MATHmaniac aka CrUNchy

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,

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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

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To all our lecturers,Dr Jesni,Prof Torla and of course,Br Supian..
Thank you so much for teaching us this whole sem,.
We also want to apologize for any mistakes that we have done,
You guys are not only our lecturers but also a brother and father to us..
TERIMA KASIH,CIKGU!!
CTS ROCK!!

Same to all friends,
Sorry,Thank You and Good Luck in everything!!

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