He doesn't call on me anymore.

And

**now**, it's time for...

*Duck's Classroom Corner*

-with Professor Duck

Today, Duck will prove to you that 2

^{n}= 0, where n is any natural number.

First off, we know that 2

^{n}- 1 = 2

^{0}+ 2

^{1}+ 2

^{2}+ 2

^{3}+ ... + 2

^{n - 2}+ 2

^{n - 1}

In binary, the i

^{th}bit of a bitstring, starting at 0 and from the rightmost bit, is equal to 2

^{i}. Therefore, the righthand side of the equation becomes the bitstring 1111...11

But! In binary, the most significant (or left-most) bit is known as the

*sign bit*, meaning that it indicates whether the number is positive or negative. Let's show our bitstring again, with the most significant bit highlighted:

1111...11

Since the most significant bit is 1, this number is clearly negative. But negative what? To negate a two's-complement binary number, you invert every bit and add 1 to the result. Let's investigate further:

1111...11 <= What we started with

0000...00 <= After inverting each bit

0000...01 <= After adding 1

So we see that the negative of 1111...11 is just 1; therefore, 1111...11 = -1.

But recalling our previous equality, we know that 2

^{n}- 1 = 1111...11, and by the transitive property of equality, we know therefore that 2

^{n}- 1 = -1.

After adding one to both sides, we arrive at our final result:

2

^{n}= 0

Next class, Duck proves that the total number of dead in the Holocaust was also 0... or, as Duck might say, 2

^{n}.

Debora, you might want to stop reading at this point.

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Nor do they seem to be making widely known the massive ignorance (or possibly denial) of Bush's constituency.

Read more...