Another proof that 2 = 1

  • Thread starter gnome
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  • #1
gnome
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another "proof" that 2 = 1

x2 = x * x
x2 = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses]
x2 = (x + x + ... + x) [again, there are "x" terms in the parentheses]
now take derivatives of both sides:
2x = (1 + 1 + ... + 1) [there are still "x" terms in the parentheses]
2x = x
divide by x:
2 = 1
 

Answers and Replies

  • #2
quartodeciman
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What does the right side look like when x = 1/2 or √2 or π and so on?
 
  • #3
gnome
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Hey -- I didn't say it was a proof, I said it was a "proof".
 
  • #4
quartodeciman
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This one is interesting because there is no obvious removal of a 0 factor. That right side development doesn't work with x=0 either, so we already know x isn't 0.
 
  • #5
STAii
333
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Yes, this one is extremly interesting.
Normally, it is easy to figure out where the error is from the first look. But this one is harder (i liked it).
now take derivatives of both sides
I think the problem starts here, we have to make sure if it is actually valid to take the derivative of both sides.
If it was me writing this "proof", i would have written it in a different way (but still, would have reached the same result).
I would have started with
d/dx[x2] = d/dx[x2]
This way, i can make sure no one will tell me that taking the derivative of both sides might be not right.

(i would like to thank Zargawee for sending me the link as somekind of emergency sms )
 
  • #6
gnome
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STAii & quartodeciman:
You're on the right track, but not quite there yet.

STAii:
I would have started with
d/dx[x2] = d/dx[x2]
This way, i can make sure no one will tell me that taking the derivative of both sides might be not right.
Are you sure about that?
 
  • #7
KLscilevothma
314
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Originally posted by gnome

x2 = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses]
x2 = (x + x + ... + x) [again, there are "x" terms in the parentheses]
now take derivatives of both sides:
2x = (1 + 1 + ... + 1) [there are still "x" terms in the parentheses]
I guess is it because we can't take derivatives on both sides like that because the number of terms on the right hand side isn't finite ?

I like this kind of "proof"! :smile:
 
  • #8
gnome
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Not quite...
 
  • #9
quantumdude
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That's a very clever "proof". I haven't seen it before.

I think the difficulty comes in when going from this:

x2=x+x+...(x times)...+x+x

to taking the derivative. The above could be written as:

x2=Σi=1xx

edit: the superscript is the upper limit of summation

When you take the derivative of the string of x's, you seem to take it for granted that as x changes, the rate at which the number of terms (which is x) changes does not matter. But that is not at all clear if we look at the limit definition of the derivative of the above series:

d(x2)/dx=limΔx-->0(Σi=1x+Δx(x+Δx)-Σixx)/Δx

I am not certain of how to explicitly evaluate the above limit on the right hand side, but I am certain that the "proof" (incorrectly) ignores the x+Δx in the upper index of the first series.
 
  • #10
StephenPrivitera
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Perhaps I'm being naive, but I think that Tom's sums simplify to (x+dx)(x+dx) and x(x) because for any given sum the numbers summed are constants. The numerator of the limit goes to (x^2+2xdx+dx^2)-x^2=dx(2x-dx). Thus the limit is (2x+dx)->2x. So we get 2x=2x, no contradiction.
 
  • #11
Hurkyl
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Additionally, there are fatal problems in that representation when x is not a nonnegative integer.
 
  • #12
quantumdude
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Originally posted by Hurkyl
Additionally, there are fatal problems in that representation when x is not a nonnegative integer.

Right, I just thought of that. If we say we can take the derivative, we are assuming that x is continuous. But if we say that the number of x's can be counted in a series, then we are assuming that x is a counting number.
 
  • #13
gnome
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It looks like a function, but is it?

Isn't it sufficient to say that what I have written is not differentiable because it is not even a well-defined function, and certainly not a continuous function?

It's clearly not an infinite series, and yet I don't specify how many terms there are. i.e.: if x=4 there are 4 terms on the right side; if x=6, 6 terms, etc.

