## 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 = t1: (a + b)(a - b) = t(a - b)2: a^2 - b^2 = ta - tb3: a^2 - ta = b^2 - tb4: a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/45: (a - t/2)^2 = (b - t/2)^26: a - t/2 = b - t/27: 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 = 8Multiply both sides by (3 - 5) [-2]1: (3 + 5)(3 - 5) = t(3 - 5)» -16 = -16FOIL (no value change)2: 3^2 - 5^2 = 8(3) - 8(5)» 9 - 25 = 24 - 40» -16 = -16Add 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 = -15Add (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 = 1Factor (no value change)5: (3 - 8/2)^2 = (5 - 8/2)^2» (-1)^2 = (1)^2`      Ahhh...`Root both sides6: 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?

1. Well, I'm not quite sure how to articulate this, but aren't you essentially disregarding sign...specifically when you square and then square-root the equation?

I'm pretty sure that is why this is false, I just wish I knew the algebraic rule that was being broken.

_kris

2. Or, more specifically: in lines 5 - 7 when you factor and then square-root...

3. "Isn't math fun?"

Um. See my post on Friday, most notably, the third section.