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-   -   Did you know? 2=1.. (http://www.discussworldissues.com/forums/general-discussion/235564-did-you-know-2%3D1.html)

PapsEdisa 01-04-2009 11:40 PM

http://www.bildschirmarbeiter.com/co...picdump-21.jpg

Tryphadz 01-05-2009 12:10 AM

I know I should (oops, I meant shouldn't) say this, all things given, but I have to say that the above emoticon is the finest creation in the whole of the universe. Never has something so simple caused so much hassle and yet so much joy to a bunch of nerds on forums...

ddwayspd 01-05-2009 12:22 AM

Quote:

I know I should say this, all things given, but I have to say that the above emoticon is the finest creations in the whole of the universe. Never has something so simple caused so much hassle and yet so much joy to a bunch of nerds on forums...
Sort of like when a female walks into a game store. All you can hear is the sound of heavy breathing and D20's hitting the floor.

UvgpXK0J 01-05-2009 12:37 AM

I love that gif.

Given how many previously banned members have been let back in, perhaps its time to welcome Gamergod back into the fold. or not.

TpDoctorOneTp 03-14-2009 11:18 AM

Did you know? 2=1..
 
http://boobers.nion.be/motivators/mi...r-calculus.jpg

How long did it take for you to crack it? http://www.discussworldissues.com/fo...es/tongue1.gif

ivandiadser 03-14-2009 11:24 AM

Too bad its messed up!

TpDoctorOneTp 03-14-2009 11:34 AM

Quote:

Too bad its messed up!
Well of course, it's not like the maker of that thing was the first one that would find 2=1 if it was true! http://www.discussworldissues.com/fo...es/tongue1.gif

The thing is, how long does it take to..crack it? [yes]

BamSaitinypap 03-14-2009 12:10 PM

Er, zero? http://www.discussworldissues.com/fo...lies/wink1.gif

shihoodiacarf 03-14-2009 04:05 PM

where is the mistake in the calculation? I've gone over it twice and it seems to be right [help]

StevenS 03-14-2009 04:16 PM

The mistake is the firs line...
a=b

The very nature of algebraic equation is to use a single letter to represent a single integer thus allowing an unknown number to be represented as said letter.
a cannot equal b because algebra states that a must only equal a and b must only equal b.

that aside... in the fourth line:
(a+b) (a-b) = b (a-b)

this suggests that (a+b) is equal to b which again at the top of the equation is stated that a=b so how can a+b = b

There's a whole sh-ugar storm of assmption disguised by the infinitely daring leap-of-faith attitude that if u put 1+2=3 as letters you can just re-write the laws of physics as you see fit. Unfortunately 1=1 which is why your gross montly paycheck didn't just double after reading neinstein's theorem above.

Good show though a+b/b for effort

BamSaitinypap 03-14-2009 04:57 PM

Quote:

The mistake is the firs line...
a=b

.... etc. etc....

There's a whole ...Unfortunately 1=1 which is why your gross ...

Good show though a+b/b for effort
No really, it's zero. The last line is just thrown in for fun. a = b = 0.

And this isn't calculus, it's algebra.

StevenS 03-14-2009 05:53 PM

I could have concluded it was zero if the last line was omitted.
I totally agree on the calculus/algebra thing.

You cannt have two distinctly different given values (x and y for example) in algebra for the same number though. You can on one occasion have two different numericals represented with one universal value (say x) however*.

*This exception being a square root...

eg.
SqRoot of 9 = x
x = either +3 or -3

or indeed as the answer is in this case 0

OWV9LSxH 03-14-2009 07:06 PM

There is still this one big mistake noone has pointed out.

When going from 4th line to 5th line

(a+b)(a-b) = b(a-b)
to
(a+b) = b

you are essentially dividing by (a-b), which equals zero (as first line states a=b).

Division by zero is not allowed, so all calcuations after that are irrelevant.

...moving back to the shadows...

DariushPetresku 03-14-2009 07:15 PM

Quote:

There is still this one big mistake noone has pointed out.

When going from 4th line to 5th line

(a+b)(a-b) = b(a-b)
to
(a+b) = b

you are essentially dividing by (a-b), which equals zero (as first line states a=b).

Division by zero is not allowed, so all calcuations after that are irrelevant.

...moving back to the shadows...
Yeah that's it, in "classic" algebra math rules the following mistake is made:

When dividing by (a-b) you're stating:

A+b = b OR a-b = 0

The last one is often forgotten, but is true in this case, since a = b

Tryphadz 03-14-2009 07:17 PM

Quote:

a cannot equal b because algebra states that a must only equal a and b must only equal b.
Not quite - the start of the 'puzzle' is perfectly valid, as it's essentially saying "one has two variables, a and b, and they are equal to each other". To argue otherwise, would be saying that the function y = x cannot be plotted - and it quite obviously can!

Quote:

this suggests that (a+b) is equal to b which again at the top of the equation is stated that a=b so how can a+b = b
Well that was the point of the 'puzzle' - i.e. you end up 'proving' that 2 =1

Quote:

Division by zero is not allowed, so all calcuations after that are irrelevant.
*Ding* And we have our winner!

Edit:
Quote:

Yeah that's it, in "classic" algebra math rules the following mistake is made:

When dividing by (a-b) you're stating:

A+b = b OR a-b = 0

The last one is often forgotten, but is true in this case, since a = b
The first one (a + b = b) is also true, as there is a number out of the set of real numbers for which there is defined answer: a = b = 0. It's the part where both sides need to be divided by (a - b) that isn't permitted, as division by zero doesn't have a defined solution in real number algebra.

DariushPetresku 03-14-2009 08:14 PM

Quote:

The first one (a + b = b) is also true, as there is a number out of the set of real numbers for which there is defined answer: a = b = 0. It's the part where both sides need to be divided by (a - b) that isn't permitted, as division by zero doesn't have a defined solution in real number algebra.
Yeah you're right, I was a little too fast.

Point is the OR statements are left out.

Same for the 2nd to last line, you're dividing by a. Either that's possible, or a = 0. That's what caused the mess.

QuidQuoPro 03-14-2009 08:36 PM

Heh noticed you removed your "or divide by 0", but that isnt possible because that will equal nothing, so as soon as you divide a or b by 0 it equals nothing (my maths is a bit sketchy, but I think thats correct). Even Windows Calculator wont allow you to divide by 0 as the answer would be nothing, which it cannot generate - look for yourself!

29clepayJainync 03-14-2009 09:38 PM

Quote:

Heh noticed you removed your "or divide by 0", but that isnt possible because that will equal nothing, so as soon as you divide a or b by 0 it equals nothing (my maths is a bit sketchy, but I think thats correct). Even Windows Calculator wont allow you to divide by 0 as the answer would be nothing, which it cannot generate - look for yourself!
It's not nothing, it's infinity.

derisgun 03-14-2009 11:07 PM

I see the point about the step where essentially there is a div/0.

Isn't there however an actual algebraic error done prior which is what allowed that situation?

I think it lies along these lines.....b does not equal the square root of b^2.

TpDoctorOneTp 03-14-2009 11:34 PM

It's a fun prank to give to someone that just started algebra that might miss that "divide by zero" situation at 4th line..[thumbup]


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