Is there a better false proof for 2=1?

Question

The (FALSE) proof that 1 = 2 goes as follows.
Let a = 1 and b = 1.
Then
a = b
If we square both sides of that equation, we get
a^2 = b^2
and if we multiply both sides of that same equation by a, we get
a^2 = ab
So we have these two equations:
a^2 = b^2
a^2 = ab
Let's rearrange both of these, to:
a^2 - b^2 = 0
a^2 - ab = 0
Since both of these equate to 0, they equate to each other.
a^2 - b^2 = a^2 - ab
Factor both sides, to obtain
(a - b) (a + b) = a(a - b)
Now that we have (a - b) on both sides, they effectively cancel out because we divide both sides by (a - b). As a result, we get
a + b = a
We initially assigned a = b = 1; therefore
1 + 1 = 1, and
2 = 1
Here, the value of "a' remained consistent throughout.
How can you divide both sides by zero which is (a-b) and get numbers appearing after that? Any better versions that can trick people?

O i quoted that from a user from a previous question.

what the park is Pero yo no se. Pero conaso este es una pregunta mas rado. Por que tu tienes este pregunta, viejo?:?

Answer 1

Here's one I made up a long time ago:

Let's say you want to solve the equation x^x^x^x*... = 2. (In other words, the left side is "x to the x to the x ...", etc., with infinite exponents.) The way to solve it is to say the exponent must also be 2, therefore x^2 = 2, or x = sqrt(2). The hard part is convincing some folks that this solving technique really is valid, but it is - calculate sqrt(2)^sqrt(2) on your calculator and keep repeating the exponent, and you will find this really does converge to the number 2.

OK, now let's change the problem slightly: x^x^x^x... = 4. Using the same technique, we get x^4 = 4, or x = the fourth root of 4. However, the fourth root of 4 is sqrt(2), i.e. the same answer we got for the first problem. Therefore, 2 = 4 ??? See if your mathematically-inclined buddy can explain that one... :)

* * * * *

OK, looking at the bogus proof for the poster ahead of me, his trick is in the last step: taking the square root in this case is invalid because one side is the square of a negative number, while the other side is the square of a positive number. Cute, though...

Answer 2

That was my proof.

Because, if a = 1 and b = 1, and

a + b = a

Then, replace each occurrance of a with 1, and each occurrance of b with 1. This will get you

1 + 1 = 1

2 = 1

Edit: To CATALISTA, who claimed I performed an incorrect step.

Suppose a = b. This BRANCHES off into two equations:

a = b. Square both sides, and you get
a^2 = b^2.

a = b. Multiply both sides by a, and you get
a^2 = ab

I'm not multiply "a" from the "a^2 = b^2" equation. I'm multiplying "a" in the original equation, a = b.

Answer 3

As the late great C F Gauss said when confronted with Fermat's Last Theorem, something like this:
I could write a million such bits of nonsense, that I could neither prove nor dispose of. What of it?
These "proofs" all depend on dividing by zero. This is why division by zero is undefined.

How about a proof that you can't divide by zero? Here's one:
A/B = C implies that BC = A
Take A and C non-zero numbers, and B = 0
Then A/B = C or A/0 = C implies that 0xC = A.
To make this proof easier to follow, try substituting any non zero numbers, for example:
8/4 = 2 implies that 2x4 = 8
8/0 = ? implies that 0 x ? = 8
Since 0 x anything = 0, there is no number that you can multiply by 0 to get 8 (or any non zero number for that matter)
This is why division by zero in general is undefined. (I know, you can prove it in particular cases involving L'hospital's rule, Squeeze Theorem, etc, but I'm talking about in general.)

You can sneak in a division by zero and "prove" 1=2, all sorts of nonsense. Notice how toward the end of your "proof" where you "effectively cancel out because we divide both sides by (a-b)." Hoping that your sorry audience doesn't remember that a = b so a-b = 0. But that's not math. That's just showing off how sophomoric you are, and how stupid you think everybody else is.

if you want to impress people, do something real.

Sorry about the tone of this post, but questions like this make math seem like something illogical and impossible. They give math a bad name.

Answer 4

http://en.wikipedia.org/wiki/False_proof has a lot of fallacies. I like this one in particular:
-20 = -20
16 - 36 = 25 - 45
4² - 4*9 = 5² - 5*9
Add 81/4 to both sides...
4² - 4*9 + 81/4 = 5² - 5*9 + 81/4
Factor both sides...
(4 - 9/2)² = (5 - 9/2)²
4 - 9/2 = 5 - 9/2
4 = 5

This must be the proof for 2+2=5 that so many people have been looking for... Can you find the wrong step?

Answer 5

The trick is that you have built a second-degree equation. This kind of equations always have 2 roots (solutions). You can see that
a^2 - b^2 = a^2 - ab is true for a=b=1 (or in general, a=b) and also for b=0 and a=any value.

So in short: you added information to the problem. In real-life problems, you can get this kind of equations and their roots can be illogical (like getting a negative number for a rectangle side). What we do is neglect this kind of results, because if our model is OK, it will give us at least one real and correct solution.

Answer 6

That is the technical answer yes I would agree. Pero yo no se. Pero conaso este es una pregunta mas rado. Por que tu tienes este pregunta, viejo?:?

Answer 7

Let a=1 and b=2 and s= their sum
s=a+b
s(a-b) = (a-b)(a+b)
sa -sb = a^2-b^2
b^2- sb = a^2 - sa
b^2- sb +s^2/4 = a^2 -sa + s^2/4 [complete the square]
(b-s/2)^2 = (a-s/2)^2
b-s/2 = a-s/2 [Take sqrt both sides]
b=a
2=1

Answer 8

if your theory of false proof is logical, then

a = b

does not stand but instead,

a + z - y = b + z - y

where z and y are balancing factors to make the theory stand out as true.

mercury of love