Proof: 1 = 2 (no division by zero!)

[DaBigCheez]

So, I came up with (or heard while not really listening and subconsciously stole, one of the two, but I think I came up with) a fun little proof that 1 = 2. Obviously, it doesn't work - disproving an axiom is kinda hard - but I was wondering if others would come to the same conclusion as to why it doesn't that I did.

This is only for real numbers, I don't feel like dealing with imaginaries.



1. Let's start with something everyone can agree on:
x^2 = x^2

2. This is equivalent, by the definition of squaring, to:
x^2 = x*x

3. This is equivalent, by the definition of multiplication, to:
x^2 = (x + x + x + ...) x times.

4. As these are the same equation, their derivative must be the same. Take the derivative of both sides:
d(x^2)/dx = d((x + x + x + ...) x times)/dx
2x = (1 + 1 + 1 + ...) x times.

5. Condense the right side:
2x = x

6. Divide through by x:
2 = 1



Step 3 looks a little odd at first, but just think about it a bit. It works for any number; positive, negative, decimal, irrational. Not sure about imaginary numbers, which is why I restricted this to reals.

Note that the ONLY division in this entire proof is in step 6, when we divide through by x. x is not by definition zero (this works for any real x), so it's not a divide by zero problem. (The /dx in step four denotes differentiation, not division.)

So, what do YOU all think about this proof, and what has to be wrong with it? I know what I think, I'm just curious if others can come up with a more subtle (or more obvious) reason it can't work.

Enjoy!

 

[perfectimo]

2. This is equivalent, by the definition of squaring, to:
x^2 = x*x

That is why it doesn't work, I understand algebra to a degree and one of the first rules that you must do the same thing to both sides.

 

[DaBigCheez]

EDIT: Whoops, double-quote

2. This is equivalent, by the definition of squaring, to:
x^2 = x*x

That is why it doesn't work, I understand algebra to a degree and one of the first rules that you must do the same thing to both sides.

That isn't an algebraic manipulation, it's just the definition of what squaring something does. By definition, x^2 is equal to x times x. (I'm not "doing anything" to either side.)

 

[Anarchemitis]

Wouldn't it be x[sup]2[/sup]=y[sup]2[/sup]?

 

[black lincon]

He wrote it wrong but he's right about why one doesn't equal two(or i think he is its a wee bit hard to comprehend), its the same reason you aren't allowed to use infinity in math(not really but their sorta similar i guess).

 

[klakkat]

The problem is with the derivative. the expansion is fine, but by the chain rule you have an extra factor of x you didn't account for; that handwaved in 'x times' is still a function of x.

 

[Gene O]

I think I see where x necessarily becomes zero.

x^2 does equal x + x + x ... + x, but only for one specific value of x. 3^2 = 3 + 3 + 3 where x = 3, but if you try to evaluate this equation over an interval of x, you'll be getting 1^2 = 1 +1 + 1 and 2^2 = 2 + 2 + 2. It doesn't work. Does that make sense?

 

[TKTom]

1. Let's start with something everyone can agree on:
x^2 = x^2

2. This is equivalent, by the definition of squaring, to:
x^2 = x*x

3. This is equivalent, by the definition of multiplication, to:
x^2 = (x + x + x + ...) x times.

4. As these are the same equation, their derivative must be the same. Take the derivative of both sides:
d(x^2)/dx = d((x + x + x + ...) x times)/dx
2x = (1 + 1 + 1 + ...) x times.

5. Condense the right side:
2x = x

6. Divide through by x:
2 = 1!

Your derivitive is incorrect. You have the equivalent of the sum using index i=1,...,x of (x). Your x is variable and dependant on your inner term so this sum is invalid, the problems with this invalidity arise in the derivation. If you were to use the definitions of derivation on that equation you would see the issue.

 

[perfectimo]

EDIT: Whoops, double-quote

2. This is equivalent, by the definition of squaring, to:
x^2 = x*x

That is why it doesn't work, I understand algebra to a degree and one of the first rules that you must do the same thing to both sides.

That isn't an algebraic manipulation, it's just the definition of what squaring something does. By definition, x^2 is equal to x times x. (I'm not "doing anything" to either side.)

Okay then, it just looked like you were only expanding one side, my eyes must be playing tricks on me. As I said though I have a basic understanding of algebra and that was all I saw different from how I used to set out algebratic equations. I know nothing about derivatives.

 

[Anarchemitis]

Guess what I learned in Math today?

• Log(S[sup]n[/sup])=(a/(1-r[sup]n[/sup]))(1-r)

Eg: Sum of Functions (S=/=Whole number, rounded to Asymptote)
5, 2.5, 1.25, 0.625, 0.3125...

S=(5/(1-(1/2)[sup]n[/sup])=Asymptote

Used for finding quadratic graphable functions from sets of organized numbers that increase or decrease, and finding the total sum thereof.

 

[jboking]

ok, so i just started calc. this year and im trying to comprehend everything you wrote.

i first thought that when you said d/dx((x+x+x...)x times) i thought "X times" referred to the number of "+x"s there were. if this was true there is no x on the outside and (x+x+x) becomes (1+1+1) which would end up with 2x=3 dividing by x would give you 2=3/x.

now assuming that the "x times" actually ment the (x+x+x...)x then when you take the derivative of that you have to use the product rule which resolves to be (X+X+X)+X(1+1+1)=2x
dividing by x you would be left with 1+1=2.

