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Prove by induction: $A(n) = 2^n \lt\ n!$

$A(n+1) = 2^{n+1} \lt\ (n+1)!$

The expression below is true for $n = 5$.

$2^n \lt\ \frac{n!}{2}$

Then, we have:

$2^n \lt\ \frac{n!}{2} \lt\ n! \lt\ (n+1)!$

Which implies that:

$2^{n+1} \lt\ (n+1)!$ for every $n \ge\ 5$

Is this proof correct?

Update: the case is true for $n = 4$. It's true for $n = 5$, too. I started the induction proof with $n = 5$.

Let be more clear:

$A(n) = 2^n \lt\ n!$ is true for $n = 5$.

For $n= 5$, the expression $2^n \lt\ \frac{n!}{2}$ is true too.

From the expression above, we can deduce: $$2^n \lt\ \frac{n!}{2} \lt\ n! \lt\ (n+1)!$$

$$2^n * 2 \lt\ n! \lt\ (n+1)!$$ Hence: $$2^{n+1} \lt\ (n+1)!$$

I would like to know if there is something wrong with this proof. Remember, I assumed the expression is true for $n=5$, not $n=4$ (which is true too).

Update 2: I think is neccessary to prove the expression below by induction first:

"For $n= 5$, the expression $2^n \lt\ \frac{n!}{2}$ is true too."

We don't know if it's true for $ n\gt\ 5$

Daniel W. Farlow
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Carlitos_30
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    Where did you use induction? Try something more like the following: $$2^{n+1}=2\cdot2^n<2\cdot n!<(n+1)\cdot n!=(n+1)!.$$ – Daniel W. Farlow Jul 29 '16 at 20:33
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    To use induction you should be careful in choosing a base case. Certainly $2^1 \not \lt 1!$. – hardmath Jul 29 '16 at 20:37
  • Please, read the update. Thanks. – Carlitos_30 Jul 29 '16 at 20:54
  • @Carlitos_30 The other answers are perfectly fine, and you should find them helpful, but I tried to address your question concerning your proof. My main issue concerning your proof is that it is completely unclear what you are assuming and where you are using your assumption. Where exactly do you use the inductive hypothesis? Where exactly do you make the condition that $n\geq5$ turn into a useful fact? Etc. – Daniel W. Farlow Jul 29 '16 at 21:11

4 Answers4

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For $n=4$ we have

$$2^4<4!$$

Now, assume for some $k>4$ we have $$2^k<k!$$

Then, note that

$$\begin{align} 2^{k+1}&=2\times 2^k\\\\ &<2\times k!\\\\ &<(k+1)k!\\\\ &=(k+1)! \end{align}$$

And we are done!

Mark Viola
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Since you are asking specifically for whether or not your proof is correct, I have to say...if I were the one grading your work, then I would probably give it 5/10 or 6/10. Why? Well, it seems you are missing the main point concerning proofs by induction. That is, you need to

  1. Prove that your result is true for some specific number $n=\,?$ (this is the base case).
  2. Fix some $k\geq\,?$ and assume (this assumption is the inductive hypothesis) that your statement holds for $k$. Then you show that, based on your assumption, your statement is also true for $k+1$.

You really should read this post in the future on how to write up clear induction proofs. To that end, consider the following, which is the core part of the inductive argument.

\begin{align} 2^{k+1}&= 2\cdot2^k\tag{exponent rule}\\[1em] &< 2\cdot(k!)\tag{by inductive hypothesis}\\[1em] &< (k+1)\cdot k!\tag{$k\geq4\implies2<k+1$}\\[1em] &= (k+1)!\tag{factorial definition} \end{align} All of the other "stuff" you can write out in that template I linked to, but the above part is the main argument.

Community
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Daniel W. Farlow
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  • Thanks for your time. About my proof, the whole question can be reduced to this: The expression is true for n = 5. Now, we suppose that $2^k \lt\ k!$ is true for some k (which obvoiusly has to be greater or equal than 5). From $2^k \lt\ k!$ we can deduce that $2^k \lt\ \frac{k!}{2}$, being assumed that the least $k$ value in this example is 5. This is the argument I am not sure if is valid in this stage of the proof. – Carlitos_30 Jul 29 '16 at 22:17
  • @Carlitos_30 Given $2^k\lt k!$, how do you deduce that $2^k\lt \frac{k!}{2}$? You can prove the latter statement (independently) but you can't automatically deduce it from the former. – Steven Stadnicki Aug 01 '16 at 19:44
  • Because $n \ge\ 5$, it holds when $n = 5$. The third step in the proof of Daniel is justified given that $k \ge\ 4$ – Carlitos_30 Aug 01 '16 at 21:51
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How to structure a proof by induction:

Base case: $n=4$

$2^4 < 4!$

Inductive hypothesis: $2^n < n!$

Show that: $2^{n+1}<(n+1)!$ based on the inductive hypothesis.

$2^{n+1} = 2 \cdot 2^{n}<2\cdot 2 (n!) < (n+1)!$

For all $n\ge 4, 2^n < n!$

Doug M
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Claim: $2^n < n!$ for $n\geq 4$.

Base case: $2^4 < 4! \implies 16 < 24$, which is true.

Inductive step: Assume that $2^n < n!$ for some $n \geq 4$. Show that $2^{n+1} < (n+1)!$ as follows:

$2^n < n!$

$2^n \cdot 2 < n! \cdot 2$

and since $n! \cdot 2 < n! \cdot (n+1)$, we have

$2^n \cdot 2 < n! \cdot (n+1)$

$2^{n+1} < (n+1)!$

KaliMa
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