And as Hurkyl pointed out, it works only for nonnegative integers so even if it qualified as a function it would not be differentiable.

Would it make sense to call my x an "unspecified constant" rather than a variable? It seems to me that I've seen something like that before, but I can't remember where.
 
  • #14
STAii
333
1
Ok, i think i have solved it (i am totally sick now, so i might be writting nonesense).
The main problem is that (1+1+1+1 .. (x times) .. +1+1) is meaningless if x is not a positive integer (iow : x=1,2,3,4,5,6...)
And since it has no meaning for values around the positive integers, it is not continuous, and we can't find its derivative (this is like finding the derivative of f(x) = x!, it is meaningless since x! is only defined for positive integers, therefore we use the gamma function instead (i think))
So what i did is that i gave it a little bit more meaning :)
Let's define [ x ] as the floor function of x, that is, if :
[ x ] = n
then :
n <= x < (n+1)

In this case, it would be more general to define multiplying as :
a*b = (a+a+a+a+...( [ b ] times )) + (b-[ b ])a
for example :
3*2.5 = 3+3+(2.5-2)3 = 3+3+(0.5*3)=6+1.5 = 7.5
Looking better, this would be :
a*b = ([sum](from k=1 to [ b ])a) + (b-[ b ])a
Now, using this information, look what i did.
d(x^2)/dx = d(x^2)/dx (i don't think anyone disagrees on this one)
2x = d(x*x)/dx
2x = d([sum](from k1= to [ x ])x + (x-[ x ])x)/dx
2x = d([sum](from k1= to [ x ])x)/dx + d((x-[ x ])x)/dx
2x = [sum](from k1= to [ x ])1 + d((x-[ x ])x)/dx
2x = [ x ] + (x-[ x ])d(x)/dx + x*d(x-[ x ])/dx
2x = [ x ] + x - [ x ] + x*(d(x)/dx - d([ x ])/dx)
If x is not an integer, then : d([ x ])/dx)=0
2x = [ x ] + x - [ x ] + x*(1-0)
2x = 2x

So there is no problem from the first place.
I would like to hear comments about this if possible, thanks :smile:.

Edit : adjusted the sums.
Edit : changed a term.
 
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  • #15
gnome
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You went through all this to prove that 2=2?
How much fever do you have???
 
  • #16
STAii
333
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Lol, not a lot of fever (Actually today i am better :P).
I went thru all of that to proove that ur proof is wrong :).
 
  • #17
benzun_1999
259
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Originally posted by gnome
x2 = x * x
x2 = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses]
x2 = (x + x + ... + x) [again, there are "x" terms in the parentheses]
now take derivatives of both sides:
2x = (1 + 1 + ... + 1) [there are still "x" terms in the parentheses]
2x = x
divide by x:
2 = 1

The mistake starts in the 4th line.
If you cut the derivatives that is x you will get x and not 2x, so the final result will be x=x

-benzun
All For God
 
  • #18
burak_ilhan
14
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3=2

X3=x2*x
X3=x2(1+1+1.......) ( x terms)
X3=x2+x2+........
Derivatives:
3x2=2x+2x+2x...... (x terms)
3x2=2x*x
3x2=2x2
3=2
 
  • #19
STAii
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Originally posted by benzun_1999
If you cut the derivatives that is x you will get x and not 2x
Can you please explain ?
 
  • #20
gnome
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Posted by STAii:
I went thru all of that to proove that ur proof is wrong :).
But we all know my "proof" is complete nonsense. The point was not to have it proven wrong; merely to show where the fallacies are.
 
  • #21
Loren Booda
3,099
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The RHS hides a transitional anomaly between a summation of continuous variables and seeming equivalent using discrete natural numbers. &delta; is therein exploded from an infinitesimal to a finite domain. The derivative of the sum on the RHS becomes undefined as its number of terms approaches infinity, unlike examples with well-behaved calculus.

Also,

x=1+1+...+1 "x times."

Taking the derivative on both sides yields

1=0+0+...0 "x times."

So 1=0
 
  • #22
benzun_1999
259
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Confused.................

