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1 + 1 = 2 ?





Shin
Hi there,

I believe everything has an exception. In order to prove this, I have to start from a proven "non-exception" stuff like Science. In this case I will use mathematics. Start from a simple one Smile
Is 1 + 1 always = 2 ?? If you don't think so, I would like to hear from you. Please tell me why you think so. Thanks a lot! Wink
EanofAthenasPrime
Shin wrote:
Hi there,

I believe everything has an exception. In order to prove this, I have to start from a proven "non-exception" stuff like Science. In this case I will use mathematics. Start from a simple one Smile
Is 1 + 1 always = 2 ?? If you don't think so, I would like to hear from you. Please tell me why you think so. Thanks a lot! Wink


Define your parameters. For example, 1 cup of flour+1 cup of flour does not always equal 2 muffins...
Indi
1 + 1 = 1 for small values of 1, and 1 + 1 = 3 for large values of 1.
carlospro7
In the truth table 1 + 1 = 1. 1 or 1 = 1
tempdbs
EanofAthenasPrime wrote:
Shin wrote:
Hi there,

I believe everything has an exception. In order to prove this, I have to start from a proven "non-exception" stuff like Science. In this case I will use mathematics. Start from a simple one Smile
Is 1 + 1 always = 2 ?? If you don't think so, I would like to hear from you. Please tell me why you think so. Thanks a lot! Wink


Define your parameters. For example, 1 cup of flour+1 cup of flour does not always equal 2 muffins...


You can define in other way -

Answer this -


1 cup of Water + 1 cup of Sugar = ?? ( 2 Cup of Sugarwater or 1 Cup of Sugar water)
Genesiz
There is a poem written by Brian Patten called 'The Sick Equation'. In it he uses his parents marriage as proof that 1+1=1. He says that in school he was taught that 1+1=2, but when his parents got married they still stayed seperate (1).

The equation 1+1=2 says that when two things are added together, they equal the sum of the two things. This is not always true in life.
dwinton
Shin wrote:
Hi there,

I believe everything has an exception. In order to prove this, I have to start from a proven "non-exception" stuff like Science. In this case I will use mathematics. Start from a simple one Smile
Is 1 + 1 always = 2 ?? If you don't think so, I would like to hear from you. Please tell me why you think so. Thanks a lot! Wink


If you mean in math, I can prove 1+1=2.

From Doctor Math:

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
EanofAthenasPrime
tempdbs wrote:
EanofAthenasPrime wrote:
Shin wrote:
Hi there,

I believe everything has an exception. In order to prove this, I have to start from a proven "non-exception" stuff like Science. In this case I will use mathematics. Start from a simple one Smile
Is 1 + 1 always = 2 ?? If you don't think so, I would like to hear from you. Please tell me why you think so. Thanks a lot! Wink


Define your parameters. For example, 1 cup of flour+1 cup of flour does not always equal 2 muffins...


You can define in other way -

Answer this -


1 cup of Water + 1 cup of Sugar = ?? ( 2 Cup of Sugarwater or 1 Cup of Sugar water)


Neither. 1<sugar water<2
DoctorBeaver
In binary notation 1+1=10 Laughing
Jakob [JaWGames]
I would guess that it you looks at the modern way to count you always get 1+1=2 because of that we have made the system in that way...
powers1983
Jakob [JaWGames] wrote:
I would guess that it you looks at the modern way to count you always get 1+1=2 because of that we have made the system in that way...


I think thats exactly it - we made the number system and one of the basic rules that we incorporated was that 1+1 does indeed equal 2. If you move away from numbers then there as loads of examples of 1 (something) + 1(something else) != 2 (anything).

David.
Soltair
Well there's a small mathematical demonstration to proof anything around you is false.

Consider two parameters, "a" and "b", which would be both equal to one.

Step one a=ab and a=b
Step two a-ab=a-b
Step three a(a-b)=(a+b)(a-b)
Step four a=a+b
Step five 0=b
Step six 0=1

Considering that 1 isn't 1, 1+1=anything. But of course this is just a derived 0 division, which as you know puts "error" on your calculator because it leads to this kind of errors. 0 truly is a nasty boy in mathematics... Wink
polis
In normal algebra 1+1=2

In binary boolean algebra 1+1=1

in Administration theory, 2+2 = 5
ThreeRight
http://www.ph.utexas.edu/~gleeson/httb/section1_3_4_4.html

makes a really great argument that 1+1=1
Arnie
Soltair, what you're doing is mathematically incorrect because (a-b) is 0, and from step 3 to 4 you're dividing by zero. Besides, from step 2 to 3 you're factoring out (a-b) and factoring out 0 is also impossible.
moeller
in the end, we will all discover that 1+1=2. No exceptions. (Math not baking)
EanofAthenasPrime
lol but in a universe where matter decayed at an infinite rate, 1+1=0
Indi
i think i did this "proof" or a similar one on Frihost a while ago, but i'm not sure. Normally it's used to prove that 2 = 1, but that's the same thing as 1 + 1 = 1, so:

