, everything you said there was true. Yes, you could have said "for $n\ge 2$, but "if $n\ge 3$" was true. The "style" (going from something you want to prove to something true, instead of the other way around) does need to change. – André Nicolas Feb 07 '13 at 02:14

1 Answers1

2

Your proof has essentially the right idea. The induction hypothesis is used to say $$(k+1)k!>(k+1)2^k>2\cdot 2^k=2^{k+1}.$$

Clayton
  • 25,087
  • 7
  • 60
  • 115