Proof: 1 = 2 (no division by zero!)

[The_root_of_all_evil]

1=2 for high values of 1 and low values of 2. Numerical Methods ftw.

 

[Lukeje]

Oh yeah, I did. Whoops!!! Mind you, try writing something -4 times. That can't be done anyways so my point that the two functions are not equal still stands.

Just because it can't be done in our limited notation doesn't mean it can't be done.
The answer of course is that his summation is equal to 2xdx between the limits 0 and x, hence meaning that the answer is 2x=2x after differentiation, as required.

Here is a proof on a similar vein that proves 3=0 (though you may have already seen this one):

If x^2+x+1=0

Then x^2=-x-1 and so x=-1-1/x (x=/=0)
Substitute x back into the original equation and we get x^2+(-1-1/x)+1=0

Therefore x^2-1/x=0 and x^2=1/x, meaning that x^3=1 and x=1

Substitute into the original equation and we get 1^2+1+1=0

so 3=0.

All you've done is prove that 1 is not a solution. The solutions are of course x = -1/2 + (i(3)^1/2)/2 and x = -1/2 - (i(3)^1/2)/2.

 

[The_root_of_all_evil]

Real problem is at line 4 where you've differentiated within BODMAS. Given how differentiation works, you've exploited the error. The error itself comes from the remainder from differentiating/integration.

For instance..working backwards

m=2n (Theory)
Multiply by x
mx=2nx
Integrate
(1/2)m(x^2)+C=nx^2+C2

Given X can be any number, C & C2 can match up to be equal...so any value(m) can equal any other value.

The other problem is that in using x=1, x=x^2=x^3.

 

[The_root_of_all_evil]

This explains it better than I could. [http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html]

 

[Lukeje]

But Lukeybabes, in maths you cannot say something can be done unless you can see it can be done. If you can't write it with any notation at the moment then it can't be done, unless you can prove otherwise.

As far as the second bit goes, well done. (though I would point out that stating "of course" may seem a little patronising to people who haven't done anything beyond standard grade maths and so all of this seems completely new).

Okay... -4, -4 times gives 16 = x^2. -4, -1 times is 4. Therefore -4, -4 times is 16.
It can be envisioned as so. If we set 0 to -16 (=-4 -4 -4 -4), then -4, -4 times would be 0. Setting -16 back to 0 (i.e. adding 16 to 0) gives 16. QED.

Oh, and for the other 'proof' its 'obvious' if you have ever studied quadratic eq'ns, and know that there can only ever be two solutions. The two solutions are the two I have given above which were found using the fact every quadratic eq'n may be written as (ax+b)^2 + c = 0, where a, b and c are constants.
This leads to the general solution that for a quadratic eq'n of the form Ax^2 + Bx + C = 0, x = [-B (plus or minus) (B^2 - 4AC)^(1/2)]/(2A).
NB. i = (-1)^1/2.

 

[Lord Beautiful]

This isn't quite my cup of tea. It would be if we were face-to-face with pencil and paper, but this is just getting a tad confusing, which is odd.

 

[Doug]

*head explodes*

I gotta stop coming into these math threads

2+2=4!

e^iQ = cos(Q) + i sin(Q)

2+2=/=4!, 4! = 4*3*2*1 = 24.


And the mistake the OP is making is that x^2 = 2xdx between the limits 0 and x, = (x + x + x + ... ), x times.
d(2xdx)/dx = 2x as req'd.

Erm, I meant ! in the 'shocking' way instead of the factoral way. Still, you're right

 

[Dr Pussymagnet]

Man, here's something you'll never see on another gaming forum.

 

[Fraught]

*head explodes*

I gotta stop coming into these math threads

Fuck man. I wanted to say EXACTLY that. This discussion here is so confusing.

 

[Marbas]

4. As these are the same equation, their derivative must be the same. Take the derivative of both sides:
d(x^2)/dx = d((x + x + x + ...) x times)/dx
2x = (1 + 1 + 1 + ...) x times.

You are telling lies.

d((x*x)/dx = x+x by the product rule.

That sum of yours isn't a standalone function, it's the result of a function.

What you did there was take the derivative your sum incorrectly.

 

[guyy]

Well, here's a relatively obvious flaw:

...

x^2 = x*x

3. This is equivalent, by the definition of multiplication, to:
x^2 = (x + x + x + ...) x times.

Step 3 here is wrong. Say X = -3; X*X = (-3)*(-3) = 9, but adding X X's together is [(-3)+(-3)+(-3)] = -9. The same problem occurs for any other negative number, so in reality:
x^2 = (|x| + |x| + |x| + ...) x times.

