**1**

# 1 = 2 Proved

### #1

Posted 25 April 2006 - 07:34 AM

H tends 2 infinity

---------------------------------

Never argue with an idiot. They drag you down to their level then beat you with experience.

### #2

Posted 25 April 2006 - 09:57 AM

**Edited by bred, 25 April 2006 - 05:42 PM.**

### #3

Posted 25 April 2006 - 09:57 AM

Edit: In reply to Bred, no they're not.

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### #4

Posted 25 April 2006 - 12:34 PM

### #5

Posted 25 April 2006 - 01:16 PM

You are differentiating different equations.

X*X = X whereas X+X = 2X

The first is a parabola and the second a straight line.

### #6

Posted 25 April 2006 - 02:36 PM

The logic is wrong.

You are differentiating different equations.

X*X = X whereas X+X = 2X

The first is a parabola and the second a straight line.

Yes but ultimately it sums up to the same thing, 4^2 = 16 and 4+4+4+4 = 16.

However, yeah they are totally different functions. You would have to use fourier analysis to build two functions from the same building blocks - sine and cosine. Ok maybe I'm over complicating things.

### #7

Posted 25 April 2006 - 02:37 PM

tomorrow doesn't matter,

turn that music up,

till the windows start to shatter,

cos you're the only one who can get me on my feet,

& i can't even dance

No Tomorrow - Orson

### #8

Posted 25 April 2006 - 04:26 PM

we say x

^{2}= x lots of x

now if we differentiate the entire equation basically its 2x on the left hand side

and on the right hand side every x is 1 aka x lots of 1 which is x

so 2x=x

divide by x

2=1

it "works" because every line makes sense and is mathmatically accurate apart from 2x = x and 2=1 obviously

If i am not here i am somewhere else

### #9

Posted 25 April 2006 - 04:35 PM

it "works" because every line makes sense and is mathmatically accurate apart from 2x = x and 2=1 obviously

That's not true - every line

**appears**to make sense! If they were all "mathematically accurate", then by implication 2=1 is mathematically accurate

I've been puzzling over this for a while now - I can think of two ways of stating what is wrong with the "proof", but I'm not completely sure yet. It's actually quite subtle, if I'm thinking on the right lines

### #10

Posted 25 April 2006 - 05:11 PM

### #11

Posted 25 April 2006 - 07:26 PM

H tends 2 infinity

---------------------------------

Never argue with an idiot. They drag you down to their level then beat you with experience.

### #12

Posted 25 April 2006 - 08:26 PM

However, yeah they are totally different functions. You would have to use fourier analysis to build two functions from the same building blocks - sine and cosine. Ok maybe I'm over complicating things.

I guess you've learned about Fourier series recently . Yes, you are over-complicating things!

### #14

Posted 25 April 2006 - 08:39 PM

I'm going for the second line is incorrect, because:

I agree with that; the second line is not an equality.

Now, here are the two thoughts I had:

- The first line is meaningless, because x is not necessarily an integer. (e.g. how can you do + + ... + "pi times"?)

I'm not really convinced by that though - The "x lots of x"
**cannot**be differentiated termwise, since the number of terms is variable (i.e. there are*x*terms).

I think that's the best explanation.

### #15

Posted 25 April 2006 - 08:45 PM

Party pooper

### #16

Posted 25 April 2006 - 08:48 PM

its in the maths textbook. the sign changes when u divide though.

Sorry, but I don't agree (or see where that would happen)!

### #17

Posted 25 April 2006 - 10:01 PM

Just that I am enjoying the debate but keeping quiet for now!!

H tends 2 infinity

---------------------------------

Never argue with an idiot. They drag you down to their level then beat you with experience.

### #18

Posted 26 April 2006 - 06:36 AM

At X=2

X = X + X = 2X (but it also = X - 2X +3X -X-6!)

Similarly, at X=3, X = X + X + X = 3X

Differentiating will give the rate of change (gradient) for each function at a discrete point only.

So, setting d/dx(X ) = d/dx(2X) implies that the gradient of X = the gradient of 2X.

So, 2X = 2 implies that the gradients of both functions = 2 at X = 1.

Similarly, when X = 3, X = 3X and the gradients are equal when 2 X = 3,

ie, gradients = 3 at X = 3/2.

### #19

Posted 26 April 2006 - 08:32 AM

As I said previously, the logic is flawed.

At X=2

X = X + X = 2X (but it also = X - 2X +3X -X-6!)

Similarly, at X=3, X = X + X + X = 3X

Yes, but if

*x*= 2, then and if

*x*= 3, then .

I don't think this is where the problem is. does equal , as my working showed. I think the problem is in the "

*x*times" when you differentiate since this is not taken care of.

If you assume that

*x*takes only positive integer values and differentiate the whole RHS, i.e. taking care of the "

*x*times", then you do not get a contradiction, so this must be where the error is ...

### #20

Posted 26 April 2006 - 12:27 PM

They are two different functions which just happen to give the same result at one particular value.

Try thinking of Higher maths questions where a circle and a straight line meet at a tangent; two very different functions but they share one set of values.

Come to think of it, does that mean that a circle is really a straight line?

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