Wednesday, September 20

All Numbers Are Equal

Theorem: All numbers are equal.

Proof: Choose arbitrary a and b, and
let t = a + b. Thena + b = t

(a + b)(a - b) = t(a - b)

a^2 - b^2 = ta - tb

a^2 - ta = b^2 - tb

a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4

(a - t/2)^2 = (b - t/2)^2

a - t/2 = b - t/2

a = b
Now, we know that all numbers are not equal. So, can you figure out where this proof breaks down?

12 comments:

Anonymous said...

You can't take say if x^2=y^2 that x=y. (line 7 of your theorem)
x could equal -y.

Luke said...

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.

Anonymous said...

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.

Luke said...

when a=3 and b=5 then you get (-1)^2=(1)^2 or 1=1 which is true and it holds.

Matthew said...

Somewhere in the middle you are dividing by 0 I think, I just haven't seen it yet. I will get back to you.

Luke said...

Matthew: That was my first thought too when I looked it over. I don't think that's it this time though.

Anonymous said...

(-1)^2=(1)^2, but your theorem then says that that means -1=1, which is not true.

Luke said...

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.

Tifani said...

My brain hurts!

Anonymous said...

I have no idea. I thought a and b were variables.

Heidi said...

I got lost after the 3rd line.

ajp said...

What about imaginary numbers, you know, when you divide by 0.