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Thread: 3=2 ???

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

    See this illustration:

    -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

    Can u find any flase??
    sorry dude .... Hash

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    i cant do math anymore






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

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    Quote Originally Posted by crayxmp View Post
    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.
    sorry dude .... Hash

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

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    lmao
    sorry dude .... Hash

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    Quote Originally Posted by Ťŋē ώү 'm View Post
    Can any one prove 3=2. I hope none. But Ramanujam had.

    See this illustration:

    -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

    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
    Last edited by sveetu; 04-22-2009 at 06:20 AM.
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    Quote Originally Posted by Ťŋē ώү 'm View Post
    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
    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

 

 

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