1)Find the continued fraction of
1st step:
2nd step:
Find the gcd(13,7)
13 = 1 x 7 +6
7 = 1 x 6 + 1
6 = 1 x 6 + 0
gcd (13,7)=1
3rd step:
Multiply and
to obtain,
4th step:
Substitute it
Thus,
2)Find the continued fraction of
1st Step:
2nd step: find the gcd (23,21)
23=1 x 21 +2
21=10 x 2 +1
2=2x1+0
3rd step:
Multiply and
to obtain
4th step:Substitute
Thus,
prove that for any integer a, 3|a^3-a
let a be an integer.when we divide by 3,
we have 3 possiblities:
a=3k
a=3k+1
a=3k+2
we want to show that 3 divides a^3-a
we know that a^3=a(a^2-1)=a(a-1)(a+1)
1st case,
if a=3k,then a^3=3.k(3k-1)(3k+1)
Thus,3|a^3-a for a=3k
2nd case,
if a=3k+1,Then a^3-a=3((3k+2)(3k+1)(k))
Thus 3|a^3-a for a=3k+1
3rd case,
if a=3k+2,then a^3-a=3((3k+2)(3k+1)(k+1))
Thus, 3|a^3-a for a=3k+2
since 3|a^3-a for all cases,then we can conclude that 3|a^3-a for any integer a
Solve .Give 3 positive solutions.
First step:
Find the continued expansion of
thus,
Second step:
Since n=2 and it is even, the solutions are where k=1,2,3...
Thus,
.
.
.
are the positive solutions.
Third step:
Calculate the value of the numerator and denominator,
So,the answers for 3 positive solutions of are
Definition:
An arithmetic function is a function that defined for all positive integers
Definition:
An arithmetic function f is called multiplicative if f(mn) = f(m) f(n) whenever m and n are relatively prime positive integers.It is called multilplicative if
f(mn) = f(m) f(n) for all positive integers m and n.
Let's see...
and
are multiplicative function.
Let say . Find
and
.
Positive divisor of n=
by multiplying P,
thus,
Get it??
Let's use number,..
thus,
Let say k=5,so