First, pick any integer x. Note that:
x = 1 + 1 + ... + 1 (x times). Or to put it in proper math terms:
Code:
     x
x =  ∑   1
   i = 1


For example:
for x = 3: 3 = 1 + 1 + 1
for x = 5: 5 = 1 + 1 + 1 + 1 + 1
and so on.

Multiplying both sides by x gives:
x = x + x + ... + x (x times) Or in other words:
Code:
      x
x =  ∑   x
    i = 1


Which means that:
for x = 3: 9 = 3 + 3 + 3
for x = 5: 25 = 5 + 5 + 5 + 5 + 5

So far so good.

Now take the derivative of both sides. The derivative of x is 2x. The derivative of x + x + ... + x is 1 + 1 + ... + 1.

So now we get:
2x = 1 + 1 + ... + 1 (x times). Or:
Code:
      x
2x =  ∑   1
    i = 1


But we already know from the beginning that 1 + 1 + ... + 1 (x times) is x, or:
Code:
  x
  ∑   1 = x
i = 1


Which means if we substitute we get:
2x = x

Which is the same as:
x + x = x

Which, if we divide by x gives:
1 + 1 = 1

Arnie
The fact that two expressions have the same derivative does not imply that they are equal. Take x and x + 3. Both have 2x as their derivative but x + 3 certainly does not equal x.
Soltair
Arnie wrote:
Soltair, what you're doing is mathematically incorrect because (a-b) is 0, and from step 3 to 4 you're dividing by zero. Besides, from step 2 to 3 you're factoring out (a-b) and factoring out 0 is also impossible.


If I identify my variables at the end of everything, it does work. This is only to show why they've decided to put "error" on most calculators when it comes to dividing by zero. And if you read that tiny little part at the end of my message, I am well aware of these facts. This demonstration is not mine but comes from a book about 0 in mathematical universe...
polis
Arnie wrote:
The fact that two expressions have the same derivative does not imply that they are equal. Take x and x + 3. Both have 2x as their derivative but x + 3 certainly does not equal x.


That's right.
EanofAthenasPrime
Indi wrote:
i think i did this "proof" or a similar one on Frihost a while ago, but i'm not sure. Normally it's used to prove that 2 = 1, but that's the same thing as 1 + 1 = 1, so:

First, pick any integer x. Note that:
x = 1 + 1 + ... + 1 (x times). Or to put it in proper math terms:
Code:
     x
x =  ∑   1
   i = 1


For example:
for x = 3: 3 = 1 + 1 + 1
for x = 5: 5 = 1 + 1 + 1 + 1 + 1
and so on.

Multiplying both sides by x gives:
x = x + x + ... + x (x times) Or in other words:
Code:
      x
x =  ∑   x
    i = 1


Which means that:
for x = 3: 9 = 3 + 3 + 3
for x = 5: 25 = 5 + 5 + 5 + 5 + 5

So far so good.

Now take the derivative of both sides. The derivative of x is 2x. The derivative of x + x + ... + x is 1 + 1 + ... + 1.

So now we get:
2x = 1 + 1 + ... + 1 (x times). Or:
Code:
      x
2x =  ∑   1
    i = 1


But we already know from the beginning that 1 + 1 + ... + 1 (x times) is x, or:
Code:
  x
  ∑   1 = x
i = 1


Which means if we substitute we get:
2x = x

Which is the same as:
x + x = x

Which, if we divide by x gives:
1 + 1 = 1



well your first fallacy is conveniently reassigning the same variable to different sides of the equation...

Your second fallacy is "The derivative of x is 2x. The derivative of x + x + ... + x is 1 + 1 + ... + 1."....
Arnie
If you're going to point out fallacies, you should explain why something is wrong, because I see no error in those steps.

Soltair, sorry I didn't read the bottom of your post. But (a-b) is always 0 when a and b are equal, making the equation illegal. If they aren't equal, your first step already is illegal since a=ab and a=b can only be true if a=b. So either way 0=1 is impossible, regardless of the moment you identify the variables.
EanofAthenasPrime
But as soon as he uses deriviates, that makes the equation invalid. In the order of operations, there ARENT deriviates. As a matter of fact, I don't even know what deriviatives are and I can tell you that math operation is invalid in the context in which it is used.
Indi
Every time i introduce this "proof", i get dozens of false "solutions". This time is no exception. ^_^

polis wrote:
Arnie wrote:
The fact that two expressions have the same derivative does not imply that they are equal. Take x and x + 3. Both have 2x as their derivative but x + 3 certainly does not equal x.


