1. ## Math Proof: 2=1

I dont know if this fourm as seen this before, but its pretty neat, i can prove that 2=1 right?

Given: a=b

2. Originally Posted by Lammalord
I dont know if this fourm as seen this before, but its pretty neat, i can prove that 2=1 right?

Given: a=b

a=b
a2 =ab
a2 -b2 =ab-b2
(a+b)(a-b)=b(a-b)
Make note: a-b=0

And math says you can't divide by zero.

3. ahh wait, let me re-do this, messed something up (its soppost to be squared)

there, think i fixed that up...

humm wow, didnt see that one, ahh not even my math teacher last year saw that one...

4. Originally Posted by Lammalord
ahh wait, let me re-do this, messed something up (its soppost to be squared)

there, think i fixed that up...

humm wow, didnt see that one, ahh not even my math teacher last year saw that one...
well, I really can't claim credit, I would've missed it too if (I think it was) Glaysher didn't respond to PH's post of this question (although PH knew the answer as well)...

This is my 800th post!!!!

5. ## completion - just for fun

Originally Posted by Lammalord
I dont know if this fourm as seen this before, but its pretty neat, i can prove that 2=1 right?...
Hi,

I can prove that every number is equal to every number: See attachement.

EB

6. Originally Posted by earboth
Hi,

I can prove that every number is equal to every number: See attachement.

EB
Again dividing both sides by c - a - b is dividing by zero and therefore proof invalid.

If proof was valid you would have derived a contradiction as you had originally assumed a does not equal b. It would not prove all numbers are equal. It would prove that you can only add equal numbers.

7. Hi.

I can prove that 1 = -1. Se below.

$\displaystyle 1 = \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i \cdot i = -1$

-Kristofer

8. Originally Posted by TriKri
$\displaystyle \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1}$

9. Originally Posted by TriKri
Hi.

I can prove that 1 = -1. Se below.

$\displaystyle 1 = \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i \cdot i = -1$

-Kristofer
This is proof that Euler shown when the theory of complex numbers was still in it's early stages of development.

10. Originally Posted by TriKri
Hi.

I can prove that 1 = -1. Se below.

$\displaystyle 1 = \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i \cdot i = -1$

-Kristofer
It hasn't been proven that $\displaystyle \sqrt{-1\cdot-1}=\sqrt{-1}\cdot\sqrt{-1}$

11. Originally Posted by Quick
It hasn't been proven that $\displaystyle \sqrt{-1\cdot-1}=\sqrt{-1}\cdot\sqrt{-1}$
I would HOPE that it hasn't been proven!

-Dan

12. Originally Posted by Quick
It hasn't been proven that $\displaystyle \sqrt{-1\cdot-1}=\sqrt{-1}\cdot\sqrt{-1}$
I suppose not, I mean, I really hope it hasn't!

13. Proof that it isn't:

√(-1 * -1) ≠ √(-1) ^ 2
√(1) ≠ i ^ 2
1 ≠ -1

Heh. That was probably obvious, but I love doing short proofs like that.

...Well, I guess I just said the opposite of what the origonal problem said. It makes my brain hurt.

14. Originally Posted by The Pondermatic
Proof that it isn't:

√(-1 * -1) ≠ √(-1) ^ 2
√(1) ≠ i ^ 2
1 ≠ -1

Heh. That was probably obvious, but I love doing short proofs like that.

...Well, I guess I just said the opposite of what the origonal problem said. It makes my brain hurt.
There are like 3 errors in that proof!

First the preservation of the not equality was not used properly
Second the use of illegal math factorizations
Third a mathematical logical error.

But yet what you said in conclusion was completely true!

15. Really?

*looks back at his work, fails to see problems*

Heh, I guess i'm a bit of a newb.

What exactly did I do wrongly again?

I have to admitt, though, if my proof really is wrong, that's pretty cool that it's right anyway.

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