Via StumbleUpon, I came across this short text page which lists three mathematical ‘proofs’ which seem to violate common sense, listed below. The first is:
The second one is:
The third one is:
Each of these proofs is (intentionally) wrong! They highlight classic fallacies in mathematical thinking. See if you can figure out where, in each of them, the proof goes wrong, and then look for the answers below the fold…
(Note: the third proof involves the ‘imaginary number’ 
The mistake in the first proof is highlighted in red below:
The mistake lies in the assumption that we can divide by 
The algebra ‘muddies the waters’ significantly, and it can be helpful to see the same proof, but with 
The next ‘proof’ is a little more subtle. Again, we highlight the problem step in red:
The problem here is that, in order to get to the step in red, there is an implicit square root taken of both sides of the equation. However, there are potentially two roots to any square; e.g. 
The third ‘proof’ suffers from essentially the same problem as the second, except that complex numbers are involved:
The step highlighted in red assumes that we break up the square roots as 
we can at best say that 
in which case we can at best conclude that
For our ‘proof’ above, we find that the statement in red should be replaced with “
There’s a nice classic book that covers such mathematical errors, and other more important mathematical paradoxes: Bryan Bunch, Mathematical Fallacies and Paradoxes (Dover, New York, 1997).
If you really want to be certain that you understand such mistakes, try writing your own ‘proof’ for 
December 12, 2008 at 3:00 pm |
Tedd Chiang wrote a really lovely speculative fiction story entitled “Division by Zero”. There is a copy of dubious legality available here. I recommend buying his short story compilation Stories of Your Life and Others. Spring for the hard back copy. This book is a keeper, and the design of the paperback version is inexplicably hideous.
December 12, 2008 at 3:06 pm |
NOOOOO!!!!!! Not math!!! Although solving proofs was one of the few things I enjoyed when I was taking high school math, I’ve forgotten just about all of it. I’m going back to doing banking on my Excel spreadsheets. Thanks…
December 14, 2008 at 4:19 pm |
These are really about the special cases of definitions. That is the curse of proofs and programs alike. You have to look at form and values together so I guess that makes them hard?
December 14, 2008 at 6:41 pm |
Thanks for the explanations. i was sick these type of things, tried to explain, but that is good.
December 14, 2008 at 9:37 pm |
Ya, thanks for the Details. I been tired of these getting passed around the internet. It known that they are math illusions, but having the details of were they are incorrect is great.
December 15, 2008 at 6:24 pm |
With the third problem. If you accept the square root as worked out . To solve 1/i you have to multipy the top and the bottom by the modulus of the bottom i.e -i which would leave you with -i/1 which when squared would give you isquared as on the other side. If I am not mistaken. Not that I know a lot, but I like maths…
December 15, 2008 at 6:54 pm |
This is very interesting. Spotting math errors come into handy in my own work, which questions mainstream assumptions, both in science and mathematics.
I find that mainstream scholars and “experts” are much more likely to commit subtle errors.
NS
http://sciencedefeated.wordpress.com/
December 15, 2008 at 11:33 pm |
The problem with some of these “proofs” is you cannot prove that one number equals another using algebra. The whole point of algebra is solving for a variable.
I was helping an algebra one student with his homework the other day and found that when there was not a variable in the end they were taught to put unsolvable because you cannot solve for anything using algebraic means.
These are more of philosophic problems. A bunch of math geeks or philosophy majors trying to show off that the world is chaotic and not the same as we know it.
Thanks for the unproofs though…
December 16, 2008 at 7:33 am |
Thanks for the laugh, was so freaking funny
December 16, 2008 at 12:57 pm |
I think you mean “late 19th century”, not “late 1900s”.
December 16, 2008 at 1:02 pm |
Jacob: Whoops! You’re right! I’ve touched it up in the post.
December 17, 2008 at 6:12 am |
que pouco sentidiño hai!!!
December 17, 2008 at 11:57 am |
Very effective explanation.
Divide by zero and taking only the positive value of a square root can fox many.
By the way, I discovered you blog through “stumble”. So the loop is complete.
Cheers!
PS: If you wish to read a real good book on math, I would recommend “Journey Through Genius (link in my blog!)
December 17, 2008 at 3:44 pm |
nice explanation… sick of getting stupid msgs frm friends ( aspiring engineers… ) and explaining them…
December 17, 2008 at 7:08 pm |
y=100 z=0
x = (y+z)/2
2x = y+z
2x(y-z) = (y+z)(y-z)
2xy-2xz = y^2 – z^2
z^2-2xz = y^2 – 2xy
z^2-2xz + x^2 = y^2 – 2xy +x^2
(z-x)^2 = (y-x)^2
z-x = y-x
z = y
0 = 100
There.
December 17, 2008 at 9:39 pm |
Nice “demonstration” Shaun. This is quite convincing, in a misleading way.
December 18, 2008 at 12:20 am |
Very interesting articles .It’s fun about math!!
