right, you can prove 1 = 2, and 3 = 4 etc. BUT...?

Question

who can prove 3 = 6?

again, my weirdo of a sis has nothing better to do than make me ask questions on here 4 her.

Answer 1

just ask my wife....when we are in an argument, somehow she can make 1=2 and 3=4....go figure!

Answer 2

what? 1=1 2=2 3=3 4=4 5=5 6=6

Answer 3

Like before, once you "prove" that 1 = 2, you can multiply both sides by 3 to get 3 = 6.

In fact, once you've "proved" that 1 = 2, you can "show" that any number = any other number, just by choosing the right combination of additions and multiplications to apply to each side. For instance, I can start with 1 = 2, multiply both sides by 63 to get 63 = 126, and subtract 59 from both sides to get 4 = 67.

You'll find more information about the "1=2 proof" in the article below. Have fun!

Answer 4

First you have to prove that anything other than 1 exists.

No, 0 does not exist. 1 eliminated 0 when it eliminated the vacuum and due to the infinite nature of 0, it has no value and therefore cannot be used in an equation.

Infinity never was and never becomes a number. It is not even a constantly changing number because for a number to change, it first has to be a number so, don't even try it.

Contrary to what you have been taught, 1+0=1 is not valid because 1 cannot exist where there is 0.

Answer 5

The proof of 1=2 is false unless 1 is located at 0 and 2 is the first positive number greater than 0 (not infinitesimal: the fact that we cannot represent it does not matter because we cannot represent many numbers using our decimal system. If we say this number does not exist, then the real numbers have holes and this is obviously untrue). In this case you can use Cauchy sequences to prove 1=2.

The proof is simple: If points 1 and 2 have no extent, then we can place them anywhere we wish on the number line remembering to preserve order (in this case we place 1 at 0 and 2 at 1).
A simple application of d(x,y) = 0 or Cauchy limit logic proves they must be the same number. In this way, you can prove (falsely) that 1 is equal to any number you like.

Mathematicians today are stupid beyond belief!

Answer 6

Same again; prove 1 = 2

then as 3 = 1 + 1 + 1, therefore 3 = 2 + 2 + 2

which means 3 = 6.

Answer 7

Let a = b.
Multiply through by a: a² = ab
Subtract b²: a² − b² = ab − b²
Factor both sides: (a − b)(a + b) = b(a − b)
Cancel out (a − b): a + b = b
Observing that a = b, we substitute: b + b = b
Combine like terms on the left: 2b = b
Mulitply by 3: 6b = 3b
Divide by b: 6 = 3

Obviously it's not a proof, it's faked by using division by zero

Answer 8

OK, how about this one? It's more direct than most, and the error is more obvious.

0 = 0, a true statement.
0(3) = 0(6), obtained by factoring both sides.
3 = 6, obtained by dividing both sides by zero.

That seems too easy, but it is the logical equivalent of most of the other proofs like this.

Answer 9

It's impossible, all of these are proved from the great fallacy that you can divide by 0 which is wrong.

Answer 10

1=2
so
1X3=2X3
3=6 ha ha ha