## Prove 2 = 3 The Ramanujan way

July 2, 2007 by Dilip Maripuri

Ramanujan’s proof!!…. Can u find any flaws??

Can U Prove 3=2??

This seems to be an anomaly or whatever u call in mathematics. It seems, Ramanujan found it but never disclosed it during his life time nd that it has been found from his dairy.

See this illustration: That’s the way………….

6 = -6

9-15 = 4-10

adding 25/4 to both sides:

9-15+(25/4) = 4-10+(25/4 )

Changing the order

9+(25/4)-15 = 4+(25/4)-10

(this is just like : a square + b square – two a b = (a-b)square. )

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)(3-5/ 2) = (2-5/2)(2-5/ 2)

Taking positive square root on both sides:

3 – 5/2 = 2 – 5/2

3 = 2

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on July 5, 2007 at 8:24 pm |Avinash MeetooWhat about this?

Let a = b

a^2 = ab

a^2 – b^2 = ab – b^2

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

Divide by a-b on both sides, you get (a+b) = b

But if, say, a=b=1, you have just proved that 2=1🙂

on July 5, 2007 at 8:28 pm |Dilip MaripuriHa ha it seems we are having a few more ramanujans way to go avi🙂

on July 5, 2007 at 8:59 pm |Dilip MaripuriLet me add Ramanujans Puzzling Partitions

A partition is a way in which a whole number can be expressed as the sum of positive integers. Number 5 can be partitioned in seven ways:

5

4 + 1

3 + 2

3 + 1 + 1

2 + 2 + 1

2 + 1 + 1 + 1

1 + 1 + 1 + 1+ 1

The number 4 has only five partitions.

4

3 + 1

2 + 2

2 + 1 + 1

1 + 1 + 1 + 1

Starting with 5, the number of partitions for every seventh integer is a multiple of 7, and starting with 6, the number of partitions for every 11th integer is a multiple of 11. Moreover, similar relationships occur where the interval between the chosen integers is a power of 5, 7, or 11 or a product of these powers.

check it out

cheers🙂

on July 8, 2007 at 5:40 pm |Dilip Maripuriby the way,

theres something i have overlooked😦

6 = -6

9-15 = 4-10

are these two steps possible

6 = -6

n

9-15 = 4 – 10 => -6 = -6 ?????????