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1 = 2 Proved
Started by Mr H, Apr 25 2006 07:34 AM
54 replies to this topic
#1
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Posted 25 April 2006 - 07:34 AM
H tends 2 infinity
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#2
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Posted 25 April 2006 - 09:57 AM
Edit: scrapping what I previously posted, for now at least.
Edited by bred, 25 April 2006 - 05:42 PM.
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Posted 25 April 2006 - 09:57 AM
There was a proof that someone showed me a few months ago, where you could prove that any integers, n, eg. 2, 3, 5, 10, 100......=1, if I can find it, I'll put it up here.
Edit: In reply to Bred, no they're not.
Edit: In reply to Bred, no they're not.
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#4
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Posted 25 April 2006 - 12:34 PM
You suddenly introduced '1' in the third line down on the right hand side. If you introduce it on the left hand side also, then the equality holds true.
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Posted 25 April 2006 - 01:16 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.
You are differentiating different equations.
X*X = X
whereas X+X = 2XThe first is a parabola and the second a straight line.
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Posted 25 April 2006 - 02:36 PM
QUOTE(duncad @ Apr 25 2006, 02:16 PM) 
The logic is wrong.
You are differentiating different equations.
X*X = X
whereas X+X = 2XThe 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.
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Posted 25 April 2006 - 02:37 PM
i've seen a proof like that before, one of my maths lecturers showed us it. can't remember why he said it worked though!
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Posted 25 April 2006 - 04:26 PM
i dont know if everyone is understanding what was said
we say x2 = 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
we say x2 = 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
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Posted 25 April 2006 - 04:35 PM
QUOTE(Dave @ Apr 25 2006, 05:26 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
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Posted 25 April 2006 - 05:11 PM
I'm going for the second line is incorrect, because:
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Posted 25 April 2006 - 07:26 PM
H tends 2 infinity
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Posted 25 April 2006 - 08:26 PM
QUOTE(dfx @ Apr 25 2006, 03:36 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!
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Posted 25 April 2006 - 08:29 PM
QUOTE(Mr H @ Apr 25 2006, 08:26 PM)
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Posted 25 April 2006 - 08:39 PM
QUOTE(ad absurdum @ Apr 25 2006, 06:11 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.
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Posted 25 April 2006 - 08:45 PM
its in the maths textbook. the sign changes when u divide though.
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Posted 25 April 2006 - 08:48 PM
QUOTE(PK Barackis @ Apr 25 2006, 09:45 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)!
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Posted 25 April 2006 - 10:01 PM
QUOTE(ad absurdum @ Apr 25 2006, 09:29 PM)
QUOTE(Mr H @ Apr 25 2006, 08:26 PM)
Just that I am enjoying the debate but keeping quiet for now!!
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#18
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Posted 26 April 2006 - 06:36 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
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.
At X=2
X
= X + X = 2X (but it also = X - 2X +3X -X-6!)Similarly, at X=3, X
= X + X + X = 3XDifferentiating 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.
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Posted 26 April 2006 - 08:32 AM
QUOTE(Steve @ Apr 25 2006, 09:26 PM)
QUOTE(duncad @ Apr 26 2006, 07:36 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 = 3XYes, 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 ...
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Posted 26 April 2006 - 12:27 PM
No, the problem lies in the fact that people are thinking of X and 2X as being two different ways of writing the same function.
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?
and 2X as being two different ways of writing the same function.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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