Abbott and Costello arithmetic

PolymerBob · Oct 17, 2012

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 .......... :laugh:

......

Dee4707 · Oct 17, 2012

Is this a game???

l'etoile · Oct 18, 2012

PolymerBob said:
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?

dorispulaski · Oct 18, 2012

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 . . . ".

PolymerBob · Oct 18, 2012

Here is the weird part. Take a look at the function 1 / X. https://encrypted-tbn3.gstatic.com/...fHbCz-EBkcbVeyRFM9HuuETjzy8E_FShq0vCe4eJlAh9Q

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. )

Mathman · Oct 18, 2012

(-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).

)

Mathman · Oct 18, 2012

PolymerBob said:
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.

l'etoile · Oct 19, 2012

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?

PolymerBob · Oct 20, 2012

Mathman said:
(-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 ) ? :laugh:

Abbott and Costello arithmetic

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