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Thread: Abbott and Costello arithmetic

  1. #1
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    Abbott and Costello arithmetic

    We are going to have some fun with a little piece of half baked arithmetic. We are going to prove that the natural logarithm of -1 = 0. Ready? Here we go.

    ( -1 ) x ( -1 ) = 1

    Ln[ ( -1 ) x ( -1 ) ] = Ln( 1 )

    Ln( -1 ) + Ln( -1 ) = Ln( 1 ) ......... since Ln( AB ) = Ln( A ) + Ln( B )

    2 x Ln( -1 ) = Ln( 1 )

    2 x Ln( -1 ) = 0

    Ln( -1 ) = 0 ....... QED .......... ......
    Last edited by PolymerBob; 10-17-2012 at 07:21 PM.

  2. #2
    Figure Skating Is A Dangerous Sport Dee4707's Avatar
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    Is this a game???

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    Simply the best. l'etoile's Avatar
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    Quote Originally Posted by PolymerBob View Post
    We are going to have some fun with a little piece of half baked arithmetic. We are going to prove that the natural logarithm of -1 = 0. Ready? Here we go.

    ( -1 ) x ( -1 ) = 1

    Ln[ ( -1 ) x ( -1 ) ] = Ln( 1 )

    Ln( -1 ) + Ln( -1 ) = Ln( 1 ) ......... since Ln( AB ) = Ln( A ) + Ln( B )
    Wrong here. You cannot divide them into product since the antilogarithms cannot have negative values.
    Hence, A>0 and B>0 should be preceding condition.

    2 x Ln( -1 ) = Ln( 1 )

    2 x Ln( -1 ) = 0

    Ln( -1 ) = 0 ....... QED .......... ......
    If this is a game, how are we gonna play it?

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    Wicked Yankee Girl dorispulaski's Avatar
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    Come up with another piece of half baked arithmetic?

    A Mathematician, a Physicist and an Engineer entered a mathematics contest, the first task of which was to prove that all odd numbers are prime. That's easy said the Mathematician; "1's a prime, 3's a prime, 5's a prime, 7's a prime. Therefore by mathematical induction, all odd numbers are prime". Then it was the Physicist's turn: "1's a prime, 3's a prime, 5's a prime, 7's a prime, 11's a prime, 13's a prime. So, allowing for experimental error, all odd numbers are prime. The engineer provided the most straightforward proof: "1's a prime, 3's a prime, 5's a prime, 7's a prime, 9's a prime, 11's a prime, 13's a prime . . . ".
    Last edited by dorispulaski; 10-18-2012 at 03:12 AM.

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    Here is the weird part. Take a look at the function 1 / X. https://encrypted-tbn3.gstatic.com/i...hq0vCe4eJlAh9Q

    OK, now let's say you want to find the area between the curve and the X axis between -5 and -2. You integrate and get Ln( X ). So the area you want is Ln( -2 ) - Ln( -5 ). This is impossible. ................... or is it?

    Since Ln( A ) - Ln( B ) = Ln( A / B ), we get Ln( -2 ) - Ln( -5 ) = Ln( -2 / -5 ) = Ln( 0.4 ) = -0.91629...., which is the correct answer. ( The answer is negative because the curve is below the X-axis. )

  6. #6
    Custom Title Mathman's Avatar
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    (-1)x(-1) = 1

    ln(-1) + ln(-1) = ln(1) = 0

    pi i + (-pi i) = 0

    Correct!

    (Two of the values of ln(-1) are pi i and -pi i. Check: e^(pi i) = -1 and e^(-pi i) = -1. (Euler). )
    Last edited by Mathman; 10-18-2012 at 08:49 PM.

  7. #7
    Custom Title Mathman's Avatar
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    Quote Originally Posted by PolymerBob View Post
    You integrate and get Ln( X ).
    You integrate and get the logarithm of the absolute value of x.

    Integral of 1/x from -5 to -2 = ln(|-2|) - ln(|-5|) = ln(2)-ln(5) = ln(.4) = -0.91629.

  8. #8
    Simply the best. l'etoile's Avatar
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    Let's prove the series of -1 and 1 converges to zero, then.

    1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+...

    =(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+...

    =0+0+0+0+0+0+...

    =0

    huh?

  9. #9
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    Quote Originally Posted by Mathman View Post
    (-1)x(-1) = 1

    ln(-1) + ln(-1) = ln(1) = 0

    pi i + (-pi i) = 0

    Correct!

    (Two of the values of ln(-1) are pi i and -pi i. Check: e^(pi i) = -1 and e^(-pi i) = -1. (Euler). )
    Quickly now .... what is arcsin( 0.75 ) - arcsin( 0.75 ) ?

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