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I have this statement:

Which I represent using the predicate P(n).

In order to prove this I have to provide a 'first' case which satisfies this expression. Which in this example is 4. Because every value greater than or equal to 4 will indeed support the statement. Then prove the fact for n+1... but I am really confused about how is the proof structure.

I need to write a proof... and in proofs the steps have to be justified, and choosing a random value of n = 4 won't be ok without its respective justification, so what could I put there?

Martin Sleziak
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JOX
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    http://math.stackexchange.com/questions/76946/proof-the-inequality-n-geq-2n-by-induction – lab bhattacharjee Nov 25 '13 at 06:02
  • ∀ n ∈ Z+, n ≥ 4 => P(n) Is this the statement I want to prove? – JOX Nov 25 '13 at 07:16
  • You're not just proving it for $n=5$. You need to show that if it's true for $n=k$, then it's true for $n=k+1$. That, together with the fact that it's true for $n=4$, shows that it's true for all $n\ge 4$. (If you chose a random value of $n=10$ to start with, your proof would show the weaker result that it's true for all $n\ge 10$. Which is also valid.) – mjqxxxx Nov 25 '13 at 08:05
  • We learned that if f(0) and f(n)->f(n+1), then for all natural numbers, f(n)

    But I don't know how to adjust the proof to satisfy an expression which will only be valid for n>=4?

     – JOX Nov 25 '13 at 19:39
  • Daniel, it may help you to consider $g(n) = f(n+4)$. So $g(0) = f(4)$, $g(1) = f(5)$, etc. Now, $f$ is true for all integers greater than or equal to 4 if and only if $g$ is true for all natural numbers. – Doug Chatham Nov 25 '13 at 20:22
  • Interesting, seems like a good approach. By the way, I am pretty confused about the symbolic representation, what would I have to prove 'symbolically'. – JOX Nov 25 '13 at 20:33

1 Answers1

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When $n=4$, we have $2^n=2^4=16$ and $n!=4!=24$, so the base case holds. Now, assume $2^k<k!$ for some $k\ge4$.

We wish to show that $2^{k+1}<(k+1)!$. So, $$2^{k+1}=2*2^k<2*k!<(k+1)*k!=(k+1)!$$

This follows from the assumption that $2^k<k!$ and the fact that $2<k+1$, since $k>4$.

Thus, by induction, $2^n\lt n!$ for all $n\ge 4$.

Now, show that this is not true when $n=3$. Thus, we have a strict base case.

Bonnaduck
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  • The problem is how do we conclude that we need to use 4 as a 'base' for the induction. – JOX Nov 25 '13 at 20:55
  • Oops. Edited the answer to reflect that. Showing that it does not hold for n=3 shows that we need n=4 to be our base case. – Bonnaduck Nov 25 '13 at 21:01
  • And what is the correct representation for the statement?

    ∀ n ∈ Z+, n ≥ 4 ∧ P(n) ?

     – JOX Nov 25 '13 at 21:05