Einstein's 2=1 proof.

Poststructural Pub
1

In the Einstein thread, someone’s image shows Einstein proving 2=1 as follows (I left out some intermediate steps):

  1. a=b
  2. a^2 = ab
  3. a^2 - b^2 = ab - b^2
  4. (a+b)(a-b) = b(a-b)
  5. a+b = b
  6. b+b=b
  7. 2b=b
  8. 2=1

Reasons:

  1. Assumption
  2. Multiply both sides by a.
  3. Subtract b^2 from both sides.
  4. Factor the above
  5. Divide both sides by common factor (a-b)
  6. Substitute b for a (since a=b)
  7. Combine common terms on both sides
  8. Divide both sides by b

Here’s a question for the math geeks out there (myself among them).
What is the problem with the above proof?
We know 2 does not equal 1, but all the algebra above looks fine.
Or does it …

Let’s see how long it takes y’all.

Have at it!

-Chris

2

in the step from 4, on the right side of the equation it says: b(a-b), in order to get to step 5, b(a-b) must be divided by (a-b). This is forbidden and not possible in mathematics, because since a=b it would mean you are dividing something by 0 (zero). Illegal action.

Is that the reason?

3

Here’s where the trickery lies.

a=b
therefore
a-b=0
thus
You cannot divide by (a-b), since that produces a division by zero.

4

very good buddy. :smiley:

5

Since a=b, the expression in step 3 equals zero.

a^2 - b^2 = ab - b^2 :: 0 = 0

6

:laughing: :laughing:

I’m surprised nobody already mentioned that if a=b then a-b = 0 and you can’t divide by 0.

7

18 minutes for Amar!
And independently from Wanderer.
Bravo.
Gary, Step 3 is fine.
A statement that is true (0 = 0) is not a problem in a proof.

Thank you for playing.

:slight_smile:

8

Yeah, but since 0 = 0 in step 3, step 4 is invalid and so the ‘proof’ ends there (you’re gonna factor “0 = 0”?).

And! Even if it didn’t, if a=b then a-b = 0 and you can’t divide by 0. Oh wait… I see Flydood got there before me. D’oh! Nice one fd!

9

We were right back to back :wink:

10

Since I posted the picture, I felt it only proper to forgo responding until others had a chance to do so. Originally, I tried to use E and mc^2 instead of a and b, but it just got out of hand –

E = mc^2
E^2 = E(mc^2)
E^2 - (mc^2)^2 = E(mc^2)- (mc^2)^2
(E + (mc^2)) * (E - (mc^2))= (mc^2) * (E - (mc^2))
((E + (mc^2)) * (E - (mc^2)))/(E - (mc^2))= (mc^2) * (E - (mc^2))/(E - (mc^2))
E + (mc^2) = (mc^2)
(mc^2) + (mc^2) = (mc^2)
2(mc^2) = (mc^2)
(2(mc^2))/(mc^2) = (mc^2)/(mc^2)
2 = 1

The parenthesis were making my head hurt, and it was becoming obvious that this would not fit around his hand, so I went with the classic “a = b”. If there is anyone out there who has more patience for this than I do, you’re welcome to check the work and see if the parens are correct.

11

Nah. I think I’ll wait for Amar and Wanderer to reply, then copy their answers :slight_smile:

12

http://www.planetvids.com/html/Fool-Me-Once.html

13

That nuke school training in the navy seems to have finally been good for something :wink:

14

I skipped over the other responses to give my answer:

Equations 2 through 4 can only be constructed if a=b, as in equation 1. For any value of a or b, the (a-b) multipliers in equation 4 must be zero. This renders them meaningless as a multiplier (division by zero is undefined), so they cannot be factored out, and equation 5 is therefore invalid.

How’s that? :smiling_imp:

15

next question,
which mathamatician postulated that 1 and 2 are the same, only viewed from angles?

16

Dr. Irving Joshua Matrix

17

Saying that 1=2 is only like telling a novice Irish musician to cut down on the excessive ornamentation as “less is more.”

Sorry to be un-serious but I feel a bit left out because I don’t know what all those little upside-down ^s stand for. :confused:

Steve

18

that’s square.

you know like the little ‘2’ above a number, like the little 2 above 9, meaning 9 times 9.

19

Ahah. All is now as clear as mud. :smiley:

20

haha…you know!!!

like this: