I hurt you with my words

In my CS 251 class, we cover various data structures of use in programming: trees, graphs, queues, stacks, etc. So oftentimes the professor will draw an example of how we might envision one of these data structures on the blackboard, and he'll occasionally ask the class to give him numbers to fill into the nodes. The first time he called on me, I said "pi". The second time, I said "e".

He doesn't call on me anymore.

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And now, it's time for...
Duck's Classroom Corner
-with Professor Duck

Today, Duck will prove to you that 2ⁿ = 0, where n is any natural number.

First off, we know that 2ⁿ - 1 = 2⁰ + 2¹ + 2² + 2³ + ... + 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 2ⁱ. 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 2ⁿ - 1 = 1111...11, and by the transitive property of equality, we know therefore that 2ⁿ - 1 = -1.

After adding one to both sides, we arrive at our final result:
     2ⁿ = 0

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

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




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