How is my 2 = 1 equation wrong?

[ShawnD1]

I was reading on a joke site how to prove that 2 = 1. After looking at the formula and picking through it, I found the divide by zero error then changed the formula does there is no divide by zero error.

first off, a ---> b
please keep in mind that when I put =, it doesn't mean that they are equal but it means they are pretty damn close to equal. It's like when you do derivatives the long way (remember [f(x) - f(a)]/(x - a))

a - b = 0
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b) * (a - b) = b * (a - b)
a + b = b
b + b = b
2b = b
2 = 1

What's wrong?

 

[Shalmanese]

from step 4 -> step 5, your dividing by zero.

 

[ShawnD1]

nope. since a and b are not exactly equal, it is not a divide by 0 error.

 

[Venix]

please keep in mind that when I put =, it doesn't mean that they are equal but it means they are pretty damn close to equal. It's like when you do derivatives the long way (remember [f(x) - f(a)]/(x - a))

....what?

 

[ShawnD1]

derivatives the long way:

derivative of x^2 at x = 5
a --> x
(x^2 - a^2) / (x - a)
= (x + a) * (x - a) / (x - a)
= x + a
= x + x
= 2x
= 2(5)
= 10

On your calculator, input x^2 then have it calculate dy/dx at x = 5. You will get 10.

Notice that both sides are divided by x - a. Since the limit of a is x but the actual value of a does not quite equal x, both sides can be divided by x - a without having a divide by 0 error

Doing derivatives the long way is just like my whacky formula in that both sides are divided by a number almost equal to 0. Since both sides are not divided by exactly 0, the format is still correct (to at least some degree).

 

[DrPizza]

Originally posted by: Shalmanese
from step 4 -> step 5, your dividing by zero.

You mean from line 5 to line 6 when he cancels (a-b) and doesn't explain how he does it? (by dividing both sides by a-b, which = 0)

 

[DrPizza]

Originally posted by: ShawnD1
nope. since a and b are not exactly equal, it is not a divide by 0 error.

if you state that a - b = 0, then a exactly = b. Period. End of discussion.

 

[ShawnD1]

So you're saying that the current way we do derivates is wrong as well? Does this mean our entire system of math is wrong?

 

[PowerEngineer]

a - b = X
a = b + X
a^2 = ab + aX
a^2 - b^2 = ab - b^2 + aX
(a + b) * (a - b) = b * (a - b) + aX
a + b = b + aX/(a-b) <----- a can not equal b
b + X + b = b + aX/(a-b)
2b + X = b + aX/(a-b)
2b = b + aX/(a-b) + X
2 = 1 + aX/(ab-b^2) + X/b <------ B can not equal 0

choose your value of X...

 

[DrPizza]

Originally posted by: ShawnD1
So you're saying that the current way we do derivates is wrong as well? Does this mean our entire system of math is wrong?

what on earth are you talking about?

The definition of a derivative is the limit as h approaches 0 of f(x+h) - f(x) all over h.

What does that have to do with the problem? The definition of the derivative is derived from finding the slope of a secant line to the curve passing through the points (x,f(x)) and (x+h, f(x+h)). As the two points get closer together, the slope of the secant line approaches the slope of the tangent line. The slope of the tangent line is the limit of the slope of the secant line as the distance between x and x+h gets smaller, thus, the limit as h ->0.

The relation between your screwy proof and your statement about derivatives is about as clear as the relationship between the price of tea in China and the score of Monday Night Football.

[but, I will admit, I love challenging students with that proof you posted, and having them find the fault. More often than not, I run into a math classroom right before the late bell rings, scribble that (or one of several others) proof on the board, and run out without saying so much as a word.... it leaves the other math teacher and students bewildered sometimes]

 

[DrPizza]

My apologies..... I see what you're getting at....

the derivative "the long way" in calculus of x^2 is (without being lazy and not writing important parts):

f'(x) = lim as h ->0 ( f(x+h) - f(x) )/h
= lim as h ->0 ((x+h)^2 - x^2)/h
= lim as h ->0 ((x^2 +2xh +h^2 - x^2)/h
= lim as h ->0 (2xh +h^2)/h
= lim as h ->0 (2x + h)
= 2x

f'(x)=2x
f'(5)=2*5 = 10

Note: I'm not going to consider for more than a moment, and that moment's already passed, proving the limit using sigma epsilon notation!

But, I can see what you mean. However, in the derivative, it's a limit, and cancelling factors is a proven method for finding the limits. Nonetheless, that doesn't mean that in algebra that division by zero is allowed, nor does it allow you to say a-b = 0 and a!=b. In terms of limits, if b is approaching the value of a, then a-b is approaching zero.

 

[ShawnD1]

What I was trying to show with the long way of doing derivatives is that you can divide each side by almost zero (h --> 0 or a --> x). Since the limit of a is b, the difference of the two is almost 0. Since the difference is not 0, I can divide by that difference without breaking any rules.

 

[CTho9305]

Originally posted by: ShawnD1
What I was trying to show with the long way of doing derivatives is that you can divide each side by almost zero (h --> 0 or a --> x). Since the limit of a is b, the difference of the two is almost 0. Since the difference is not 0, I can divide by that difference without breaking any rules.

