whats wrong with the proof?

[gimp]

this proof shows that 2 can seemingly equal 1, where's the mistake?

x = y
xx = xy
xx - yy = xy - yy
(x + y)(x - y) = y(x - y)
x + y = y
x + x = x
2x = x
2 = 1

[pawq]

you cant divide by 0 i guess -.-

x = y | *x
xx = xy | -yy
xx - yy = xy - yy
(x + y)(x - y) = y(x - y) | /(x-y)

(x-y)=0 (obvious, anyone not understanding why?)
so
/(x-y) = /0 = fail

x + y = y
x + x = x
2x = x
2 = 1

i dont know if its right, but i think so anyway took a while to find it

[totem]

[A implies B] equals [not A or B]

so we have :

[(x=y) implies (2=1)] equals [not(x=y) or (2=1)]

and as we all know that 1 is not equal to 2, the answer is that x is not equal to y :/

you use logic as politicians do

[pawq]

wot

[totem]

now you have the choice :

either 1 equals 2
or x is not equal to y

you choose

[VT]

pawq is right, because the task was it to find the mistake in the proof. totem only shows that there is a mistake in the proof but this is also quite obvious without the equivalent interpretation of the implication

[pawq]

hey VT! i didnt even know you are registered here 5th post in such topic, gz

[gimp]

yes, pawq is right! the problem lies within the division which takes place between the 4th and 5th equations. since x = y, x - y is zero, and you can't divide by zero. good job pawq.

how about another one?

this proof shows that 10 seemingly equals 9.999999.... wheres the mistake?

x = 9.999999...
10x = 99.999999...
10x - x = 90
9x = 90
x = 10

[pawq]

i remember this prof from maths lesson in secondary school, then our teacher told us its right.
but now i dont really think that if
x = 9.(9)
then
10x = 99.(9)
hmm.. maybe it is? but please give more of them, its pretty interesting

[VT]

your teacher was right. there is no mistake in the proof. otherwise there would be a number greater than 9,9999..... and smaller then 10 (for example the arithmetic mean of both numbers), which is not true.

[pawq]

hurray good teacher

[welle]

Yes, VT is as right as the statement. This is exactly the same proof:

Hi VT o/

[totem]

pawq is right, because the task was it to find the mistake in the proof

indeed !

to make my point more clear, i meant the mistake was the first statement : x=y
isn't this correct ?

EDIT : my fault, i'm just a nab in maths

[milagros]

solving some equation means that you do a set of equivalence ending up in the final solution A<=>B<=>...<=>C
for example 2x+3=15 <=> 2x=12 <=> x=6

however in many cases you can do only set of implications(+some equivalences) A=>B=>...=>C and in that case the solutions of C have to be checked in original A, if it's correct
for example all equations with sqrt(x) like sqrt(x) = x-1 => x = x^2-2x+1 <=> ...

in some cases if you need any one solutions and not all, it may be enough to do set of opposite implications A<=B<=...<=C
if you fins some solutions in C, they are solutions of A, however you won't find all of them
for example cos(x) = 1 <= x = 0

however if you mix set of <= and set of =>, you don't really derive anything and can't say anything about result you got
so let's see

x = y => xx = xy (other implication doesn't hold)
(x + y)(x - y) = y(x - y) <= x + y = y (other implication doesn't hold)

[gimp]

pawq is right yet again, there is no mistake in the proof, nothing denies that 9.99999.... does equal 10.

[milagros]

this proof shows that 10 seemingly equals 9.999999.... wheres the mistake?

well; it is equal
there is no problem with that

[Bismuth]

I prove you that there are as many rationals than naturals.

Just make a table.

