MATHmaniac aka CrUNchy: 2009

CLICK HERE FOR BLOGGER TEMPLATES AND MYSPACE LAYOUTS �

Continued Fraction

Labels: Exercises | at 11:25 PM

1)Find the continued fraction of $\frac{21}{39}=\frac{7}{13}$

1st step:
$\frac{7}{13}=\frac{1}{\frac{13}{7}}$

2nd step:
Find the gcd(13,7)
13 = 1 x 7 +6 $\to 1$
7 = 1 x 6 + 1 $\to 2$
6 = 1 x 6 + 0
gcd (13,7)=1

3rd step:
Multiply
$\frac{1}{7}\rightarrow 1$ and $\frac{1}{6}\rightarrow 2$
to obtain,
$\frac{13}{7}=1+\frac{6}{7}$
$\frac{7}{6}=1+\frac{1}{6}$

4th step:
Substitute it
$\frac{7}{13}= 1 + \frac{1}{\frac{13}{7}}$
$\frac{7}{13}=\frac{1}{1+\frac{6}{7}}$
$\frac{7}{13}=\frac{1}{1+\frac{1}{1+\frac{1}{6}}}$

Thus, $\frac{7}{13}=\left [ 1;\overline{1,6} \right ]$

2)Find the continued fraction of $\frac{67}{23}$
$\frac{67}{23}= 2+ \frac{21}{23}$

1st Step:
$\frac{21}{23}=\frac{1}{\frac{23}{21}}$

2nd step: find the gcd (23,21)
23=1 x 21 +2 $\to 1$
21=10 x 2 +1 $\to 2$
2=2x1+0

3rd step:
Multiply $\frac{1}{21}\rightarrow 1$ and $\frac{1}{2}\rightarrow 2$ to obtain
$\frac{23}{21}=1+\frac{2}{21}$
$\frac{21}{2}=10+\frac{1}{2}$

4th step:Substitute
$\frac{21}{23}=\frac{1}{\frac{23}{21}}$
$\frac{21}{23}=\frac{1}{1+\frac{2}{21}}$
$\frac{21}{23}=\frac{1}{1+\frac{1}{10+\frac{1}{2}}}$

Thus, $\frac{67}{23}=[2;\overline{1,10,2}]$

Read Comments

Proof

Labels: Exercises | at 10:58 PM

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

Read Comments

XM mud....argggghhh

| at 10:27 PM

For 082,,,,gud lux weh,
make sure we scor *NUMBER THOERY*

STUDY=X FAIL
X STUDY=FAIL
(STUDY+X STUDY)=(FAIL+X FAIL)
STUDY(1+X)=FAIL(1+X)
HENCE,
STUDY=FAIL

WARNING!!!!
(DUN TRY THIS DURING XM)

Read Comments

THANK YOU,SIR

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!!

Read Comments

Pell's Equation

Labels: Exercises | at 8:28 PM

Solve $x^{2}-18y^{2}= 1$ .Give 3 positive solutions.

First step:
Find the continued expansion of $\sqrt{18}$

$x_{0}:\sqrt{18}=4+(\sqrt{18}-4), a_{0}=4$
$x_{1}:\frac{1}{\sqrt{18}-4}.\frac{\sqrt{18}+4}{\sqrt{18}+4}=\frac{8+(\sqrt{18}-4)}{2}=4+\frac{\sqrt{18}-4}{2},a_{1}=4$
$x_{2}:\frac{2}{\sqrt{18}-4}.\frac{\sqrt{18}+4}{\sqrt{18}+4}=\frac{2(\sqrt{18}+4)}{2}=8+\sqrt{18}-4,a_{2}=8$
thus, $\sqrt{18}=\left [ 4;4,8 \right ]$

Second step:
Since n=2 and it is even, the solutions are
$x=P_{2k-1}, y=q_{2k-1}$ where k=1,2,3...
Thus,
$x= P_{1}, y=P_{1}$
$x= P_{3}, y=P_{3}$
$x=p_{5},y=q_{5}$
.
.
.
are the positive solutions.

Third step:
Calculate the value of the numerator and denominator,
$C_{0}=\frac{P_{0}}{q_{0}}=4$
$C_{1}=\frac{P_{1}}{q_{1}}=a_{0}+\frac{1}{a_{1}}=4+\frac{1}{4}=\frac{17}{4}$
$C_{2}=\frac{P_{2}}{q_{2}}=\frac{a_{2}p_{1}+p_{2}}{a_{2}q_{1}+q_{2}}=\frac{8(17)+4}{8(4)+1}=\frac{140}{33}$
$C_{3}=\frac{P_{3}}{q_{3}}=\frac{a_{3}p_{2}+p_{1}}{a_{3}q_{2}+q_{1}}=\frac{4(140)+17}{4(33)+4}=\frac{577}{136}$
$C_{4}=\frac{P_{4}}{q_{4}}=\frac{a_{4}p_{3}+p_{2}}{a_{4}q_{3}+q_{2}}=\frac{8(577)+140}{8(136)+33}=\frac{4756}{1121}$
$C_{5}=\frac{P_{5}}{q_{5}}=\frac{a_{5}p_{4}+p_{3}}{a_{5}q_{4}+q_{3}}=\frac{4(4756)+577}{4(1121)+136}=\frac{19601}{4620}$

So,the answers for 3 positive solutions of $x^{2}-18y^{2}= 1$ are

$\frac{17}{4}$ , $\frac{577}{136}$ , $\frac{19601}{4620}$

Read Comments

Fermat's Theorem

Labels: Example, Theorem | at 7:28 PM

THEOREM 1
Let P be a prime number and suppose that P is not divisible by a.Then, $a^{p-1}\equiv 1(mod P)$

THEOREM 2
If P is a prime number,then $a^{p}\equiv a(mod P)$ for any integer a.

Now,Using Fermat's Theorem we're going to proof:

a) $1^{P-1}+2^{P-1}+3^{P-1}+...+(P-1)\equiv -1(mod P)$
Theorem 1 said, $a^{p-1}\equiv 1(mod P)$ so,
$1^{P-1}\equiv 1(modP)$
$2^{P-1}\equiv 1(modP)$
$3^{P-1}\equiv 1(modP)$
.
.
.
$(P-1)^{P-1}\equiv 1(modP)$
thus,
$1^{P-1}+2^{P-1}+3^{P-1}+...+(P-1)^{P-1}\equiv (P-1)(modP)$
simplify it,
$1^{P-1}+2^{P-1}+3^{P-1}+...+(P-1)^{P-1}\equiv -1(modP)$
How (P-1) become congruence to -1??
Ok,let's say P as 19
so,when (19-1) become 18,then
$18\equiv -1(mod 19)$

b) $1^{P}+2^{P}+3^{P}+...+(P-1)^{P}\equiv 0(modP)$
THEOREM 2 said,
$a^{p}\equiv a(mod P)$ so,
$1^{P} \equiv 1(modP)$
$2^{P} \equiv 2(modP)$
$3^{P} \equiv 3(modP)$
.
.
.
$(P-1)^{P} \equiv (P-1)(modP)$
thus,
$1^{P}+2^{P}+3^{P}+...+(P-1)^{P}\equiv ((P(P-1))\div (2) )(mod P)$
simplify it,
$1^{P}+2^{P}+3^{P}+...+(P-1)^{P}\equiv 0(mod P)$
How come $((P(P-1)\div (2))\equiv 0(mod P)$
Again P as 19,thus $((19(18)\div (2))\equiv 0(mod P)$

Read Comments

Subscribe to: Posts (Atom)