Question

when n > 5. Prove that for all k. we have

Answer 1

\[(n-1)! +1\neq n ^{k}\]

Answer 2

so we are showing that these two expressions are not equal?

Answer 3

yea

Answer 4

my first thought is find the contrapositive of your statement

Answer 5

would we prove it by contradiction? we make the equation equal, then prove it somehow. but not sure how to go about it

Answer 6

maybe...

Answer 7

if we sub in n = 6, we would get 121 = 6^k, which doesnt work, but surely its not that simple ?

Answer 8

well we want to show it doesn't = for all k

Answer 9

let me see if zarkon is here...
one sec..

Answer 10

ok thanks

Answer 11

Induction?

Answer 12

not sure induction works its the k which confuses me

Answer 13

Show there exist k such that \[(n-1)!+1=n^k\], then \[n \le 5\].
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So if k=1, then
\[(n-1)!+1=n^1=n\]
Does (n-1)!+1=n for any n?
for n=2 it works: (2-1)!+1=1+1=2=2

Answer 14

thanks. yea i get what youve done, it works for all n. next step put in n = n+1 ? and not sure how to prove it for all k appreciate the help

Answer 15

now let's assume it is true for some integer j
so there is some n such that \[(n-1)!+1=n^j\]

now let's show the expression is true for some n if j=k+1
so we have
\[(n-1)!+1=n^{j+1}\]
\[(n-1)!+1=n^j \cdot n \]
(n-1)!+1=[(n-1)!+1] \cdot n\]
wouldn't that mean n=1
but i'm totally sure how that shows n will be less than 5

Answer 16

k=j+1*

Answer 17

sorry whats the last formula?

Answer 18

\[(n-1)!+1=[(n-1)!+1] \cdot n\]
1=n

Answer 19

thanks

Answer 20

but gummage this n is less than 5
but i don't see how this proves what we want to prove
we just showed it is true for some n

Answer 21

we proved that when n < 5 the equation is true, so by contradiction when n > 5 the equation is incorrection, which has proved the question surely? or is there more to it

Answer 22

i was trying to prove the contrapositive of your statement
if i can do then that means the orginal statement is true
i think there is more to it

We showed the expression is true for all k for some n
we didn't show for which n it is true
so the part that is left is to show n will always be less than or =5

Answer 23

james what do you think?

Answer 24

are you sure we're on the right tracks here? lost with this proof now

Answer 25

I'm thinking about it. (btw, notice it is also false for n = 4.)

Answer 26

I'm just posting so I will get a notice when this get's solved. I wanna see how it's done :)

Answer 27

@gummage: what course is this question from?

Answer 28

I gotta get to school. I hope I can see the solution when I get back. Later.

Answer 29

sorry back now. still working on it, it's from a Proofs module. It's revision for a test on proofs.

Answer 30

any progress?

Answer 31

think ive done it eventually... proving using contrapositive.

if you assume k is a natural number and (n−1)!+1= n^k , and sub in n = 6.

then you get 121 = 6^k

take logs from that log(121) = k*log(6)

log(121)/log(6) = k therefore k is not a natural number so it doesn't work.

not sure about this but yeah... what you guys think?