Code: Select all

___|__1__|__2__|__3__|__4__|__5__|__6__|__...
   |  1  |  2  |  3  |  4  |  5  |  6  |  ...
1  |  _  |  _  |  _  |  _  |  _  |  _  |  _  
_  |__1__|__1__|__1__|__1__|__1__|__1__|__...
   |  1  |  2  |  3  |  4  |  5  |  6  |  ...
2  |  _  |  _  |  _  |  _  |  _  |  _  |  _  
_  |__2__|__2__|__2__|__2__|__2__|__2__|__...
   |  1  |  2  |  3  |  4  |  5  |  6  |  ...
3  |  _  |  _  |  _  |  _  |  _  |  _  |  _  
_  |__3__|__3__|__3__|__3__|__3__|__3__|__...
   |  1  |  2  |  3  |  4  |  5  |  6  |  ...
4  |  _  |  _  |  _  |  _  |  _  |  _  |  _  
_  |__4__|__4__|__4__|__4__|__4__|__4__|__...
   |  1  |  2  |  3  |  4  |  5  |  6  |  ...
5  |  _  |  _  |  _  |  _  |  _  |  _  |  _  
_  |__5__|__5__|__5__|__5__|__5__|__5__|__...
   |  1  |  2  |  3  |  4  |  5  |  6  |  ...
6  |  _  |  _  |  _  |  _  |  _  |  _  |  _  
_  |__6__|__6__|__6__|__6__|__6__|__6__|__...

...

And then you go in diagonal and give a natural for each rational. For example: 1/1 is 1, 1/2 is 2, 2/1 is 3, 1/3 is 4... 1/5 is 11, 3/6 is 18, etc.

You will be able to give a number to every rational. There are as many naturals as there are rationals.

Edit: of course, you have to find the mistake

[Xiphias]

http://imgur.com/faz5t.gif

Haha that was fucking hilarious Jappe DDDDDD made me cry xD

[pawq]

indeed very good one

[jaytea]

cute

i've come across all of those, except the torus one, in classes and seen some proofs and i find them quite fascinating. but i'm not a fundie mathematician, i fucking hate it after studying it for so many years

[Bismuth]

Jappe found the mistake: there is none

[ribot]

this proof shows that 2 can seemingly equal 1, where's the mistake?

x = y
xx = xy
xx - yy = xy - yy
(x + y)(x - y) = y(x - y)
x + y = y
x + x = x
2x = x
2 = 1

nothing is wrong, dont worry, dont panic

the clue is infinity, transcendental vs prime numbers

[welle]

[pawq]

Not allowed:
- Big pictures in signature (also very big ones in the posts)
this is broken quite often, not in sigs but in posts. this is ancient rule though and people dont have dial ups anymore

eod.

[milagros]

nothing is wrong, dont worry, dont panic
the clue is infinity, transcendental vs prime numbers

or maybe something is wrong ^^
transcendental & prime numbers have completely nothing to do with it

[welle]

this is ancient rule though and people dont have dial ups anymore

[8-ball]

this is ancient rule though and people dont have dial ups anymore

[Bismuth]

Jappe becomes Abula, Pawq is jealous.

[gimp]

Hello again and good luck.

-2 = -2
4 - 6 = 1 - 3
4 - 6 + 9/4 = 1 - 3 + 9/4
(2 - 3/2)^2 = (1 - 3/2)^2
2 - 3/2 = 1 - 3/2
2 = 1

[Tigro]

-2 = -2 fine, equals
4 - 6 = 1 - 3 fine, equals
4 - 6 + 9/4 = 1 - 3 + 9/4 fine, equals
(2 - 3/2)^2 = (1 - 3/2)^2 here it starts getting tricky, as the power of 2 makes us see same results (1/4 on each side of equation), however, it is the power of 2 which forces the sign flag go to the positive numbers, even if left side (2-1,5) is positive and right side (1-1,5) is negative.
2 - 3/2 = 1 - 3/2 doing this is not an equivalent alteration (is that the word?). Reason is that since you are playing with +/- signs, it is important to mention that sqrt(x^2) doesnt equal x, but |x|.
2 = 1 Q.E.D.

[badyl]

What about this?

[kuchitsu]

That's like saying that 2 = -2 because 2 * 2 = 4 and -2 * -2 = 4. Just because a number has two sixth roots doesn't mean that they are equal lol.

[Tigro]

What about this?

The very same reasoning as the gimp´s one above:
sqrt(x^2) =/= x
but
sqrt(x^2) = |x|