MHB Is \( n! > 2^n \) for \( n \ge 4 \)?

  • Thread starter Thread starter karush
  • Start date Start date
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\tiny{2.3.3}$
Prove $n! > 2^n$ for $n\ge4$
ok well the only thing I know to do is just try some numbers

$n=4, \quad 4\cdot3\cdot2\cdot1=24 \quad 2^4=32 \quad \therefore 32\ge24$
$n=5, \quad 5\cdot4\cdot3\cdot2\cdot1=120 \quad 2^5=32 \quad \therefore 120\ge32$

not sure just what the proof would be and looks like it must just be intergers

[DESMOS]advanced: {"version":5,"graph":{"showGrid":false,"xAxisStep":1,"squareAxes":false,"viewport":{"xmin":-2.621396030848641,"ymin":-129.26632180463903,"xmax":26.406631626246817,"ymax":78.78768164675861}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"x!\\ \\left\\{x\\ge0\\right\\}","style":"SOLID"},{"type":"expression","id":"2","color":"#388c46","latex":"2^x\\ \\left\\{x\\ge0\\right\\}","style":"SOLID"}]}}[/DESMOS]uld be
 
Physics news on Phys.org
karush said:
$\tiny{2.3.3}$
Prove $n! > 2^n$ for $n\ge4$
ok well the only thing I know to do is just try some numbers

$n=4, \quad 4\cdot3\cdot2\cdot1=24 \quad 2^4=32 \quad \therefore 32\ge24$
$n=5, \quad 5\cdot4\cdot3\cdot2\cdot1=120 \quad 2^5=32 \quad \therefore 120\ge32$

not sure just what the proof would be and looks like it must just be intergers

uld be
Are you allowed to use induction? Let k be the lowest integer such that [math]k! > 2^k[/math]. Then show that [math](k + 1)! > 2^{k + 1}[/math]

-Dan
 
Re: aa2.3.3 Prove n! > 2^n for n\ge4

topsquark said:
Are you allowed to use induction? Let k be the lowest integer such that [math]k! > 2^k[/math]. Then show that [math](k + 1)! > 2^{k + 1}[/math]

-Dan

I thot that is what i did?

- - - Updated - - -
 
You showed a couple of examples which show a definite trend but didn't give an actual proof. This is a simple induction proof and basically copies what you have shown.

-Dan
 
Since $n\ge4=2^2$,
$$n!\ =\ \underbrace{n}_{\ge2^2} \cdot \underbrace{(n-1)}_{>2} \cdot \cdots \cdot \underbrace{3}_{>2} \cdot \underbrace{2}_{\ge2} \cdot 1\ >\ 2^n.$$
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top