That's right.

Nope. i'm afraid that's not right. Read the "proof" more carefully. At no point do i say that because two derivatives are equal, their anti-derivatives must be equal to.

EanofAthenasPrime wrote:
well your first fallacy is conveniently reassigning the same variable to different sides of the equation...

Your second fallacy is "The derivative of x is 2x. The derivative of x + x + ... + x is 1 + 1 + ... + 1."....

Neither of those are fallacies.

To make things more clear, perhaps i will make a MathML page to show the actual "proof" in real mathematical terms. In the forum, it's hard to read. It might help. Or it might not.
Arnie
Oh yeah, sorry. It's the other way around in this mindgame: two apparently equal statements producing different derivatives. That's impossible, so the trick must be somewhere before the derivative is taken (or in the way it's taken). I'm betting the real derivative of
Code:
  x
  ∑  x
i = 1
is 2x as well. Since I'm too lazy to dive into literature on differentiating ∑'s, I'm not going to prove it though. It's likely that the x "on top" has to be taken into account as well because it determines how many times the x "on the right" has to be added up. That basically makes it a factor.

In other words, the statement in the code block above equals:
Code:
     x
x *  ∑  1
   i = 1
which equals x*x which equals x.
EanofAthenasPrime
I think the problem lies in never being taught anything about deriviatives...perhaps once I acquire the knowledge I can better defeat your equation.

In calculus, a branch of mathematics, the derivative is a measurement of how a function changes when the values of its inputs change. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. In higher dimensions, the derivative of a function at a point is a linear transformation called the linearization.[1]

The process of finding a derivative is called differentiation. The fundamental theorem of calculus states that differentiation is the reverse process to integration.


Again...I have never been taught derivatives (am 16 and haven't yet taken calculus), however, it seems to me you can't use them in common equations.

In order for an equation to remain stable both sides must undergo the same change, using division, multiplication, addition, or subtraction. What you did was say

In equation 3^2=3+3+3, using this magic word called "Derivatives" I can disobey the laws of mathematics and say (3^2)/(6)=(9)/(3)

First off all, what I do know of derivatives is that they are used for FUNCTIONS, not EQUATIONS or EQUALITIES. NO WHERE IN THE ORDER OF OPERATIONS DOES IT SAY THAT YOU CAN USE DERIVATIVES...

You always have to modify both sides of the exact same way if you wish to avoid fallacies. Indi, using derivatives, you essentially subtracted 3 from one side of the equation and subtracted 6 from the other. That is not mathematics. That is just being silly.
Indi
i have made a MathML page with the "proof" here. It should be a lot clearer.

Arnie wrote:
Oh yeah, sorry. It's the other way around in this mindgame: two apparently equal statements producing different derivatives. That's impossible, so the trick must be somewhere before the derivative is taken (or in the way it's taken).

Excellent deducing. ^_^ You have eliminated half of the false proof from suspicion.

Arnie wrote:
I'm betting the real derivative of
Code:
  x
  ∑  x
i = 1
is 2x as well. Since I'm too lazy to dive into literature on differentiating ∑'s, I'm not going to prove it though.

Dont be blinded by notation! ∑ x just means (x + x + ... + x) for some number of x's. And you know how to differentiate that. (x + x + ... + x) for N x's differentiates to (1 + 1 + ... + 1) for N 1's. That's just good old term-by-term differentiation, which works because differentiation is a linear operation.

Arnie wrote:
It's likely that the x "on top" has to be taken into account as well because it determines how many times the x "on the right" has to be added up. That basically makes it a factor.

Correct. It would mean that for what i wrote above, N is x. No trick there. Or is there?

Arnie wrote:
In other words, the statement in the code block above equals:
Code:
     x
x *  ∑  1
   i = 1
which equals x*x which equals x.

Yes. The way the proof goes is exactly like that. First i say x = 1 + 1 + ... + 1 for x 1's. Then i multiply both sides by x to get x = x + x + ... + x for x x's. Then i differentiate both sides to get 2x = 1 + 1 + ... + 1 for x 1's. And from the first equation, you know that the right side is equal to x. Therefore 2x = x.

Don't let the notation blind you. i can promise you that it is not the summation that causes the problem (finite and infinite summations are differentiated all the time). But as you said, it's impossible for two equal functions to have different derivatives. Something is fishy...!

