[Pickett]

I've seen all of these "proofs" being added recently for 2 = 0, 1 < 0, etc, etc...so I figured I'd add my two proofs for 2 = 1, and proof that magic exists...

I apologize if these have already been posted...I haven't seen them yet though, so I thought I would put them up. These aren't difficult, and I fully expect them to be solved very quickly, but they're fun.

This first one is the golden oldie (I learned this one quite a few years ago and still like it):

let a = b

a² = ab (multiply both sides by a)

a² - b² = ab - b²(subtract b² from both sides)

(a + b)(a - b) = b(a - b) (factor)

(a + b) = b (divide both sides by (a - b) )

b + b = b (substitution)

2b = b

2 = 1

Then this following one uses calculus (fun one at first when learning calculus):

x = 1 + 1 + ... + 1 (x times)

x² = x + x + ... + x (multiply through by x)

2x dx = (1 + 1 + ... + 1) dx (take deriviative of both sides)

2x = (1 + 1 + ... + 1)

2x = x

2 = 1

and my last one is to show that something TRULY can come from nothing:

0 = 0 + 0 + 0 + ...

0 = (1 - 1) + (1 - 1) + (1 - 1) + ... (since 0 = 1 - 1)

0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... (associative law of addition)

0 = 1 + 0 + 0 + 0 + ... (since -1 + 1 = 0)

0 = 1

So, as if by magic something appeared out of nothing! YAY!

[Guest]

Well, the first one is easy. (a-b) is equal to 0, so when you divide by (a-b), you're dividing by zero, which causes your error.

I'm not sure about the second one, but I think the error for the third one lies in the infinite series. Basically you started with a finite series of zeros, then treated it as an infinte series later.

[Guest]

It's been a long time since calculus...

an error with the derivative. the derivative of x^2 isn't 2xdx but 2dx. You're treating 2*dx as 2*x*dx.

I think???

[Pickett]

It's been a long time since calculus...

an error with the derivative. the derivative of x^2 isn't 2xdx but 2dx. You're treating 2*dx as 2*x*dx.

I think???

nope, the deriviative of x² is in fact 2x dx. getting somewhat closer though.

[Pickett]

Well, the first one is easy. (a-b) is equal to 0, so when you divide by (a-b), you're dividing by zero, which causes your error.

I'm not sure about the second one, but I think the error for the third one lies in the infinite series. Basically you started with a finite series of zeros, then treated it as an infinte series later.

yes on the first one!

yes and no on the third one: Yes it has to do with the infinite series...but 0 does equal 0 + 0 + 0 + ... as an infinite series...so, there's something else wrong with it...

[Guest]

classic, a-b=0 since a=b, therefore you cannot divide by 0

x=1+1+1...+1 (x times)

your treating x as an actual number, not a variable,

therefore the derivative of x^2 is 0 and the derative of (x+x+x...+x) is 0+0+0+0...+0

to elaborate

if x=4

x=1+1+1+1 (x times)

x^2=x+x+x+x

or

4^2=4+4+4+4

d/dx(4^2)=d/dx(4+4+4+4)

0=0

this problem does not state what value x equals, but the first statement implies that x is equal to a number and is not a variable

using the associative property

0= 1+(-1+1)+(-1+1)...+(-1+1)+(-1)

when using the associative property, the last set of (1+(-1)) will also shift over, leaving a lone (-1) to cancel out the first lone (+1)

[Guest]

You're missing a one

0 = (1 - 1) + (1 - 1) + (1 - 1) + ... (since 0 = 1 - 1) - six 1s

0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... (associative law of addition) - seven 1s

So if we worked backwards a step

0 = (1 - 1) + (1 - 1) + (1 - 1) + (1 error we're missing a 1

you have that extra 1 you'd need an additional -1 to equal another 0... an equal number of 1s and -1s

[Guest]

The problem is that the series 1-1+1-1+1-1... does not converge, it just oscillates between 1 and 0 as more terms are added. Non-convergent series can't be evaluated, and in this case attempts to evaluate one result in different answers. Just goes to show how slippery infinite series can be...

[Guest]

x = 1 + 1 + ... + 1 (x times)

x^2 = x + x + ... + x (multiply through by x)

2x dx = (1 + 1 + ... + 1) dx (take deriviative of both sides)

2x = (1 + 1 + ... + 1)

2x = x

2 = 1

it's not really have an x on the right side that's screwing this up for me.

needs to be more of:

f(x) = 1x

then

x = x all the way through

x^2=x^2 ...

...

not really sure how to explain that

[Guest]

The trick here is that by expressing x as (1+1+...1) you've concealed the fact that the RHS is still a function of x (ie. if x increases by 1 you need to add an extra 1 to the other side). So it looks like the (1+1+...1) is constant when of course it's not. Multiplying by x confuses the picture, so when differentiating you differentiate each x but make no consideration for the fact that the number of x's also varies with x.