mat luar biasa

7/27/2019 Mat Luar Biasa

1/23

CHONG HUI ERN

(n+1)^2 = n^2 + 2*n + 1

Bring 2n+1 to the left:

(n+1)^2 - (2n+1) = n^2

Substract n(2n+1) from both sides and factoring, we have:

(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)

Adding 1/4(2n+1)^2 to both sides yields:

(n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2

This may be written:

[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2

Taking the square roots of both sides:

(n+1) - 1/2(2n+1) = n - 1/2(2n+1)

Add 1/2(2n+1) to both sides:

n+1 = n


2/23

CHONG KOON KEAN

Let say a = b

a(a) = b(a) or a = b(a)

add a (-b) to both sides

a-b = ab-b

factor both sides out:

(a+b)(a-b) = b(a-b)

divide both sides by (a-b)

(a+b)(a-b) = b(a-b)

(a-b) (a-b)

The (a-b)s on both sides will cancel out only leaving:

a+b = b

now let us consider this statement:

a = b = 1

with respect to our new formula we will find that:

1+1 = 1


3/23

EMILY WONG

let say 1 = 1

41 40 = 61 60

16 + 25 40 = 36 + 25 60

4 + 5 2 * 4 * 5 = 6 + 5 2 * 6 * 5

(4

5) = (6

5)

4 5 = 6 5

4 = 6

2 = 3

1 + 1 = 3


4/23

GOH HSIA CHEE

0 = 0 + 0 + 0 + 0 ...

0 = (1-1)+(1-1)+(1-1)...

0 = 1 + (-1+1) + (-1+1) + (-1+1)...

0 = 1 + 0 + 0 + 0 + 0...

0 = 1


5/23

GOH HUI SAN

substitute "tan x":

assume u = sec x and dv = sin x dx:

but cos x * sec x = 1 so:

then:


6/23

HENRY GOH

Let x = 1

x^2 = x

x^2-1 = x-1

(x+1)(x-1) = (x-1)

(x+1) = (x-1)/(x-1)

x+1 = 1

x = 0

0 = 1


7/23

HIDIE KONG

assume:

rewrite -1 two different ways:

take the square root of both sides:

using laws of square roots, rewrite both sides:

multiply both sides by and reduce:

the square root of a number squared equals the number itself, so:


8/23

KIU KWONG XIAN


9/23

LAI SZE MAY

Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides:

Step 3: Simplifying:

Step 4: In other words, i/1 = 1/i.

Step 5: Therefore, i/ 2 = 1 / (2i),

Step 6:i/2 + 3/(2i) = 1/(2i) + 3/(2i),

Step 7:i(i/2 + 3/(2i) ) = i( 1/(2i) + 3/(2i) ),

Step 8: ,

Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

Step 10: and this shows that 1=2.
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10/23

LAU UNG HONG

x

0.

10 x 9.

10 x x 9. 0. 9 x = 9

x = 1


11/23

LAU YONG SIONG

x*x = x + x + x + ... + x (x times)

d/dx (x*x) = d/dx ( x + x + x + ... + x )

d/dx (x*x) = d/dx (x) + d/dx (x) + ... d/dx (x)

2x = 1 + 1 + ... + 1

2x = x

2 = 1


12/23

LEE SZE YIN

-20 = -20

16-36 = 25-45

4^2 (4x9) = 5^2 (5x9)

4^2 (4x9) + (81/4) = 5^2 (5x9) + (81/4)

4^2 (2x4x(9/2)) + (9/2)^2 = 5^2 (2x5x(9/2)) + (9/2)^2

(4 (9/2)) (4 (9/2)) = (5-(9/2)) (5-(9/2))

(4 (9/2))^2 = (5-(9/2))^2

4 (9/2) = 5-(9/2)

4

(9/2) + (9/2) = 5-(9/2) + (9/2)

4 = 5


13/23

NG YAN MEI

We will start with to simple equations:

a) 3 a - 2 a = a

b) a = b + c

Now, by inserting (b) into (a), we get:

3 a - 2 a = 3 ( b + c ) - 2 ( b + c )

After multiplication:

3 a - 2 a = 3 b + 3 c - 2 b - 2 c

Now let us clarify the structure of the equation by shifting all terms including the

number 3 on the left side, and all terms including the number 2 on the right side:

3 a - 3 b - 3c = 2 a - 2 b - 2 c

Eq may be written even more clearly by the use of brackets:

3 ( a - b - c ) = 2 ( a - b - c )

Now it is very easy to see, that indeed

3 = 2


14/23

PHOR ZHI YING


15/23

SIAW MEI YEE

2*e = f

2^(2*pi*i)e^(2*pi*i) = f^(2*pi*i)

e^(2*pi*i) = 1

Therefore:

2^(2*pi*i) = f^(2*pi*i)

2=f

Thus:

e=1


16/23

SII TUONG SIENG

Prove

=1


17/23

TIONG CHIONG YEW


18/23

WEE WEANG WEANG

a. log[(-1)^2] = 2 * log(-1)

On the other hand:

b. log[(-1)^2] = log(1) = 0

Combining a) and b) gives:

2* log(-1) = 0

Divide both sides by 2:

log(-1) = 0


19/23

WINNIE TAN

We can re-write the infinite series 1/(1*3) + 1/(3*5) + 1/(5*7) + 1/(7*9)

+...

as 1/2((1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + ... ).

All terms after 1/1 cancel, so that the sum is 1/2.

We can also re-write the series as (1/1 - 2/3) + (2/3 - 3/5) + (3/5 - 4/7)

+ (4/7 - 5/9) + ...

All terms after 1/1 cancel, so that the sum is 1.

Thus 1/2 = 1.


20/23

WONG LING JIE

2519 Mod n means the reminder of 2519/n, here / is the integer division.

2519 Mod n

2519 Mod 2 = 1

2519 Mod 3 = 2

2519 Mod 4 = 3

2519 Mod 5 = 4

2519 Mod 6 = 5

2519 Mod 7 = 6

2519 Mod 8 = 7

2519 Mod 9 = 8

2519 Mod 10 = 9

Sequential Numbers with 2519

1259 x 2 + 1 = 2519

839 x 3 + 2 = 2519

629 x 4 + 3 = 2519

503 x 5 + 4 = 2519

419 x 6 + 5 = 2519

359 x 7 + 6 = 2519

314 x 8 + 7 = 2519

279 x 9 + 8 = 2519

251 x 10 + 9 = 2519


21/23

WONG YI LING

Sequential Inputs of numbers with 8

1 x 8 + 1 = 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

123456 x 8 + 6 = 987654

1234567 x 8 + 7 = 9876543

12345678 x 8 + 8 = 98765432

123456789 x 8 + 9 = 987654321


22/23

WOO POOI KEH


23/23

YII MING ING

Let (right) = 1

Since wrong is the opposite of right,

Let (wrong) = -1

(wrong)(wrong) = (right)

Two wrongs make a right*

*as long as you multiply them.

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