Question

Prove or Disprove: Given k; n belong to N If k/n! then k/i for some 0 <= i <= n.

Answer 1

consider \(n=0\). clearly \(n!=1\); if \(k|1\) it follows that \(k=1\) but \(0\le 1\not\le 0\)

Answer 2

but 0 does not belong to N

Answer 3

so that disproves it in the case that \(0\in\mathbb{N}\)

Answer 4

in our class 0 does not belong to N

Answer 5

so do you think that the proof will be by induction?

Answer 6

take \(n=4\) so \(n!=24\); clearly \(12|n!\) so we can take \(k=12\). it's easy to check, however, that \(12\) does not divide any of \(0,1,2,3,4\)

Answer 7

go smaller...

6|3!

Answer 8

\(n=4\) is just a sufficiently easy yet dramatic counterexample that I found it more useful but you're correct, \(n=3\) with \(k=6\) also works

Answer 9

so i can be 0

Answer 10

\[0\le i\le n\]

Answer 11

for any m in the natural numbers

m|0 since 0=0m

Answer 12

i got it, i was dividing!so i tought is true. Thanks!

Answer 13

if \(i\) can be zero then the problem is booring

Answer 14

here \(m\in\mathbb{N}=\{1,2,3,\ldots\}\) so no problem

Answer 15

since \(m\in\mathbb{N}\) then

\(m|0\) since \(m0=0\)

Answer 16

oops, misread

Answer 17

then @Zarkon is correct, so it follows the \(i=0\) possibility makes this trivially true for all \(k,n\)

Answer 18

you are not looking at it correctly.

for this problem \(k\in\mathbb{N}\) so it can't be zero

then I can find an integer (in this case 0) such that


k0=0

Answer 19

like i said...this problem is boring if \(i\) can take the value 0

Answer 20

oh so it is true since i equal to 0 but how can we proof it then?