24 replies to this topic

[shiv]

[note=szupie]Copied from http://hishamthu.blo...-any-flaws.html[/note]

Ramanujam'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, Ramanujam found it but never disclosed it during his life time and that it has been found from his dairy.

See this illustration:

-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)(B) + 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:

3 - 5/2 = 2 - 5/2

3 = 2 ????????

Edited by szupie, 21 March 2006 - 11:05 PM.

[mastercomputers]

Well everything seems to check out right, except I found those algebraic equations hard to interpret, here's something that may help interpret why this maybe so:

a^2 - 2ab / c + ( b^2 ) / ( c^2 ) = ( c - ( b / c ) )^2

Where

a = 3, b = 5 , c = 2 and ^x is used to represent powers by x (a number)

So in that above equation, what you're trying to prove is that:

a = c

And it's quite obvious, in that relationship, considering there's no 'a' equivalent on the RHS, so in solving this equation to it's simpliest form, it's possible, you'll get a = c.

I haven't proved this, however I've looked at the above post and can see why it would be considered this, Ramanujam was an excellent Mathematician, could say he was obsessed/artistic with numbers.


Cheers,


MC

[mastercomputers]

If this is the case, then 'a' can be any number except 0 or 1 and possibly negative numbers although I never checked:

e.g.

If a = 5 then c = a - 1 and b = a + c

a = 5, b = 9, c = 4

or

If a = 4, c = a + 1, b = a + c

The equation I'm using is:

a^2 - ab + b^2/c^2 = c^2 - bc + b^2/c^2

5^2-5*9+9^2/4^2 = 4^2-9*4+9^2/4^2

I don't know how to put it, but 'a = c' does not work with my equation other than 'a' could be replaced with 'c' as long as 'c' follows the rules of 'a - 1' or 'a + 1' and that 'b' remains 'a + c', in quadratic equations you can have two answers, although I'm trying to think how this came about, could it be a graphed on a 3D cube?

Hopefully someone can make sense of this, but that's it for me, basically without the additional added to it, we get a^2 - ab = c^2 - bc which evaluates to true as well.


Cheers,


MC

[pyost]

If you have this:

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

it doesn't mean that the square roots of those two are equal. Here's an example:

(-2)^2=(2)^2

but this one is not true

-2=2

However, if you have a=b, then a^2 is equal to b^2.


It is a simple mathematical rule:

sqrt((x)^2) = b b >= 0

This one has to solutions:
x=b
x=-b

On the other hand, if b is negative, the equasion has no solution, because a square root cannot give a negative number.
sqrt((3-5/2)^2) = sqrt((2-5/2)^2) = -1/2

After all, I am going to a mathematical school

~edit~

The approach was a bit wrong, here's a better one.
N.B. |x| = -x if x<0 or |x|=x if x>=0

Here's the solution to the equasion:
sqrt(x^2)=sqrt(y^2)
|x|=|y| <-- this is always so, that's how you do it.

Now when we replace those, we get:
|3-5/2|=|2-5/2|
|1/2|=|-1/2|

And that get's us to 1/2=1/2 Simple as that, just had to find a glitch (not in the matrix, lol)

Edited by pyost, 22 March 2006 - 09:26 AM.

[demolaynyc]

Well everything seems to check out right, except I found those algebraic equations hard to interpret, here's something that may help interpret why this maybe so:

a^2 - 2ab / c + ( b^2 ) / ( c^2 ) = ( c - ( b / c ) )^2

Where

a = 3, b = 5 , c = 2 and ^x is used to represent powers by x (a number)

So in that above equation, what you're trying to prove is that:

a = c

And it's quite obvious, in that relationship, considering there's no 'a' equivalent on the RHS, so in solving this equation to it's simpliest form, it's possible, you'll get a = c.

I haven't proved this, however I've looked at the above post and can see why it would be considered this, Ramanujam was an excellent Mathematician, could say he was obsessed/artistic with numbers.
Cheers,
MC

Aahhhh! Such math! I like Math but those confusing equations just made me dissy. But I did get this poster about the "a=c". But it's still weird.

[BDIT]

In your post you mention "Here a = 3, b=5/2 for L.H.S and a =2, b=5/2 for R.H.S.", but in my little knowledge in math I know that if you set the value of a = 3 in left hand side, you must set same value (a=3) in right hand site. On the other hand if you set the value a=2 for right hand site, you must set the same value (a=2) for left hand site. You can not put different value of a in the same equation.
So, according to my little knowledge, your calculation is not a valid mathematical calculation.

[sparkx]

I know these tricks with math you bend 1 or 2 rules that are so basic that most people forget about them and don't realize what has happened. An example and simple variation of this trick uses division by 0 which cannot work because anything times by 0 is 0. So the lesson works but there are exceptions to them. All the lessons before work in real math except division by 0.

5=2
add 1 to each side
5+1=2+1 --> 7=3
still false but wait divide both sides by 0.
7/0=3/0 --> 0=0
Therefore
5=2

I believe I found the problem: (3-5/2) 2 = (2-5/2)2. This equation is flawed: a2 – 2(a)(B) + b2 = (a-B)2 try reducing in you get: b=2a-2b. I don’t think that’s correct.

Now I’m not the best at math but it makes since that the equation doesn’t work.
Thanks,
Sparkx

Edit: Smilies are anoying...

Edited by sparkx, 30 April 2008 - 08:48 PM.

[sparkx]

Replying to Feedbacker

I’m confused division by zero is exactly what I said; you just said it in a different way.

I said: Any number times 0 is 0. (true use your calculator).
You said: Any number divided by 0 is infinity. (also true).

Unless there is something I am missing (which might be because I haven't taken calculus in a formal class yet) I think both things are correct.

still false but wait divide both sides by 0.
7/0=3/0 --> 0=0

This was simply to show a common mistake made by people who often forget basic rules of math, similar to the mistake that people may make in this problem.

Thanks,
Sparkx

[LegallyHigh]

Haha, I hated Proofs in High school. I think I'll pass at attempting this one.

[Madamathic]

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