1. ## 3=2 ???

Can any one prove 3=2. I hope none. But Ramanujam had.

See this illustration:

-6 = -6

9-15 = 4-10

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

Can u find any flase??

2. i cant do math anymore

3. I had proved that 1 = 2 but i forgot the calculations!

4. Originally Posted by crayxmp
I had proved that 1 = 2 but i forgot the calculations!
Proof 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 non-zero 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 non-zero 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 non-integer 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.

5. Lolz, tu to alladin ka genie hai - bas bola aur chhaap diya!

6. lmao

7. Originally Posted by Ťŋē ώãү ¡'m
Can any one prove 3=2. I hope none. But Ramanujam had.

See this illustration:

-6 = -6

9-15 = 4-10

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

Can u find any flase??
well yeah i do find it false after this step :
(3-5/2)(3-5/2) = (2-5/2)(2-5/2)

because first of all you can't even take under root of both sides because you can only take square root of a-b if a>b in this case a-b = -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 = 2

8. Originally Posted by Ťŋē ώãү ¡'m
Proof 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 non-zero 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 non-zero 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 non-integer 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.
lol i even wanted to reply this ryte now.. but it seems computer language to me i will figure out error on weekend i love math lol

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