Question

Prove the statement is true using mathematical induction:

2^n-1 ≤ n!

Answer 1

Is \[n \ge 0\]?

Answer 2

do you know steps when you have to prove by induction?

Answer 3

Yea kind of

Answer 4

@pgpilot326 we consider the base of case, if n=0 satisfy, then, basic step at 0. if not, at 1

Answer 5

ok, show me step 1. (we must follow)

Answer 6

\[2^{0}=1=0!\]

Answer 7

use exactly notation <=

Answer 8

ok, too many cook will spoil the food, i let other help you

Answer 9

\[2^{(0-1)}=2^{-1}=1/2 \le 0!=1\]

Answer 10

what???? who are they???

Answer 11

I will do this in steps. Step 1.
\[2^{n-1} \le n!\] Consider the base step n = 0
\[2^{0-1} \le 0!\]
\[1/2 \le 1\] Base step holds.

Answer 12

\[2^{1-1}=2^{0}=1\le1!=1\]

Answer 13

\[2^{2-1}=2^{1}=2\le2!=2\]

Answer 14

I will let this guy continue, let's see what he comes up with cos typing up equations is annoying

Answer 15

pgpilot this math discerte

Answer 16

Although I am not sure what he is doing :))

Answer 17

Assume true for k, where k > 2. Then show it's true for k+1

Answer 18

you mean k >= 0 don't you?

Answer 19

i assume \[n \in \mathbb{Z}\]

Answer 20

Anyway here is the full proof but i'm not typing this up in pretty symbols
2^n-1 ≤ n!
n = 0 holds.
Assume 2^n-1 ≤ n! is true and prove it for n+1
2^n-1 x2 ≤ n! x 2 Multiplied both sides by 2
2^(n+1)-1 ≤ n! x 2

Now we need to show that n! x 2 ≤ (n+1)! which is easy.
(n+1)! = n!(n+1) since n >=0 we can say that n! x 2 ≤ (n+1)!
Hence:
2^(n+1)-1 ≤ (n+1)! as required.

Answer 21

@pgpilot326 0 is an integer.