I am very very sorry i am confused thouroughly, though this question lies in the basics of maths, i am confused. Can anyone explain what is derivatives?

-benzun
 
  • #23
ahrkron
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It is a measure of how much a function changes when you change the value of its argument.

How much you pay for gas at the pump is a simple function of the ammount of gas you get. There, you know that every drop will be charged at a certain ammount per gallon. The ratio of price to gallons is the derivative of that function.

In more complicated situations, the ratio may vary depending on the value of the argument. For example, the ammount of light in a given day can be seen as a function of time: in the morning, you get more light with every passing minute (i.e., the change in your function is positive as time passes), but in the afternoon, the ratio is negative (the ammount of light decreases as time goes by).

You can surely get more info using google.
 
  • #24
STAii
333
1
Originally posted by gnome
But we all know my "proof" is complete nonsense. The point was not to have it proven wrong; merely to show where the fallacies are.
No offense, but did you bother reading my post ?
What i did is that i showed where the fallacy is in your proof, and actually corrected it, then i showed that after this correction was made, there is no problems anymore (iow, 2=2), which shows that there are no other fallacies in your original proof (unless i made a mistake in my post, which is why i am asking for people to read it, and see if it is right or not).
Thanks.

EDIT:
Originally posted by Loren Booda

x=1+1+...+1 "x times."
Taking the derivative on both sides yields
1=0+0+...0 "x times."
So 1=0
You know what you are somehow doing (from my POV) ?
It is like if you take the following function :
Code:
f(x)= 1   , x=1
      1/2 , x=2
      1/3 , x=3
      ....
And derivate it at x=1.
If you apply the derivation rules you will obviously get f'(1)=0, but also, if you examine the function, you will easily notice that it is not continuous, and therefore not derrentiable (sp?)
On the other hand L(x)=x is a continous function, but h(x)=1+1+1+1+...+1 (x times) is not a continuous function anywhere, therefore you can't take take its derivative.
To solve the problem, find a function that that equals L(x) at every x, and still is continuous.
The function that i was able to figure out (it might be wrong, and there might be other functions) was (adjusted to match L(x)) :
h(x)=1+1+1+...+1 ([ x ] times) + (x-[ x ])
For example, take h(5.5)=1+1+1+1+1 + (5.5-5) = 5+0.5 = 5.5
It is obvious (but i can't proove it so far) that h(x)=L(x) at every x.
Therefore, it is logical that L'(x)=h'(x)
1 = 0+0+0+...+0 ([ x ] times) + (1-0)
1 = 1 (iow, there is no problem)
(BTW, the last 2 lines are correct only when x is not and integer, when x is and integer, you will have to take the derivative from right and left, i haven't tried that yet)

What do you think Loren ?
 
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  • #25
gnome
1,036
1
I believe that your proof is no more valid than mine. This is my interpretation of what you are doing.

We start off with
L(x) = R(x) , both valid continuous functions of x [identical, in fact, as L(x)=R(x)=x^2]
Now you replace R(x) with S(x) + T(x) where S(x) is a sum of a string of integers of unspecified length, and T(x) is a string (also of unspecified length) of fractions. I think this is already fatally flawed because I don't think it adequately defines a domain. But even if it did, both of those "functions" (if they are functions) S(x) and T(x) are discontinuous. You are arguing that that's OK because when added together they form a continuous function, and so you conclude that it's OK to take derivatives of them.

But in doing so, you are taking advantage of the property of derivatives that says that:
if f(x) and g(x) are both differentiable, then d/dx[f(x)+g(x)] = d/dx f(x) + d/dx g(x)
But neither S(x) nor T(x) is differentiable, so you can't go through the mechanics of differentiating them, add the results together, and say it's OK. The mere fact that, after going through all those gyrations you eventually end up with a true statement (2x=2x), doesn't make the gyrations legitimate.

Look at it another way:
you say d((from k1= to [ x ])x)/dx = [x]
You got that by applying simplified tools or techniques that we learned for finding derivatives. But try to get the derivative of
S(x) = d((from k1= to [ x ])x)/dx using the definition of a derivative. You know, the limit definition.