Like i said i just started calc i may be doing something wrong, but whatever.

 

[TKTom]

Like i said i just started calc i may be doing something wrong, but whatever.

In a way you are correct. This is the issue, but that x+x+...+x is not (x+x+...+x)*x it is the sum of x xs (ecks eckses). You see?

If I could write the notation on this forum this would be a lot clearer to most people.

 

[jboking]

Like i said i just started calc i may be doing something wrong, but whatever.

In a way you are correct. This is the issue, but that x+x+...+x is not (x+x+...+x)*x it is the sum of x xs (ecks eckses). You see?

If I could write the notation on this forum this would be a lot clearer to most people.

i sort of see. Hum, too bad you can't write the notation on this forum, it would make a lot more sense to me.

 

[DaBigCheez]

I think I see where x necessarily becomes zero.

x^2 does equal x + x + x ... + x, but only for one specific value of x. 3^2 = 3 + 3 + 3 where x = 3, but if you try to evaluate this equation over an interval of x, you'll be getting 1^2 = 1 +1 + 1 and 2^2 = 2 + 2 + 2. It doesn't work. Does that make sense?

Yeah, that's what I was pretty sure the issue was, I was just curious if others could see any other reason or if they could state it more clearly; the right-hand function is itself a function of x, so it's a different function at every x (even though, if you evaluate at any given x, it works, you just can't extrapolate).

Note to others: I wrote the notation a little wrong, but (x + x + x + ...) implies a series, not specifically 3x. I should have done (x + x + x ... + x), my bad. The (x + x + x + ...) x times would mean that, say, if x = 4, it's (4 + 4 + 4 + 4), if x = 2 it's (2 + 2), etc.; it is, as you said, the number of x inside the parentheses. If I could put a sigma in there, it'd be a little easier; index function, you said, tktom?

 

[jim_doki]

*head explodes*

I gotta stop coming into these math threads

 

[klakkat]

Like i said i just started calc i may be doing something wrong, but whatever.

In a way you are correct. This is the issue, but that x+x+...+x is not (x+x+...+x)*x it is the sum of x xs (ecks eckses). You see?

If I could write the notation on this forum this would be a lot clearer to most people.

The sum notation is the easiest way to see that (x+x+...x) still has an x dependence in the number of terms, which was mentioned earlier in this thread. As pointed out, the sum has a hidden x dependence, which has a nontrivial derivative; it's invalid to just take the derivative of the terms in the sum without also taking the derivative of the growth function of the sum, which is a function of x. Doing so would result in getting the derivative 2x again, exactly what you would expect.

Sidenote: derivatives of sums with respect to a variable in the sum's limits are a pain. I recommend series expansions that terminate at some finite point not dependent on the variables involved instead.

 

[TKTom]

Yeah, that's what I was pretty sure the issue was, I was just curious if others could see any other reason or if they could state it more clearly; the right-hand function is itself a function of x, so it's a different function at every x (even though, if you evaluate at any given x, it works, you just can't extrapolate).

If I could put a sigma in there, it'd be a little easier; index function, you said, tktom?

Yeah, I've been banging on about the index. Try working it out for youself now on a bit of paper using sigma notation:

Calculate (using the definition of derivative being limit as h tends to zero of (f(x+h)-f(x))/h) ) the derivative of the sum of x xs. And you will see that you are trying to exact a continuous limit on something that is discrete, which doesn't work.

(Wait ignore that last bit, this can be done if you exact the limit correctly and you can find both xs are there as they should be.)

This limit is as h tends to zero (sum(i=1:a)[f(x)] means the sum of index over i=1,...,a of f(x) );

limit{(sum(i=1:x+h)[x+h] - sum(i=1:x)[x])/h}

Theres some cancellation going on here, so this can become:

limit{(sum(i=1:h)[x+h] +sum(i=1:x)[h])/h}

The splitting the numerator and cancelling the h in the right side summation leaves a term that is not dependant on h, so I will remove it form the limitand the summation on the left is the same as h*(x+h), so this is:

limit{(h*(x+h) )/h} + sum(i=1:x)[1]

Which will give:

x + x=2x

Which is the same as the derivative of x^2.

 

[klakkat]

Nice. I didn't feel like doing that one explicitly.

 

[SilentHunter7]

Guess what I learned in Math today?

• Log(S[sup]n[/sup])=(a/(1-r[sup]n[/sup]))(1-r)

Eg: Sum of Functions (S=/=Whole number, rounded to Asymptote)
5, 2.5, 1.25, 0.625, 0.3125...

S=(5/(1-(1/2)[sup]n[/sup])=Asymptote

Used for finding quadratic graphable functions from sets of organized numbers that increase or decrease, and finding the total sum thereof.

I'm going back to Calc II, where it's safe...

 

[klakkat]

Nah. Become a Math major. Complex analysis is your friend!*

*I am not actually a math major. Close to it, though