Mathmetical proof that 2 is equal to 1

This is 2(0) = 1(0) which is 0 = 0.

Not at all.

I like the infinite series scheme on one side, or the use of e^(pi*j).

How do you cancel the term? If you divide both sides by it, you're dividing by zero.

Gosh, really? So 2 doesn't equal 1? ;)

Cheers

Phil Hobbs

I was confusing people with that when I was in high school. ...Jim Thompson

The other fun one is demonstrating to a schoolboy that he never goes to school at all. You start with 365 days, then subtract all the time he sleeps and watches TV and stuff, then subtract weekends, then subtract holidays, and the number winds up negative, because of all the double counting.

My father pulled that on me when I was in about grade 4.

Cheers

Phil Hobbs

So that's what "mathmetical" means...

Bill Bowden schrieb:

Thou must not divide by "0"!

Ciao

Reinhard

Simpler than that...

Divide each side by zero and you get

It does... for large values of 1 or small values of 2.

Yes, and in this one you don't have to divide by zero.

(2-3)^2 = (4-3)^2

The root of both sides is 2-3 = 4-3

Add 3 to both sides and 2 = 4 or 1 = 2

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So can I take the root of 4 to prove -2 = 2? Both are roots of 4!

-1 +1

Ed

That's an incorrect (incomplete) statement, and that's the one which leads this argument into error.

You cannot say "*The* root of (2-3)^2 is (2-3)". Rather, "*A* root of (2-3)^2 is (2-3))".

(2-3)^2 is a positive number. Therefore, it has two square roots.

The same is true for (4-3)^2. Positive, thus it has two square roots.

If you're going to try to take the square root of each side symbolically (by just erasing the ^2 from each side) you must also add a symbol such as +/- to each side to acknowledge the presence of two roots... and since there are two possible values on either side, you can no longer claim equality in all cases.

And that appears to work, only because you implicitly assumed that there's only one root available on each side. Since you (in effect) chose the negative-valued square root in one case, and the positive-valued square root in the other case, the assertion that the two roots are equal is incorrect, and the "adding 3 to both sides to prove 2 = 4" is likewise incorrect.