3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

[iGuest 3]

Replying to pyostYa it is rightIt is my suggestiononly Its just simple but it is true 3-52=2-5/2 is not correctBut modolus of 3-5/2=modolous 2-5/2 is correct-feedback by arunraj

[iGuest 3]

Replying to pyostYa it is rightIt is my suggestiononly Its just simple but it is true 3-52=2-5/2 is not correctBut modolus of 3-5/2=modolous 2-5/2 is correct by arun raj-feedback by arunraj

[iGuest 3]

the flaw is in the place where he took square root 

the value should have been reeversed as square root can't be negative

[Madamathic 0]

the flaw is in the place where he took square root

 

the value should have been reeversed as square root can't be negative

There is no question of reversing the value, the moment you reverse, both sides become inequal!! The basics says that while taking the sq.root, LHS & RHS must have the 'positive' sign. In the present case, the LHS is positive while RHS is negative, which is the flaw.

 

I have another poser, can you prove: TWO TOW'S ARE FIVE !

Madamathic

[iGuest 3]

try to prove ramanujams proof wrong3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

 Sir as we all know, square roots of a no. Can be positive or negative so we cannot say that (3-5/2) is paoitive or nagetive...This is all a probability also same in rhs side

-reply by Vineet Paharia

 

[iGuest 3]

3 cant be equal to 23=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

yes 3 cant be eqaul to 2 because the problem is where he took square root. To me may b lhs is possitive ane rhs is negative. So equation cannot be equal.

-reply by Atif

[iGuest 3]

I will explain it3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

always take mod when taking sqr root.

Same problem generates, when we start with -20=-20, then

16-36=25-45

=>16-36+(81/4)=25-45+(81/4)

=>(4)^2 - 2.4.(9/2) - (9/2)^2 = (5)^2- 2.5.(9/2) + (9/2)^2.

Now according to formula: (a-^2= a^2 - 2ab+ b^2,

the above eqauation reduces to:

=> (4-(9/2))^2=(5-(9/2))^2---------------------equation XXX

taking square roots(without mods, like a 5th grade kid)):

4-9/2=5-9/2

=>4=5 or 2+2=5.

But the problem is for an integer (-x)^2= +ve numbers, but sqrt(-x) is not necessarily equal to +ve number.

So, Continuing from equation XXX,

taking sqrt of both side:

|4-9/2|=|5-9/2|

=> |-1/2|=|1/2|

=>1/2=1/2

which is the correct solution for all the above problems of this type.

See all these type of problems can prove any 2 consecutive numbers equal, such as 4=5(this prob), 3=2( as by ramanuj), 6=7 or anything you want.

But when you solve it correctly, using modulus, and remembering square root rules, you will get 1/2=1/2. 

Mail me with more problem, I will explain you guys.

And, world won't end on 2012, but the chain of ending starts!

-reply by Nilesh Mahant

[iGuest 3]

3=2 ! Ramanujam3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?Taking positive square root on both sides: ---this is where the actual error liesWe can only take square root on both sides...If (-2) square= (2) square-2 cannot be equal to 2Because when we try taking the square roots, we sayEither (3-5/2) = (5/2-2)—(1)Or(3 - 5/2) = (2 - 5/2)—(2)Hence for our calculation equation (1) holds good…-reply by sony

[iGuest 3]

Its wrong3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

-6 = -6 9-15 = 4-10 Adding 25/4 to both sides: 9-15+(25/4) = 4-10+(25/4 )(this is just like : a2 – 2(a)( + b2 = (a-B)2 ) Here a = 3, b=5/2 for L.H.S and a =2, b=5/2 for R.H.S. So it can be expressed as follows: (3-5/2) 2 = (2-5/2)2 Taking positive square root on both sides: mod (3-5/2) = mod (2-5/2)

 

 

-reply by RONAK

[iGuest 3]

Notice from szupie:

Copied from http://forums.xisto.com/no_longer_exists/

are you kidding me?
(2-5/2)^2 = (3-5/2)^2
is true, because it is
(-1/2)^2 = (1/2)^2.
doesn't mean
-1/2 = 1/2.

it's all just made to confuse you. he could have just as well said:
(-1)^2 = 1^2
"taking positive root (which spoils the proof):"
-1 = 1