EanofAthenasPrime wrote:
I think the problem lies in never being taught anything about deriviatives...perhaps once I acquire the knowledge I can better defeat your equation.

Yes. Frankly you have no chance of figuring this problem out. It requires an understanding of differentiation to grok where i cheated. Of course, the problem is obvious to anyone that understands differentiation after someone has pointed it out... not so easy to spot it yourself.

EanofAthenasPrime wrote:
Again...I have never been taught derivatives (am 16 and haven't yet taken calculus), however, it seems to me you can't use them in common equations.

What is a common equation?

EanofAthenasPrime wrote:
In order for an equation to remain stable both sides must undergo the same change, using division, multiplication, addition, or subtraction.

Both sides were differentiated, then one side was simplified. Nothing strange was done in that department.

EanofAthenasPrime wrote:
What you did was say

In equation 3^2=3+3+3, using this magic word called "Derivatives" I can disobey the laws of mathematics and say (3^2)/(6)=(9)/(3)

No. What i did was say that one side was f(x), the other side was g(x)... and f(x) = g(x)... therefore f'(x) = g'(x) (taking the derivative of both sides)... and then i simplified. Except one minor gotcha (which is really the key to it all), there is nothing wrong with what i did.

EanofAthenasPrime wrote:
First off all, what I do know of derivatives is that they are used for FUNCTIONS, not EQUATIONS or EQUALITIES. NO WHERE IN THE ORDER OF OPERATIONS DOES IT SAY THAT YOU CAN USE DERIVATIVES...

i am not sure what order of operations you are referring to, but you are over thinking the problem. Functions can be equations and equations can be functions... equations can be made up of functions and functions can be made up of equations. Really they're both the same thing looked at a different way. Equalities are just a special case of equations. y = x + 3 is an equation that is also an equality that contains any number of functions (depending on how you want to count them). You could say f(x) := x + 3 and that y = f(x). Or you can say g(x) := x and h(x) := 3 and y = g(x) + h(x). It's all the same. Nothing changes except your perspective.

EanofAthenasPrime wrote:
You always have to modify both sides of the exact same way if you wish to avoid fallacies. Indi, using derivatives, you essentially subtracted 3 from one side of the equation and subtracted 6 from the other. That is not mathematics. That is just being silly.

That is not what i did. Arnie's objection was dead wrong. He didn't read the proof properly. He is right that even if f'(x) = g'(x), then it is not necessarily true that f(x) = g(x). But that's not what i did. If f(x) = g(x), then f'(x) = g'(x). And that is what i did. And, that is mathematically sound (with one caveat! ^_-).
EanofAthenasPrime
lol, if only I had been taught derivatives...
nilsmo
If you define + as addition over real numbers and assume 1 means the positive integer 1, and 2 means the positive integer 2, then 1 + 1 = 2 is true. If you have other assumptions/definitions 1 + 1 = 2 may not be true.
DjinniFire
Indi wrote:
1 + 1 = 1 for small values of 1, and 1 + 1 = 3 for large values of 1.

ok i hate that, it's funny but it really doesn't work unless you round -.- which technically doesn't give you equal sign

tempdbs wrote:
EanofAthenasPrime wrote:
Shin wrote:
Hi there,

I believe everything has an exception. In order to prove this, I have to start from a proven "non-exception" stuff like Science. In this case I will use mathematics. Start from a simple one Smile
Is 1 + 1 always = 2 ?? If you don't think so, I would like to hear from you. Please tell me why you think so. Thanks a lot! Wink


Define your parameters. For example, 1 cup of flour+1 cup of flour does not always equal 2 muffins...


You can define in other way -

Answer this -


1 cup of Water + 1 cup of Sugar = ?? ( 2 Cup of Sugarwater or 1 Cup of Sugar water)


define cup for both, if they are the same size cup then you can add them together and you'll end up with 2 cups of sugar water assuming all sugar is dissolved in water. x] but whatever adding labels to numbers always makes math weird.


Indi wrote:
i have made a MathML page with the "proof" here. It should be a lot clearer.

Arnie wrote:
Oh yeah, sorry. It's the other way around in this mindgame: two apparently equal statements producing different derivatives. That's impossible, so the trick must be somewhere before the derivative is taken (or in the way it's taken).

Excellent deducing. ^_^ You have eliminated half of the false proof from suspicion.

Arnie wrote:
I'm betting the real derivative of
Code:
  x
  ∑  x
i = 1
is 2x as well. Since I'm too lazy to dive into literature on differentiating ∑'s, I'm not going to prove it though.

