ganeshie8
  • ganeshie8
show that \[\large \dfrac{3^n-1}{2^n-1}\] is never an integer for \(n\gt 1\) and \(n\in\mathbb{Z}\)
Discrete Math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can u make it clear pls
ganeshie8
  • ganeshie8
hey please check now @chethus
anonymous
  • anonymous
still not clear

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
basically the problem is about showing that expression always evaluates to some non integer
anonymous
  • anonymous
i should know the statement rite?
anonymous
  • anonymous
*
dan815
  • dan815
even/odd?
dan815
  • dan815
thats a start
ganeshie8
  • ganeshie8
looks like a good start! i know \(3\mid (2^n-1)\) when \(n\) is even
dan815
  • dan815
|dw:1435130862127:dw|
anonymous
  • anonymous
It will be a rational number if \[2^n-1\neq0\] as both numerator and denominator will be integers clearly \[2^n \neq 1\]\[n \log2 \neq \log1\]\[n \log 2 \neq 0\]\[n \neq 0\] which is satisfied
dan815
  • dan815
|dw:1435130938137:dw|
dan815
  • dan815
ok i see why 3 divides it now
ganeshie8
  • ganeshie8
woah you always find a way to use binomial theorem !
ikram002p
  • ikram002p
when n is even :- 3|2^n-1 thus 2^n-1=3k bur 3 do not divide 3^n-1 3^n-1 cant be of the form 3f
dan815
  • dan815
lols its all i know
anonymous
  • anonymous
ganeshie can u check my post?see above
ganeshie8
  • ganeshie8
im pretty sure you can derive below identity with binomial thm too \[4^n-1^n = (4-1)(4^{n-1}+4^{n-2}+\cdots+1) = 3(stuff)\]
anonymous
  • anonymous
no wait it seems wrong XD
ganeshie8
  • ganeshie8
@Nishant_Garg yes the given expression is always a rational number for \(n\gt 1\) but that doesn't prove it can never be an integer right
dan815
  • dan815
i think this means we solved it for even atleast
ikram002p
  • ikram002p
huh duh
dan815
  • dan815
we dont need u here U OS HATER
anonymous
  • anonymous
hmmmm
ganeshie8
  • ganeshie8
haha when ppl say i hate myself, they don't literally mean it
ikram002p
  • ikram002p
in general k^n-1 do not divide g^n-1 for gcd(k,g)=1 both primes but that need primitive thingy
ikram002p
  • ikram002p
no no no dan i dont hate os :O
dan815
  • dan815
|dw:1435131204394:dw|
ikram002p
  • ikram002p
ok i have other idea
ganeshie8
  • ganeshie8
so are you saying \[\large \dfrac{g^n-1}{k^n-1}\] is never an integer when \(\gcd(g,k)=1\) ?
ganeshie8
  • ganeshie8
@ikram002p
ikram002p
  • ikram002p
yes
ikram002p
  • ikram002p
well no -.-
dan815
  • dan815
|dw:1435131317714:dw|
ikram002p
  • ikram002p
only when k is 2 (sorry not focused )
ganeshie8
  • ganeshie8
okay odd case looks a bit more involved
dan815
  • dan815
the bottom is 2 mod 3
dan815
  • dan815
|dw:1435131616477:dw|
ganeshie8
  • ganeshie8
top is 2 mod 3 bottom is 1 mod 3
ikram002p
  • ikram002p
cool
ganeshie8
  • ganeshie8
i didn't get how that helps
dan815
  • dan815
oh nvm thats just silly
dan815
  • dan815
|dw:1435131948602:dw|
dan815
  • dan815
|dw:1435132054057:dw|
dan815
  • dan815
|dw:1435132120162:dw|
dan815
  • dan815
|dw:1435132232942:dw|
dan815
  • dan815
does any of this help
ikram002p
  • ikram002p
can we try mod 5 -.-
dan815
  • dan815
or 7
dan815
  • dan815
can we do induction maybe
UsukiDoll
  • UsukiDoll
@chethus this is a proof problem (pure math) which means we have to use different definitions or theorems to prove or disprove a statement. At first, it's hard to grasp because pure math is way different than computation math which is easily found in elementary, middle, or high schools
dan815
  • dan815
what does fermats little theorem say again
UsukiDoll
  • UsukiDoll
If p is a prime number then for any integer a , the number \[a^p-a\] is an integer multiple of p. IN the notation of modular arithmetic. this is expressed as \[a^p \equiv a (\mod p) \]
ikram002p
  • ikram002p
i give up, have no clue to start with.
UsukiDoll
  • UsukiDoll
If a is not divisible by p, Fermat's Little Theorem is equivalent to the statement that \[a^{p-1} -1 \] is an integer multiple of p or in symbols \[a^{p-1} \equiv 1 (\mod p)\]
ikram002p
  • ikram002p
bbye every one :)
UsukiDoll
  • UsukiDoll
\[\large \dfrac{3^n-1}{2^n-1} \] looks similar to the a not divisible by p . I feel like letting n =p-1 but that's just silly
dan815
  • dan815
did u find a solution to this yetq
UsukiDoll
  • UsukiDoll
no. T_T
ganeshie8
  • ganeshie8
me neither, but i feel there exists a simple way for the odd case too
UsukiDoll
  • UsukiDoll
who came up with this T_T

Looking for something else?

Not the answer you are looking for? Search for more explanations.