You can't do it, because S(x) is not a continuous function. Same goes for T(x). So the whole proof goes out the window.
 
  • #26
gnome
1,036
1
By the way, I don't want to appear to be claiming credit for inventing this stupid proof. I called it "mine" only because I posted it. It was shown to me by my physics professor, who said it was shown to him by a classmate in his undergraduate days, & who knows where the classmate found it. For all we know, it was invented 300 years ago.
 
  • #27
STAii
333
1
Originally posted by gnome
both of those "functions" (if they are functions) S(x) and T(x) are discontinuous.
First, yes, they are functions.
Second, they are actually continuous for every value of x that is not an integer, this is why when i took the derivative i said that x is not an integer, cause otherwise neither of those two functions will be continuous (but their sum will still be continuous).
if f(x) and g(x) are both differentiable, then d/dx[f(x)+g(x)] = d/dx f(x) + d/dx g(x)
But neither S(x) nor T(x) is differentiable ...
Again, it seems that you didn't look well into the functions when x is not an integer.
When x is not an integer, [ x ] can be considered a constant when finding the derivative (and this is proovable).
... after going through all those gyrations you eventually end up with a true statement (2x=2x), doesn't make the gyrations legitimate.
I totally agree in this point, having a right result does not always mean having a right way of reaching the result, but so far, i don't see how this is related to my post.
Look at it another way:
you say d([sum](from k1= to [ x ])x)/dx = [x]
You got that by applying simplified tools or techniques that we learned for finding derivatives. But try to get the derivative of
S(x) = d([sum](from k1= to [ x ])x)/dx using the definition of a derivative. You know, the limit definition.
Weird ... this is the exact same thing that is in your original post !! I don't see that u tried to reach the same result using the definition of the derivative (the only difference between my version and yours, is that i used a sum symbol, and my sum ends at [ x ] instead of x, but it is the same concept; derivating the sum of a known number of x.
Here is your version of the same concept :

... = (x + x + ... + x) [again, there are "x" terms in the parentheses]
After finding derivative:
(1 + 1 + ... + 1) [there are still "x" terms in the parentheses]
= x

or am i wrong ?

... So the whole proof goes out the window.
Well, i have nothing against this happening, but when this happens, i want to know why my proof went out of the window.

You know, if sth wasn't solved for 300 years, it doesn't mean that it won't be solved forever :) (i don't mean that i solved it, i was just pointing out this fact).

EDIT: adjusted tag.
 
Last edited:
  • #28
STAii
333
1
If you still insist on finding that derivative using the definition of the derivative, i think this can help.
For ease of reading, let [sum] mean : [sum](from k=1 to [ x ])
S(x) = [sum](x)
S'(x) = lim(h->0) ( (S(x+h)-S(h))/h )
S'(x) = lim(h->0) ( ([sum](x+h) - [sum](x))/h )
S'(x) = lim(h->0) ( ([sum](x) + [sum](h) - [sum](x))/h )
S'(x) = lim(h->0) ( ([sum](h))/h )
Since h is not dependent on k, then the [sum](h) = [ x ]*h
S'(x) = lim(h->0) ( [ x ]*h/h ) = lim(h->0) ([ x ])
If x is not an integer, than the limit above has a value of [ x ]
S'(x) = [ x ]

#
 
  • #29
gnome
1,036
1
I don't see that u tried to reach the same result using the definition of the derivative (the only difference between my version and yours, is that i used a sum symbol, and my sum ends at [ x ] instead of x, but it is the same concept; derivating the sum of a known number of x.
There's another difference. My version was a joke, posted here to amuse you. You are taking my joke, changing it, and trying to construct a proof around your version of the joke to prove that my version of the joke was mathematically incorrect. How ridiculous is that?

But OK, if you want to keep beating this dead horse...

What exactly is
&sum;(from k=1 to [ x ])x + (x-[ x ])x) ?

when x = 4.5 for example?

