n joy :)

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ha ha ha ha sachchai funny !!!

Theorem : 3=4
Proof:
Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3

Theorem : All numbers are equal to zero.
Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0

Theorem: 1$(dollar) = 1c(cent).
Proof:
And another that gives you a sense of
money disappearing...
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c

Theorem: 1 = -1 .
Proof:
1/-1 = -1/1
sqrt[ 1/-1 ] = sqrt[ -1/1 ]
sqrt[1]*sqrt[1] = sqrt[-1]*sqrt[-1]
ie 1 = -1

Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5

hehe provin some theorem... challenge ur professors :)

Theorem 0/0 = 4
0/0 = (4-4)/(2-2) = ((2+2)(2-2))/(2-2)
since 4 = 2^2 and a^2-b^2 = (a+b)(a-b)
= (2+2) = 4
Proved :)

jhyap your logic of 3=4 is wrong.


because from the question it is given a+b= c i.e a+b-c=0


while solving the equation you did (a+b-c)/(a+b-c) =1 which is wrong . as 0 divide by 0 can never be equal to 1. while making the eqaution you can never give such equations as it is against the rule of algebra.

3=4
Proof:
Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3


This is wrong because u cannot divide by a+b-c which is 0 (since a+b=c)


Theorem : All numbers are equal to zero.
Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0


This is wrong because u cannot divide by a-b coz a-b=0 (u assumed a=b)

Theorem: 1$(dollar) = 1c(cent).
Proof:
And another that gives you a sense of
money disappearing...
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c

This is wrong because (10c)^2 is not equal to (0.1$)^2


Theorem: 1 = -1 .
Proof:
1/-1 = -1/1
sqrt[ 1/-1 ] = sqrt[ -1/1 ]
sqrt[1]*sqrt[1] = sqrt[-1]*sqrt[-1]
ie 1 = -1


This is wrong because sqrt[ x/y] is not equal to sqrt[x]/sqrt[y]

Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5

This is wrong because (a-b)^2 i=(x-y)^2 doesnot mean a-b=x-y if, for example. a-b is positive and x-y is negative

0/0 = 4
0/0 = (4-4)/(2-2) = ((2+2)(2-2))/(2-2)
since 4 = 2^2 and a^2-b^2 = (a+b)(a-b)
= (2+2) = 4
Proved :)


This is wrong because u cannot 0/0 is not 1 ( as u canceled (2-2)/(2-2))

oops , read last line as

This is wrong because 0/0 is not 1 (as u canceled (2-2)/(2-2))

Jhyap,

very funny. Keep it coming.
You must be smoking/drinking some good stuff :)

Rein,

brilliant. now i know that math can be deceitful

Jhyap on Theorem: 1$(dollar) = 1c(cent).
Proof:
And another that gives you a sense of
money disappearing...
1$ = 100c
= (10c)^2 <<<<<<<<<<<<<<<<< FLAW>>>(10c)^2 = 100c^2 != 100c
= (0.1$)^2
= 0.01$
= 1c <<<<<<<<<<<
Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2 <<<<<<<<<<>>>>>>> 2*4*9/2 =72/2 !=81/2
4 - 9/2 = 5 - 9/2
4 = 5 <<<<<<<<<<<<<<<<<<



heorem 0/0 = 4
0/0 = (4-4)/(2-2) = ((2+2)(2-2))/(2-2) <<<0/0 is undefined (its not infinity) use <<<<<<<<<<<<<<<<<<<<<<<<<<< since 4 = 2^2 and a^2-b^2 = (a+b)(a-b)
= (2+2) = 4 <<<<<<<<<<<<<<<<< but not = 0
Proved :)

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Theorem 0/0 = 4
0/0 = (4-4)/(2-2) = ((2+2)(2-2))/(2-2) <<< use L'Hopitals rule
since 4 = 2^2 and a^2-b^2 = (a+b)(a-b)
= (2+2) = 4
Proved :)

l'hopital's rule :) take derivitives :)

We are here for some fun! we have all seen those funny math in our high school. We do not want to understand those math, we just want to have some fun. Who cares!!!

Sara maja kir kira bhayo...

Really Funny. Specially # 3 is hilarious.

Thanks a lot.