Interesting Paradox

On an off topic note here is another math paradox I love.

It's a false calculus proof, so obviously I make an error at some point. See if you can spot it.

assume x > 0
x^2=x^2
x^2 = x + x + x + x + x + x + ... + x (x instances of the variable x)
d/dx [x^2] = d/dx [x + x + x + x + x + x + ... + x]
2x = 1 + 1 + 1 + 1 + 1 + ... + 1 (x instances of the number 1)
2x = x
2 = 1

The last division can be done because we assume x > 0 from the start. Consider this as differentiating the function f(x)=x and g(x)=x^2 over any interval of positive numbers.

Where's the error? Obviously 2 does not equal 1.

;-)

Chris
 
To: cmhardw

Our definition of x assumed that x was an integer; this equation is not meaningful for non-integer real numbers. Functions are only differentiable on a continuous space such as the real numbers, not on integers. For any particular integer x, you get a true equation. But to differentiate both sides you need an equation of functions, not an equation of integers.

Also, when taking the derivative in line 5 the derivative is taken with respect to each of the terms individually, but not with respect to the numbers of terms. This is erroneous, as the number of terms is x, the variable of differentiation. The chain rule is incorrectly not applied on the right-hand side of the equation.

Is this the error?
 
Thanks...

Anyone want to take a crack at this paradox?

Proof that 0 = 1

Start with the addition of an infinite succession of zeros
0=0+0+0+...

Then recognize that 0=1-1
0=(1-1)+(1-1)+(1-1)+...

Applying the associative law of addition results in
0=1+(-1+1)+(-1+1)+(-1+1)+...

Of course -1+1=0
0=1+0+0+0+...

And the addition of an infinite string of zeros can be discarded leaving
0=1
 
You can't move the bracets this way. If we have a finite amount of numbers, the last digits would look like this:
... + (-1 + 1) + (-1 + 1) - 1 = 0
or
... + (-1 + 1) + (-1 + 1) + (-1 = error ;)

I'm just guessing now: If something doesn't work for a finite amount of numbers, it won't work for infinite ones.
 
Excellent answer cin.

As for watermelon's paradox:

I don't think that there is a way to assign either of the statements a consistent truth value.
 
Hey Mahajarama,

Our definition of x assumed that x was an integer; this equation is not meaningful for non-integer real numbers. Functions are only differentiable on a continuous space such as the real numbers, not on integers. For any particular integer x, you get a true equation. But to differentiate both sides you need an equation of functions, not an equation of integers.
Click to expand...

Yes the error I intended was that the identity in line 3 does not hold for any interval of real numbers, and only holds for the natural numbers.

As far as differentiating I meant for that part to be accurate, in the "false proof" sense at least. I applied the linearity property to the right hand side in line 5.

d/dx [x+x+x+x+x+...+x] for x instances of x I applied as

d/dx (x) + d/dx (x) + d/dx (x) + d/dx (x) + ... + d/dx (x) = 1 + 1 + 1 + 1 + 1 + ... + 1

Obviously this doesn't work though because I cannot have x instances of the variable over a continuous function as you pointed out (let x=pi or x=e). So yes I agree the right hand side is not differentiable, but I meant to apply the derivative correctly in the "false proof" sense by applying the linearity property of the derivative.

The "proof" is completely 100% eroneous, but I have found most of the time calculus students do not catch the error and are irked by the fact that they can't see what has gone wrong. That's why I like this false proof, it really tests one's knowledge of the basic definitions of calculus and is a great riddle/puzzle for a first or second semester calculus student, in my opinion.
 
I have an interesting question/paradox that I think about sometimes. I haven't thought about it too hard, so I don't know if I've come to the correct answer, but I find it interesting anyway.

Assume you live in a 10 story building, and you live on the 8th floor. If you get on the elevator with someone the probability that that person reaches their floor before you is 6/9. The probability you live on the same floor is 1/9 and the probability you get off before them is 2/9. This assumes that everyone living on the 2nd-10th floors rides the elevator anytime they want to get to their floor of course.

The interesting part is if you take the stairs. You can view it the same way. If you start up the stairs with someone, the probability they reach their floor before you is 6/9. The probability you reach the same floor is 1/9 and the probability you reach your floor before them is 2/9.

But what if you think of it like this. Going up the stairwell, upon reaching the 2nd floor there is a 50-50 chance the person either continues up the stairs with you, or stops at their floor.

So the probability the person lives on the 2nd floor is 1/2. Say they continue with you. Upon reaching the third floor the probability that they continue with you is 1/2 and the probability that they stop at that floor is 1/2. So the probability that they live on the 3rd floor is 1/4. The probability that they live on a higher floor is also 1/4.

Continue like this and the chance that the person lives on a floor below you is
1/2+1/4+1/8+1/16+1/32+1/64 = 63/64

The chance they live on the same floor as you is (1/2)^8 = 1/128
The chance they live on a higher floor is also 1/128.

The way I resolve this paradox is that in the first case, you know you live in a 10 story building. In the second case let's assume that if you consider the chance 50-50 that upon reaching the second floor the person can continue on with you or stop at that floor that you have no idea how many floors there are beyond 8 floors (since your destination is the 8th floor).

I don't know if this is the correct solution. Again I haven't tried to really formalize this paradox or to spot my error in reasoning in using the 50-50 chance at each floor approach, but I find it interesting.

Anyone have any thoughts on this?

Chris
 
I can't remember if there is a different way of numbering for floors, I think there is. The way I numbered was this. The ground floor, that you walk into from outside, is the 1st floor. The floor above that is the 2nd, etc.

If there is a different naming convention the one I am using I guess is American. The ground floor (street level) I am calling the 1st floor.

Chris
 
I believe that the assumption that a person gets off at each floor is wrong. It cannot be 50-50... "He either picks it or he doesn't" does not mean 50-50...

Statistically speaking, the way you are depicting it is a Bernoulli trial (yes, that info is correct, I took statistics last year)... But p here does not equal 0.5.
 
ExoCorsair said:
I believe that the assumption that a person gets off at each floor is wrong. It cannot be 50-50... "He either picks it or he doesn't" does not mean 50-50...
Click to expand...
Exactly! Every floor (2-10) has the same probability, 1/9. Just ignore the fuzzy 50-50 logic and the paradox becomes clear.
 
You must log in or register to reply here.

Similar threads

Renslay
Shortest or interesting identical moves?
Replies
8
Views
210
acrystal
acrystal
someone.
Is any cuber interested in chess?
Replies
3
Views
252
Fred
Fred
artless1der
Interesting new treatment process to make cubes AWESOME
Replies
0
Views
203
artless1der
artless1der
M
Duckies ultimate and interesting and fascinating or not very good progress stories
2
Replies
33
Views
1,965
Mr.Ducky
M
Dan the Beginner
Interesting Lockup Test
Replies
3
Views
471
Dan the Beginner
Dan the Beginner
Back
Top