Sum of what? Sum of k from k=1 to k=4.5?
That's
[1+2+3+4 ]* 4.5 + (4.5 - 4)*4.5 = 10 * 4.5 + .5 * 4.5 = 47.25

I guess that's not what you had in mind, so please explain.
 
  • #30
STAii
333
1
Well, i thought that if it was a joke, it should belong to "General discussion", not "Calculus"

Ok, about the sum.
Note that the variable of the sum (k) is not related to the variable inside the sum (x).
So, in the case that you took x=4.5
Entering this value into the floor function, [ x ] = 4

so :
&sum;(from k=1 to [ x ])x + (x-[ x ])x)
is :
&sum;(from k=1 to 4)4.5 + (4.5-4)4.5)
=(4.5+4.5+4.5+4.5) + 0.5*4.5
=18 + 2.25
=20.25
which equals 4.5*4.5

Clearer now ?

EDIT : adjusted Sum symbol
 
Last edited:
  • #31
gnome
1,036
1
I don't think we are entitled to invent our own meaning for a summation.

Summation has a very precise meaning which you can look up. My understanding of it is:
&sum;k=1m f(k) means you must specify some function of k (which may be just k itself, or k2, or sin(k), etc), and you evaluate that f(k) at each of the integer values from k=1 to k=m (and hence you must specify what m is) and then add up all of those values of f(k).

That doesn't appear to be what you are doing.
 
  • #32
STAii
333
1
Well, the function after the sum does not necessarily have to be a function of k (the variable that the sum is changing). Examples include :
&sum;(from k=1 to 3)1
which equals :
1+1+1=3
Or even, suppose that in your example, f(k) was a constant function, what will you have ? (well, you will have sth like what is above).
Mainly, if the function after the sum is not a function of the changing variable (k in our case), the value of the sum becomes as follows :
&sum;(from k=n to m)f(x1)=(m-n+1)f(x1)

And secondly, the value of m in my sum is specified, it has the value of [ x ].

Anything else
 
  • #33
quartodeciman
372
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Even a joke can have more than one use.

More fun!

The rightside function is indeed an infinite series of functions. Just look at the known values:

f(0) = 0 + 0 + 0 + 0 + ...
f(1) = 1 + 0 + 0 + 0 + ...
f(2) = 2 + 2 + 0 + 0 + ...
f(3) = 3 + 3 + 3 + 0 + ...
f(4) = 4 + 4 + 4 + 4 + ...

and so on.

So, how does one write this function in a closed form as an infinite series (paying attention to the values between those positive integer argument values)?

Enter the positive unit stepup function Sn(x). It is equal to 0 when x < n and it is equal to 1 when n<=x.

-Sn(x) is equal to 0 when x < n and it is equal to -1 when n<=x.

1-Sn(x) is equal to 1 when x < n and it is equal to 0 when n<=x. So, 1-Sn(x) is a positive unit stepdown function.

Check this out.
Sn(x) (1-Sn+1(x)) is equal to 0(0) = 0 when x < n; it is equal to 1(1) = 1 when n<=x<n+1;it is equal to 1(0) = 0 when n+1<=x.
This is a positive unit impulse function between two integer values.

Using these, one can construct the terms of the infinite series for all integer argument values, including also the negative ones:

f(-0) = 0 + 0 + 0 + 0 + ...
f(-1) = 1 + 0 + 0 + 0 + ...
f(-2) = 2 + 2 + 0 + 0 + ...
f(-3) = 3 + 3 + 3 + 0 + ...
f(-4) = 4 + 4 + 4 + 4 + ...

and so on.

Here is my expression for the infinite series function:

f(x) = &Sigma;n=0&infin; {n(1-S-n(x) + Sn(x)) + (x2-n2)((1-S-n(x))(S-n-1(x))+ ((Sn(x))(1-Sn+1(x))}

I assert that this series will reproduce the values at integer arguments, both negative and positive, just like the above examples.
Alas, the infinite series converges pointwise, but not uniformly. So differentiation term by term isn't justified.
Finally, despite the hopeless appearance of discontinuity and undifferentiability at those integer argument values, I assert that f(x) = x2 for all real x.
 

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