That isn't how limits work. Try doing what you're doing with the function sin(x)/x (or any similar function with x on the bottom).
The limit of a is not b, the limit of a constant is itself. And if the difference is almost 0, then a != b, and your initial assumption is wrong.

 

[ShawnD1]

The limit of sin(x)/x where x approaches 0 is 1 (http://www.ies.co.jp/math/java/calc/LimSinX/LimSinX.html). The problem with saying that 'a' is a constant is that 'a' is not a constant. The point of the relation was to show that any number is equal to double that number. Both 'a' and 'b' are varibles.


I'll admit that the initial assumption is wrong but I can easily fix it and we end up in the same mess we started with. I don't remember the symbol for almost equal to so I'll just use ~

a ~ b
a^2 ~ ab
a^2 - b^2 ~ ab - b^2
(a + b) * (a - b) ~ b * (a - b)
a + b ~ b
b + b ~ b
2b ~ b
2 ~ 1

 

[CTho9305]

Originally posted by: ShawnD1
The limit of sin(x)/x where x approaches 0 is 1 (http://www.ies.co.jp/math/java/calc/LimSinX/LimSinX.html). The problem with saying that 'a' is a constant is that 'a' is not a constant. The point of the relation was to show that any number is equal to double that number. Both 'a' and 'b' are varibles.

Sure, but they aren't going to be functions... run the proof using 2 and 1. Note that they are constants.

 

[ShawnD1]

Originally posted by: CTho9305

Originally posted by: ShawnD1Sure, but they aren't going to be functions... run the proof using 2 and 1. Note that they are constants.

That's the problem with calculus. When doing problems with limits you MUST do the algebra before you start filling in values. Look at the long way to derive stuff, let's use the function x^2
h --> 0
(x + h)^2 - x^2 / h

If I immediately start filling in values, I end up with x^2 - x^2 / 0. That means I get 0/0. If you do the algebra of the formula first then fill in values later, it turns into 2x. Remember, with calculus you have to do the algebra first.

 

[AEB]

thing about it is equals means equals. proving that a number may act like another number doesnt actually prove anything. Try proving it using induction, your universe would hae to be REAL numbers for your example to work(div 0 problem)

 

[Peter]

What's wrong is that with your = meaning "is almost equal", all the rules of solving EQUATIONS do not apply. Your calculus steps are all invalid, then. And the result then of course is nonsense.

 

[uart]

Call the "near zero" term z ok, then this is what you get.

a - b  =  z
a  =  b + z
a^2 = ab  + az 
a^2  -  b^2  =  ab +  az -  b^2
(a + b) * (a - b)  =  b * (a - b)  + az
(a + b)  =  b + a z / (a - b)
(a + b) =  b + a z/z
(a + b) =( b + a)
1    =   1

 

[rjain]

The limit we use when we calculate derivatives approaches 0/0 for a continuous region of a function. The derivative is the ratio between how FAST the numerator (function value) and denominator (input value) approach to 0.

 

[rgwalt]

Think about it like this... We all know 2 does not equal 1. If you have 2 objects, you don't have just 1 object, or almost one object. Integers are finite representations. Thus, there is either an error in your proof, or a fundamental error in our system of mathematics. I have a feeling the error is in your work. Even if you make the argument that a is almost equal to b, but not quite, then replacing a with b in step 7 would not be valid.

R

 

[blahblah99]

You can't just "redefine" the equal sign to be "close to equal". It doesn't work that way. If in step two, you said a = b, then dividing by (a-b) is undefined, but if you take the limit as (a-b) approaches 0, then you get INFINITY, or negative infinity. PERIOD.


If you still insist on 2 = 1, then lets try an experiment. I give you 1 dollar, you give me two. Repeat 1000000 times.


Only engineers can assume 2 ~ 1.

 

[Chaotic42]

This:

(a + b) * (a - b) = b * (a - b)
a + b = b

Is division by zero, since you defined (a-b) as zero in your first statement.

Edit: Ah, I see what you are trying to do. I didn't know what a----->b was supposed to mean.

 

[PrinceXizor]

You can't use substitution and transformation on inequalities, just as you can't divide by zero. That's why it doesn't work. And, in calculus, you don't "do the algebra first" and then your calcululs. The assumptions on the nature of the f(x) you are working with are stated ahead of time.

P-X

 

ok, both are fine ways to get derivatives, fact of the matter is that Lim[a->x] of (f(a)-f(x))/(a-X) is the same as the Lim[H->0] of ((f(x+h)-f(x))/h) h is just equal to a-x so both methods are just fine. Try anything you want, it gets more obvoius the more you do it.

lim[b->a] of a-b =
lim[b->a] of a*a-b*a =
lim[b->a] of a*a-b*b+b*b-b*a =
lim[b->a] of (a+b)(a-b)-b(a-b) =
lim[b->a] of (a-b)(a)

and we see clearly that it is 0
with limits there are no two sides to the equasion, that is where your problem is.