Dont be blinded by notation! ∑ x just means (x + x + ... + x) for some number of x's. And you know how to differentiate that. (x + x + ... + x) for N x's differentiates to (1 + 1 + ... + 1) for N 1's. That's just good old term-by-term differentiation, which works because differentiation is a linear operation.

Arnie wrote:
It's likely that the x "on top" has to be taken into account as well because it determines how many times the x "on the right" has to be added up. That basically makes it a factor.

Correct. It would mean that for what i wrote above, N is x. No trick there. Or is there?

Arnie wrote:
In other words, the statement in the code block above equals:
Code:
     x
x *  ∑  1
   i = 1
which equals x*x which equals x.

Yes. The way the proof goes is exactly like that. First i say x = 1 + 1 + ... + 1 for x 1's. Then i multiply both sides by x to get x = x + x + ... + x for x x's. Then i differentiate both sides to get 2x = 1 + 1 + ... + 1 for x 1's. And from the first equation, you know that the right side is equal to x. Therefore 2x = x.

Don't let the notation blind you. i can promise you that it is not the summation that causes the problem (finite and infinite summations are differentiated all the time). But as you said, it's impossible for two equal functions to have different derivatives. Something is fishy...!

EanofAthenasPrime wrote:
I think the problem lies in never being taught anything about deriviatives...perhaps once I acquire the knowledge I can better defeat your equation.

Yes. Frankly you have no chance of figuring this problem out. It requires an understanding of differentiation to grok where i cheated. Of course, the problem is obvious to anyone that understands differentiation after someone has pointed it out... not so easy to spot it yourself.

EanofAthenasPrime wrote:
Again...I have never been taught derivatives (am 16 and haven't yet taken calculus), however, it seems to me you can't use them in common equations.

What is a common equation?

EanofAthenasPrime wrote:
In order for an equation to remain stable both sides must undergo the same change, using division, multiplication, addition, or subtraction.

Both sides were differentiated, then one side was simplified. Nothing strange was done in that department.

EanofAthenasPrime wrote:
What you did was say

In equation 3^2=3+3+3, using this magic word called "Derivatives" I can disobey the laws of mathematics and say (3^2)/(6)=(9)/(3)

No. What i did was say that one side was f(x), the other side was g(x)... and f(x) = g(x)... therefore f'(x) = g'(x) (taking the derivative of both sides)... and then i simplified. Except one minor gotcha (which is really the key to it all), there is nothing wrong with what i did.

EanofAthenasPrime wrote:
First off all, what I do know of derivatives is that they are used for FUNCTIONS, not EQUATIONS or EQUALITIES. NO WHERE IN THE ORDER OF OPERATIONS DOES IT SAY THAT YOU CAN USE DERIVATIVES...

i am not sure what order of operations you are referring to, but you are over thinking the problem. Functions can be equations and equations can be functions... equations can be made up of functions and functions can be made up of equations. Really they're both the same thing looked at a different way. Equalities are just a special case of equations. y = x + 3 is an equation that is also an equality that contains any number of functions (depending on how you want to count them). You could say f(x) := x + 3 and that y = f(x). Or you can say g(x) := x and h(x) := 3 and y = g(x) + h(x). It's all the same. Nothing changes except your perspective.

EanofAthenasPrime wrote:
You always have to modify both sides of the exact same way if you wish to avoid fallacies. Indi, using derivatives, you essentially subtracted 3 from one side of the equation and subtracted 6 from the other. That is not mathematics. That is just being silly.

That is not what i did. Arnie's objection was dead wrong. He didn't read the proof properly. He is right that even if f'(x) = g'(x), then it is not necessarily true that f(x) = g(x). But that's not what i did. If f(x) = g(x), then f'(x) = g'(x). And that is what i did. And, that is mathematically sound (with one caveat! ^_-).


I really suck at calculus but since you have a variable in the summation taking the derivative is not the same as normal summations because summation from i=1 to x of x is the same as x(x) so multiplying x step made it a degree of 2. I know if you have summation i=1 to i=5 of x then taking derivative is summation i=1 to i=5 of 1. 5 is constant so unaffected by a derivative; it's still 5(1) since derivative of x is 1. =.= Yeah that's my take at it. The derivative of the summation is calculated incorrectly because of the i=1 to x of the summation of x.
Arnie
Yes, that's what I meant with:
Quote:
It's likely that the x "on top" has to be taken into account as well because it determines how many times the x "on the right" has to be added up. That basically makes it a factor.
Bondings
@Indi, x is an integer and not a real number. This means that the function is not continuous and hence not differentiable, which means you aren't allowed to take the derivative of both sides of the equation.
nilsmo
Edit: Bondings posted while I was typing this... Ah, that's a good point Bondings since summation notation only applies for a whole number of terms (x must be a whole number).

