Upload
micky-yii
View
223
Download
0
Embed Size (px)
Citation preview
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
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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:
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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:
7/27/2019 Mat Luar Biasa
8/23
KIU KWONG XIAN
7/27/2019 Mat Luar Biasa
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.
http://www.math.toronto.edu/mathnet/falseProofs/guess9.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess10.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess11.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess12.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess13.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess14.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess15.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess16.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess17.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess18.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess18.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess17.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess16.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess15.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess14.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess13.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess12.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess11.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess10.htmlhttp://www.math.toronto.edu/mathnet/falseProofs/guess9.html7/27/2019 Mat Luar Biasa
10/23
LAU UNG HONG
x
0.
10 x 9.
10 x x 9. 0. 9 x = 9
x = 1
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
14/23
PHOR ZHI YING
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
16/23
SII TUONG SIENG
Prove
=1
7/27/2019 Mat Luar Biasa
17/23
TIONG CHIONG YEW
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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.
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
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
7/27/2019 Mat Luar Biasa
22/23
WOO POOI KEH
7/27/2019 Mat Luar Biasa
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.