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.
And
now, it's time for...
Duck's Classroom Corner-with Professor DuckToday, 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 - 1In 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
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