Arnie is correct. To be more specific, here is the fallacy.

Indi wrote:
That's just good old term-by-term differentiation


The amount of terms varies in

Code:
  x
  ∑  x
i = 1


OK, that should be enough. Here's some more elaboration that's probably confusing and redundant and useless.

The addition rule of derivatives says that (f+g)' = f'+g'. This is true. One is tempted to say therefore that derivative of (f_1+f_2+f_3+...f_x) equals the sum of the derivatives of all the functions inside. But this is not true because x is not constant. The amount of terms varies.

In other words, the derivative of a sum of functions is equal to the sum of the derivatives of each function but the derivative of a sum of different sets of functions depending on x is not equal to the sum of the derivatives.

I don't really know how to explain this. It really is extremely simple. I haven't even taken calculus - but catching the error is not hard.

Another way to explain the fallacy is by the product rule I suppose. It says the derivative of the product of two functions is not always equal to the product of the derivatives of the two functions.

Code:

    g(x)
     ∑  f(x) = g(x)*f(x)
   i = 1

Derivative of g*f != Derivative of g * Derivative f as Indi tried tricking us into thinking.

It is true though that the derivative of k*f = k * the derivative of f where k is a constant which makes this tricky.

Code:

    k
     ∑  f(x) = k*f(x)
   i = 1
Indi
DjinniFire wrote:
Indi wrote:
1 + 1 = 1 for small values of 1, and 1 + 1 = 3 for large values of 1.

ok i hate that, it's funny but it really doesn't work unless you round -.- which technically doesn't give you equal sign

That depends on how and when you round. ^_^

It's actually a joke we use to remind ourselves not to round off prematurely.

DjinniFire wrote:
I really suck at calculus but since you have a variable in the summation taking the derivative is not the same as normal summations because summation from i=1 to x of x is the same as x(x) so multiplying x step made it a degree of 2. I know if you have summation i=1 to i=5 of x then taking derivative is summation i=1 to i=5 of 1. 5 is constant so unaffected by a derivative; it's still 5(1) since derivative of x is 1. =.= Yeah that's my take at it. The derivative of the summation is calculated incorrectly because of the i=1 to x of the summation of x.

Nope.

It is perfectly legal to do what i did (with one caveat). A summation is the same thing as an integration (see here), and there's nothing wrong with differentiating an integration term.

Arnie wrote:
Yes, that's what I meant with:
Quote:
It's likely that the x "on top" has to be taken into account as well because it determines how many times the x "on the right" has to be added up. That basically makes it a factor.

Nope, sorry. As i explained above, there is nothing wrong with having the x on top of a summation - it's the same as having the x on top of an integration, which is perfectly legal.

nilsmo wrote:
Edit: Bondings posted while I was typing this... Ah, that's a good point Bondings since summation notation only applies for a whole number of terms (x must be a whole number).

Bingo. And that where the problem is. There is nothing wrong with the summation, the summation just gives the clue as to where the real problem is.

nilsmo wrote:
Arnie is correct. To be more specific, here is the fallacy.

Indi wrote:
That's just good old term-by-term differentiation


The amount of terms varies in

Code:
  x
  ∑  x
i = 1


OK, that should be enough.

Nope, sorry.

nilsmo wrote:
The addition rule of derivatives says that (f+g)' = f'+g'. This is true. One is tempted to say therefore that derivative of (f_1+f_2+f_3+...f_x) equals the sum of the derivatives of all the functions inside. But this is not true because x is not constant. The amount of terms varies.

In other words, the derivative of a sum of functions is equal to the sum of the derivatives of each function but the derivative of a sum of different sets of functions depending on x is not equal to the sum of the derivatives.

I don't really know how to explain this. It really is extremely simple. I haven't even taken calculus - but catching the error is not hard.

Nope, sorry. It is perfectly legal to differentiate a varying number of terms. The problem is not the varying number of terms, it is that the number can only be natural numbers. A continuous summation - an integration - solves the problem.

nilsmo wrote:
Another way to explain the fallacy is by the product rule I suppose. It says the derivative of the product of two functions is not always equal to the product of the derivatives of the two functions.

Code:

    g(x)
     ∑  f(x) = g(x)*f(x)
   i = 1

Derivative of g*f != Derivative of g * Derivative f as Indi tried tricking us into thinking.

i tricked you in no such way. What i did was:

f = g
hf = hg
(hf)' = (hg)'

where f, g and h are all functions of x.

And that is perfectly valid.

Bondings wrote:
@Indi, x is an integer and not a real number. This means that the function is not continuous and hence not differentiable, which means you aren't allowed to take the derivative of both sides of the equation.

