Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and
let t = a + b. ThenNow, we know that all numbers are not equal. So, can you figure out where this proof breaks down?
How Not to Respond to Suffering
15 hours ago
Subscribe to:
Post Comments (Atom)
12 comments:
You can't take say if x^2=y^2 that x=y. (line 7 of your theorem)
x could equal -y.
Anon:
So, you're saying that, if the final conclusion said a = +-b, then it's true?
That's still known to be untrue.
So, while you have a valid point, it still leaves problems.
No, it identifies an intermediate problem which is then used to build further on a false foundation. Try it with a=3 and b=5 and you will see the invalidity of the problem.
when a=3 and b=5 then you get (-1)^2=(1)^2 or 1=1 which is true and it holds.
Somewhere in the middle you are dividing by 0 I think, I just haven't seen it yet. I will get back to you.
Matthew: That was my first thought too when I looked it over. I don't think that's it this time though.
(-1)^2=(1)^2, but your theorem then says that that means -1=1, which is not true.
anonymous: Sorry for my thickheadedness (is that a word?). I get ya now. Good call on that.
What's your take on line 4 not really being a quadratic since you don't really have a variable? If you solve it like a quadratic then you have to assume a and b are variables...and t is a constant.
My brain hurts!
I have no idea. I thought a and b were variables.
I got lost after the 3rd line.
What about imaginary numbers, you know, when you divide by 0.
Post a Comment