Question

show that

Answer 1

\(d\mid n\) and \(c\mid(n/d)\) if and only if \(c\mid n\) and \(d\mid(n/c)\)

Answer 2

\(d\mid n\)->this means d is divisible by n right?

Answer 3

yes

Answer 4

\(n/d\) is the regular division

Answer 5

If we consider c=7
n=2
d=16
d is divisible by n
and c/(n/d) = 56

c is not divisible by n but still the condition holds true

Answer 6

sorry i was wrong earlier

Answer 7

\(d\mid n\)->this means d is divisible by n right?

this is wrong

Answer 8

\(d \mid n\)

means \(d\) divides \(n\)

Answer 9

example :
\(2\mid 6\)

\(3\mid 21\)

Answer 10

when reading you may replace "\(\mid\)" with "divides"

Answer 11

\[ \frac{ n }{ d} \div c =(n \div c) \div d\]

Answer 12

thats right, but how is that relevant here ?

Answer 13

the Rhs has to be a whole number
so (n/c) is whole or c|n
and n/c should also be divisible by d, or d|(n/c)

Answer 14

ohk.. got you :)

Answer 15

ok :)
\(\Large\frac{ n }{ d }=a\) and \(\Large\frac{ n }{ dc }=b\)

\(\large{a~and~b}\) are integers

now we look at this-> \(\Large\frac{ n }{ dc }=b\)

We know that d will divide n but then b is integer so c has to divide n or we can also say c has to divide n/d

we had to prove \(c\mid n\) which is done
and \(d\mid(n/c)\) is same as \(\Large\frac{ n }{ dc }=b\) :)

Answer 16

Nice!

Answer 17

This is part of a very beautiful proof.. I can share the proof if you're interested... because my next questions are based on this and I don't want to attempt them alone :)

Answer 18

yes, do share :)

Answer 19

yes share \(\Huge\ddot\smile\)

Answer 20

i see you're very good with \(\tau\) and \(\sigma\) functions

Answer 21

there is another function that is needed for the theorem that i am talking about

Answer 22

It is called mobius mu function : \(\mu\) function

please see the attached quick for the definition of this function

Answer 23

examples :

1) \(\mu(3^2*5)=0\) because \(3^2\) divides \(3^2*5\)

2) \(\mu(7)=(-1)^1 = -1\)

3) \(\mu(7*11) = (-1)^2 = 1\)

Answer 24

it is a very simple function, but is also very powerful

Answer 25

Ok :)

Answer 26

just to understand \(\mu\) function better, maybe lets attempt that problem

Answer 27

okay (:

Answer 28

we still on main problem ?
or have another question ?

Answer 29

ok for a part

the number n will be either divisible or 4 or it won't be

if \(\mu(n)\) is divisible by 4 or we can say \(2^2\) then its proved because then \(\mu(n)=0\)

if the number is not divisible by 4
it will be of the form 4k+1 or 4k+2 or 4k+3
its sure that if the n is of this ^ form them one of (n+1) / (n+2) / (n+3) will be divisible by 4
so for that \(]mu(n)\) will be 0

Answer 30

you're using the fact that one integer will be divisible by 4 in a set of four consecutive integers. nice

Answer 31

yes :)

Answer 32

main question is solved @ikram002p

Answer 33

that was easy, part b can be a bit tricky..

Answer 34

ok so it will be like this-\[\mu(1!)+\mu(2!)+\mu(3!)+\mu(4!)........\]
\(\mu(4!)=0\) because it is divisible by 4

all the \(\mu(n!)\) after \(\mu(4!)\) will be 0 because its sure they will be divisible by 4
so all the part from \(\mu(4!)\) becomes 0

we are left with this-\(\mu(1!)+\mu(2!)+\mu(3!)\) this is equal to 1+(-1)^1 +(-1)^2 =1

Answer 35

well part two we only need to know that
\(\mu (n)=0 ~if~ p^2|n \) and we know for n>=4, 2^2|n!
so
\(\Large \sum_{k=1}^n \mu(k!)=\\\Large \sum_{k=1}^4 \mu(k!)= \mu(1!)+\mu(2!)+\mu(3!)+\mu(4!)=1+(-1)+(1)+0=1\)

Answer 36

oh always late xD

Answer 37

lol

Answer 38

well i just not i put 4 instead of 3 =D . but nothing would change.

Answer 39

\(\mu(n!) =0\) for n>=4
so it should be
\(\Large \sum_{k=1}^n \mu(k!)= \Large \sum_{k=1}^3 \mu(k!)\\= \Large\mu(1!)+\mu(2!)+\mu(3!) \\\Large =1+(-1)+(1) =1\)

Answer 40

suppose \(g(n) = \sum\limits_{d\mid n}f(d)\),

is there a way to find the value of \(f(n)\) ?

Answer 41

if gn is defined yes there is, think of a question so we could try.