Correct.

Everyone always focuses on the summation. That's good because the summation is the clue to where the problem is. Unfortunately, it is very, very rare for someone to look any further - they almost always get hung up on the mechanics of differentiating a summation with a varying number of terms. But while the summation is the clue to the problem, it is not the problem itself.

For those who have been pondering this, there is nothing illegal about differentiating a summation. There is nothing invalid about the way i did it (term by term). There is also nothing illegal about differentiating over a summation with a varying number of terms - it's done all the time.

The problem is not the summation. The problem is in the first line of the proof. If you look here, the problem is right here:

x ∈ ℕ⋆

The equation i gave in the first line is valid only when x is a natural number. That means that it is not a continuous function of x. That means that it cannot be differentiated with respect to x. That means the differentiation - while it was performed correctly - was meaningless.

If, in fact, i had used an integration instead of a summation, x would not have to be a natural number, it could be any real number. That would mean that it would be a continuous function of x. That would mean that it could be differentiated with respect to x. And it would work out.

You can see for yourself soon - i will make another page showing how it works fine with an integration.
ThreeRight
Indi wrote:

i am not sure what order of operations you are referring to, but you are over thinking the problem. Functions can be equations and equations can be functions... equations can be made up of functions and functions can be made up of equations. Really they're both the same thing looked at a different way. Equalities are just a special case of equations. y = x + 3 is an equation that is also an equality that contains any number of functions (depending on how you want to count them). You could say f(x) := x + 3 and that y = f(x). Or you can say g(x) := x and h(x) := 3 and y = g(x) + h(x). It's all the same. Nothing changes except your perspective.


LOL! i learned how to find a function by placing a ruler on a graph vertically and seeing if more then 1 parts of the line touch the ruler.
carlospro7
DoctorBeaver wrote:
In binary notation 1+1=10 Laughing


haha nice one. (10 base 2) as in (2 base 10)
DjinniFire
Indi wrote:

It is perfectly legal to do what i did (with one caveat). A summation is the same thing as an integration (see here), and there's nothing wrong with differentiating an integration term.

Arnie wrote:
Yes, that's what I meant with:
Quote:
It's likely that the x "on top" has to be taken into account as well because it determines how many times the x "on the right" has to be added up. That basically makes it a factor.

Nope, sorry. As i explained above, there is nothing wrong with having the x on top of a summation - it's the same as having the x on top of an integration, which is perfectly legal.



Ok ignoring that there is a fallacy in the beginning because you have natural numbers and is not continuous. You are wrong in what you just said.

IT DOES Make a difference if X is on top. Notice what they said before such as Arnie which I kept in the quotes. "how many times the x "on the right" has to be added up". This is a factor. It is not wrong to take the derivative of a summation but the way derivative that you got was incorrect because of the x. You can't possibly believe that the derivative of the summation from i=1 to x of x is the the summation from i=1 to x of 1. Because that is x(1) which is NOT the same as 2x. Because you have that error you for sure know that the derivative was wrong.
nilsmo
Indi wrote:
nilsmo wrote:
Arnie is correct. To be more specific, here is the fallacy.

Indi wrote:
That's just good old term-by-term differentiation


The amount of terms varies in

Code:
  x
  ∑  x
i = 1


OK, that should be enough.

Nope, sorry.


I believe we saw different fallacies. I applied the addition rule to show the derivative could not be the right derivative, while you just showed you can not take the derivative. I guess your/Bonding's explanation are "better" since they are a bit shorter. My way of thinking was indirect - I assumed your step was correct and found it was incorrect.

I thought "Let's say the step is right... What in the world was your leap of logic in saying the derivative of x summed together x times is derivative of x summed together x times?" The only sane response I found is "because of the addition rule".. And the addition rule doesn't apply when you sum varying sets of functions.

This brings up the question: what was the justification in that the derivative of ( x summed together x times ) is 1 summed together x times? Was it just randomly saying a derivative?
DjinniFire
nilsmo wrote:
This brings up the question: what was the justification in that the derivative of ( x summed together x times ) is 1 summed together x times? Was it just randomly saying a derivative?