Some people earlier said some stuff about "x times" meaning "-3 times" in a case like this, which is, as they said, silly. Counting something a negative number of times is a really bizarre definition. How do you take the derivative of -3 functions?

...I'm not actually sure how to take the derivative of an absolute value function, but I doubt (d/dx)|x| = 1. That function doesn't even have a continuous slope. Done informally, its derivative would be -1 for X 0. This is probably where the 2 comes from.

 

[Samirat]

Okay... -4, -4 times gives 16 = x^2. -4, -1 times is 4. Therefore -4, -4 times is 16.
It can be envisioned as so. If we set 0 to -16 (=-4 -4 -4 -4), then -4, -4 times would be 0. Setting -16 back to 0 (i.e. adding 16 to 0) gives 16. QED.

First, I don't think this is good math. "0" is a very special number, with some very specific rules. I don't think it can be reassigned, or that the number line can be shifted, so lightly.

Secondly, it's a ridiculously complicated explanation for something absurdly obvious. -4, -4 times is this:

-(-4)-(-4)-(-4)-(-4)

Not only is this possible with our notation, it's very easy.

 

[Lukeje]

Okay... -4, -4 times gives 16 = x^2. -4, -1 times is 4. Therefore -4, -4 times is 16.
It can be envisioned as so. If we set 0 to -16 (=-4 -4 -4 -4), then -4, -4 times would be 0. Setting -16 back to 0 (i.e. adding 16 to 0) gives 16. QED.

First, I don't think this is good math. "0" is a very special number, with some very specific rules. I don't think it can be reassigned, or that the number line can be shifted, so lightly.

Secondly, it's a ridiculously complicated explanation for something absurdly obvious. -4, -4 times is this:

-(-4)-(-4)-(-4)-(-4)

Not only is this possible with our notation, it's very easy.

Yes, but to actually visualise subtracting a number, we have to have that number present to start with... Otherwise the problem of subtraction means nothing to us (this is the same reasoning that leads us to not really comprehend what 'half' of something is... as we only ever see a whole; try cutting a cake in half, what do you see? Not half a cake, but a WHOLE half of a cake).

 

[Samirat]

Okay... -4, -4 times gives 16 = x^2. -4, -1 times is 4. Therefore -4, -4 times is 16.
It can be envisioned as so. If we set 0 to -16 (=-4 -4 -4 -4), then -4, -4 times would be 0. Setting -16 back to 0 (i.e. adding 16 to 0) gives 16. QED.

First, I don't think this is good math. "0" is a very special number, with some very specific rules. I don't think it can be reassigned, or that the number line can be shifted, so lightly.

Secondly, it's a ridiculously complicated explanation for something absurdly obvious. -4, -4 times is this:

-(-4)-(-4)-(-4)-(-4)

Not only is this possible with our notation, it's very easy.

Yes, but to actually visualise subtracting a number, we have to have that number present to start with... Otherwise the problem of subtraction means nothing to us (this is the same reasoning that leads us to not really comprehend what 'half' of something is... as we only ever see a whole; try cutting a cake in half, what do you see? Not half a cake, but a WHOLE half of a cake).

I'm just saying, this is actually how you notate negative multiplication. Your problem is that you can't actually visualize a negative number? That's true. It's very hard to subtract something that doesn't already exist. The only way to really conceptualize negative numbers is in their positive opposites. For instance, to visualize subtracting 4, you have to have 4 things to take away.

I'm not entirely sure what you mean with the "half" thing. Half of a cake is still half of a cake, even if you see it as a whole. It's like if you take half of 4. It's a whole 2. But it's still half of 4.

 

[Lukeje]

I'm just saying, this is actually how you notate negative multiplication. Your problem is that you can't actually visualize a negative number? That's true. It's very hard to subtract something that doesn't already exist. The only way to really conceptualize negative numbers is in their positive opposites. For instance, to visualize subtracting 4, you have to have 4 things to take away.

Yes, I know.. thats what I was trying to do with having 4, -4s to start off with.

I'm not entirely sure what you mean with the "half" thing. Half of a cake is still half of a cake, even if you see it as a whole. It's like if you take half of 4. It's a whole 2. But it's still half of 4.

With regard to the only ever seeing 'wholes', its quite an interesting theoretical concept; we as humans cannot actually create half of something in the real world; it is a purely abstract idea that is taken for granted by most (and by me until I read about it in a book a while back... maybe by Roger Penrose?). Note how you had to have recourse to numbers?

I like how this threads now gone completely off-topic now... and seems to be about the visualisation of abstract mathematical concepts.

 

[CoziestPigeon]

Yeah, you didn't come close to thinking of this yourself, it's been floating around in chain emails and stupid internet forums for decades.