Friday, July 29, 2005

Math Problem

Saw this on Bash today:
Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
0: a + b = t
1: (a + b)(a - b) = t(a - b)
2: a^2 - b^2 = ta - tb
3: a^2 - ta = b^2 - tb
4: a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
5: (a - t/2)^2 = (b - t/2)^2
6: a - t/2 = b - t/2
7: a = b

So all numbers are the same, and math is pointless.
I went over it twice, and I couldn't find the error.

Let's go through it again, with actual numbers.
Let a=3, b=5. Then, t=8
0: 3 + 5 = 8
» 8 = 8
Multiply both sides by (3 - 5) [-2]
1: (3 + 5)(3 - 5) = t(3 - 5)
» -16 = -16
FOIL (no value change)
2: 3^2 - 5^2 = 8(3) - 8(5)
» 9 - 25 = 24 - 40
» -16 = -16
Add 5^2 and subtract 8(3) from both sides [+25-24=+1]
3: 3^2 - 8(3) = 5^2 - 8(5)
» 9 - 24 = 25 - 40
» -15 = -15
Add (8^2)/4 to both sides [+16]
4: 3^2 - 8(3) + (8^2)/4 = 5^2 - 8(5) + (8^2)/4
» 9 - 24 + 64/4 = 25 - 40 + 64/4
» 1 = 1
Factor (no value change)
5: (3 - 8/2)^2 = (5 - 8/2)^2
» (-1)^2 = (1)^2
Ahhh...
Root both sides
6: 3 - 8/2 = 5 - 8/2
» 3 - 4 = 5 - 4
» -1 = 1
Umm...
Add 8/2 from both sides [+4]
7: 3 = 5
Eep.

So wonder_yak is correct, the problem is with rooting both sides. Step 6 should be:
6: ±(3 - 8/2) = ±(5 - 8/2)
Isn't math fun?

Monday, July 11, 2005

Post Humorous

Almost ready to go live. Just need to find an American accomplice...

With a U.S. address...I'll be unstoppable!

Tuesday, July 05, 2005

Phoenix Has An Effect

We're doing the "What I want to be when I grow up" section of the English book. The teacher asked me to give a short speech about what I wanted to be, so the students could practice their listening comprehension.

After that, the students filled out worksheets on what they wanted to be.

8 (out of less than 30) decided they wanted to be programmers. Maybe we should have had them do the worksheets before my speech.