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Thread: 3=2 ???

04162009, 02:39 AM #1
 Join Date
 Mar 2008
 Posts
 5,309
T3=2 ???
Can any one prove 3=2. I hope none. But Ramanujam had.
See this illustration:
6 = 6
915 = 410
Adding 25/4 to both sides:
915+(25/4) = 410+(25/4 )
Changing the order
9+(25/4)15 = 4+(25/4)10
[This is just like : a square + b square  two a b = (ab)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
(35/2)(35/2) = (25/2)(25/2)
Taking positive square root on both sides.
3  5/2 = 2  5/2
3 = 2
Can u find any flase??
sorry dude .... Hash

04162009, 02:57 AM #2

04162009, 03:07 AM #3
I had proved that 1 = 2 but i forgot the calculations!

04162009, 03:21 AM #4
 Join Date
 Mar 2008
 Posts
 5,309
TProof that 2 = 1
By the common intuitive meaning of multiplication we can see that
4 \times 3 = 3 + 3 + 3 + 3 \,
It can also be seen that for a nonzero x
x = 1_1 + 1_2 + \cdots + 1_x
Now we multiply through by x
x^2 = x_1 + x_2 + \cdots + x_x
Then we take the derivative with respect to x
2x = 1_1 + 1_2 + \cdots + 1_x
Now we see that the right hand side is x which gives us
2x = x \,
Finally, dividing by our nonzero x we have
2 = 1 \,
Q.E.D.
This proof is false because the differentiation is ignoring the 'x' in the subscript (off x_1 + \cdots + x_x). As you are differentiation with respect to x, it clearly cannot be held constant. Once this x is accounted for in the differentiation, using the Liebniz rule, the correct answer is obtained:
x^2 = x_1 + x_2 + \cdots + x_x
We take the derivative with respect to x
2x = (1_1 + 1_2 + \cdots + 1_x) + (x_1 + \cdots + x_1)
2x = x + x
As expected.
It is often claimed that the original proof is false because both sides of the equation in line 3 represent an integer, and so after differentiating you should get 0 = 0 (as the derivative of a constant function is 0). This is fundamentally incorrect on several levels. Firstly, a function that is only defined on the integers is *not* necessarily a constant function; secondly, the derivative of such a function is *not* 0, it is undefined (picture a graph of the function; it would consist of 'dots'; so have no meaningful slope); and finally, the equation works perfectly well for noninteger values (for example, \textstyle\pi = \{1 + \cdots + 1\} \, \pi \, \text{times} = 1 + 1 + 1 + 0.141...) as evidenced by the fact that, when the differentiation is done correctly, the paradox is eliminated.sorry dude .... Hash

04162009, 03:34 AM #5
Lolz, tu to alladin ka genie hai  bas bola aur chhaap diya!

04162009, 04:34 AM #6
 Join Date
 Mar 2008
 Posts
 5,309
Tlmao
sorry dude .... Hash

04212009, 07:24 AM #7
well yeah i do find it false after this step :
(35/2)(35/2) = (25/2)(25/2)
because first of all you can't even take under root of both sides because you can only take square root of ab if a>b in this case ab = ve term.. and you can't take under root of ve terms.. can you take square root of 2 or 4? they are imaginary units
then secondly even if a was greater en b which in this case is not.. still you just can not take positive square root with your wish.. now square root of 4 can be both +ve 2 or ve 2 answers ..so whenever you take underroot you have to take +ve nd ve both just like you do in quadratic formula usually.. if you have worked with quadratic formula you will know what i am talking about
and it can be many more reasons but i am just doing calculus.. gr 12. lol.. so this is all i was able to figure out though yeah! 3 can never = 2Last edited by sveetu; 04222009 at 05:20 AM.
When you were born, everyone around you was smiling and you were crying. Live your life so that when you die, you're smiling and everyone around you is crying

04212009, 07:28 AM #8When you were born, everyone around you was smiling and you were crying. Live your life so that when you die, you're smiling and everyone around you is crying