There is no justification for saying the derivative of x summed together x times is 1 summed together x times. It was an error in simplifying the derivative. He assumed the rule in which you take the derivative of a summation is the same as taking the derivative of the term then summing them up would work, without realizing that the x above the sigma actually has an impact. The only way I can think of in determining the derivative of this would be to actually compute the summation then take the derivative which would mean derivative of x^2.
LeatherRose
the answer to your "1+1=2?" question technicly can not be answered with out specific details.

numbers can mean more than one thing as quoted by a great amount of people above. so i won't repeat it all..
i will save myself the typing time
Laughing
but i will state a few things..



tempdbs wrote:


You can define in other way -

Answer this -


1 cup of Water + 1 cup of Sugar = ?? ( 2 Cup of Sugarwater or 1 Cup of Sugar water)




1 cup of water + 1 cup of sugar = 1 cup of sugar water because the sugar would eventually disolve into the water making it just 1 sweet sugar water...

but that answer is only assuming that a measuring cup was used..and not just a random drinking glass...and mixed the two together


but if you mean like
sally mae had one pen for school that she brought from home and then her teacher gave every body including her in her class a pen. How many pens would sally mae have?
well obviously it is 2 pens.

or
if you are state x+x=y
x=x
and there are 2 x's
so 2x=y
and if x is 1
2 times 1 = 2

and if

2x=y
divide 2 and 2x
and you end up with just x
but if you do that you have to divide y by 2 also
so you end up with
x=y/2
and if x=1 and y=2
then it is
1=2/2
and that means
1=1

unless
2x is the meaning of 1 times 1
then it would be
1 times 1 = 1
and
1 is not = to 2

although..if mathimatics was just an example and you are looking for a loop hole in life..
like..when your heart stops..you die..
when you sneeze your heart stops for a second
heavan and hell...eturnal life??
so on and so forth..


so actually and technicly speaking..the answer is most likely infinate..
there are to many answers fo 1+1 to equal just one thing...

so i say if you want answers search for them..
but be prepared to search forever..
chasbeen
When adding one and one how can you be sure the total comes to exactly two? It depends what you are trying to put together. If your adding a new cow to your single cow collection then you can say I have exactly 2 cows. If you have a pint of water in a bucket and you add another pint of water to it how can you be sure there is EXACTLY two pints of water in the bucket?
What about the cows in your cow collection. How do you know one of them isn't pregnant?
Mathematics attempt to model real life which is the cause of the confusion. Rolling Eyes
Arnie
1 pint of water in that notation means it can be 0.5 to 1.5 pint. If you write down 1.0 it can be 0.95 to 1.05. The more digits, the more accurate you've measured it.
Indi
DjinniFire wrote:
nilsmo wrote:
This brings up the question: what was the justification in that the derivative of ( x summed together x times ) is 1 summed together x times? Was it just randomly saying a derivative?


There is no justification for saying the derivative of x summed together x times is 1 summed together x times. It was an error in simplifying the derivative. He assumed the rule in which you take the derivative of a summation is the same as taking the derivative of the term then summing them up would work, without realizing that the x above the sigma actually has an impact.

Incorrect. ^_^

The function:
Code:
         x
f(x) =   ∑ x, x ∈ ℕ⋆
       i = 1


Is just a short hand way of saying:
Code:
       ┌ x, x = 1
       │ x + x, x = 2
f(x) = │ x + x + x, x = 3
       │
       └ x + + x (N times), x = N


Can that be differentiated? No. Why not? Because it is not continuous anywhere.

What if we made it partially continuous - aka, piecewise continuous? Then you'd get this:
Code:
       ┌ x, 0 > x ≥ 1
       │ x + x, 1 > x ≥ 2
f(x) = │ x + x + x, 2 > x ≥ 3
       │
       └ x + + x (N times), N - 1 > x ≥ N


Can that be differentiated? Yes. But the derivative is only defined for the places where it is continuous (ie, not at the breaks). And what would happen if we differentiated that? We'd get:
Code:
        ┌ 1, 0 > x > 1
        │ 1 + 1, 1 > x > 2
f'(x) = │ 1 + 1 + 1, 2 > x > 3
        │
        └ 1 + + 1 (N times), N - 1 > x > N

Which is perfectly legal. (But, that means that we can't write f with the summation shorthand anymore.)

The trick in the false proof was that the function is not differentiable because it is discontinuous everywhere. i did the differentiation "correctly" in that if the function were continuous, it was correctly differentiated (as in the bit just above). But the entire differentiation was undefined from the start, so even though it was done "correctly", it is still invalid.

DjinniFire wrote:
The only way I can think of in determining the derivative of this would be to actually compute the summation then take the derivative which would mean derivative of x^2.

NO! Goodness, no! >_<

Alright, suppose you took the original summation:
Code:
         x
f(x) =   ∑ x, x ∈ ℕ⋆
       i = 1


Then you simplified it to get:
Code:
f(x) = x, x ∈ ℕ⋆


YOU STILL CANNOT DIFFERENTIATE THIS FUNCTION!!!

And the reason why you can't is that little bit at the end there.

And that was the entire point.
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