December 18, 2008 at 4:48 am |
for shaun’s question
(z-x)^2 = (y-x)^2
z-x = y-x
///
when you square root the answer, there are 2 possible solutions for each side +/-.. it might be :
z-x = y-x OR
x-z = x-y
this might cause discrepancy in the answer
(very similar to question 2 as posted by the author)
December 18, 2008 at 6:18 pm |
lobster have right. Shaun the awnsers are:
(z-x)^2 = (y-x)^2
z-x = y-x
z = y
0 = 100
or
(z-x)^2 = (y-x)^2
x-z = y-x
50 – 0 = 100 – 50
50 = 50 then….you sux
December 18, 2008 at 6:25 pm |
oh sorry lobster took bad anwser, my is right:P because form hiw anwser is
x-z = z-y
50 = -50
December 18, 2008 at 7:02 pm |
No need to demean Shaun. He was just doing what skulls suggested at the end. Though it would be nice if somebody could come up with a different false proof that looks realistic (as in, one that used a different error besides the two shown).
December 19, 2008 at 5:30 am |
integral( tgx dx ) =
integral( sinx * 1/cosx dx ) =
integration by parts: | u’ = sinx dx u = -cosx, v = 1/cosx v’ = (sinx dx)/cos^2x | =
-1 + integral(sinx/cosx dx) =
-1 + integral( tgx dx )
finally:
integral( tgx dx ) = -1 + integral( tgx dx )
0 = -1
December 19, 2008 at 11:19 am |
quermit: Nicely done! You’ve got a future in ‘bad’ math!
December 19, 2008 at 5:21 pm |
POZDRAWIAM POLSKE I WYKOP.PL
December 20, 2008 at 5:36 am |
I found all of them
Now I have circles in coordinate system on math.
Również pozdrawiam wykop.pl i Polske ;D
December 20, 2008 at 12:55 pm |
Good Job
Pozdrowienia dla wykop.pl
December 21, 2008 at 8:37 pm |
Matthew,
I think you are just a dumbass jealous of those that can understand basic math. Go to school and stop ‘dreaming’, math doesn’t try to show a chaotic world, it is YOU that think on it that way. Math is just math.
What happened is that: you cannot understand clearly math and goes to ‘philosophy’ to try to explain. Next you are going to say that a god wanted it to be that way…
Those are fake ‘proofs’ and here was shown why they are wrong (and by the way, it is not that difficult to spot such errors).
December 21, 2008 at 9:35 pm |
quermit,
I know where is the point of your ‘proof’ =)
When you do integration by parts, you come from the supposition that (uv)’ = u’v + uv’
if you do u = -cosx and v = 1/cosx
(u*v)’ = 0
the error comes when you do the opposite passage in integration of both sides, creating solutions that do not match, (in integration by parts is done this integration implicitly)
and therefore creating that fake proof. ^^
deeply, the same kind of mistake commited in the more simple algebric problems demonstrated, assumpting a ‘probable’ solution that is not the correct ^^
I dont know if you posted having it in mind, but I have output it anyway…
========
I will post a similar problem for demonstrating how that kind of mistake can be generated in integration;
integral((e^x dx)/2)
| assumpt u = e^(x/2) , v’ = (e^(x/2))/2
u’ = (e^(x/2))/2 , v = e^(x/2) |
then,
integral((e^x dx)/2) = e^x – integral((e^x dx)/2)
e^x = integral(e^x dx)
That is not true always. Note that integral(e^x dx) = e^x + constant, so that is only true if constant = 0 . This is an example of how false solutions are generated through integration by parts.
=======
Mostly, i believe that kind of error occurs if:
(u*v)” = (u*v)’ (not sure of it, i have to proof yet… but seemed to be it. Damn, i am not even a mathematic, wth am I doing here??!?!?)
December 22, 2008 at 6:13 am |
Zhar, you are not mathematic or mathematician ?
December 22, 2008 at 11:56 am |
pitch
well, sorry for the error,
english is not my main language.
I am not a MATHEMATICIAN.
December 23, 2008 at 2:01 am |
What??!!!! No Asymptotes???!!!!!
December 23, 2008 at 5:21 am |
Ok try this one:
x=1+1+1+…+1(x ones)
x^2=x+x+…+x (multiply both side by x)
2x=1+1+…+1 (Differentiate both side)
2x=x (d 1st assumtion)
2=1 !??
Hav fun
December 23, 2008 at 12:12 pm |
glubot
x=1+1+1+…+1(x ones)
2x=1+1+…+1 (Differentiate both side)
Nobody said that number of 1s is equal in both statemens.
More, we can’t compare infinites, infinite isn’t a number
December 23, 2008 at 12:23 pm |
oh no =))))))
my fault, there is nothing infinite =))))
x=1+1+1+…+1(x ones) = constant
x^2=x+x+…+x (multiply both side by x) =
=(1+1+..+1)+(1+1+..+1)+…+(1+1+..+1) = constant
2x=1+1+…+1 (Differentiate both side)
Here comes bad, because on the left and on the right are constants
Should be 0 = 0 after diff
In fact, x isn’t a function, only a number. thus diff(x^2)!=2x, x doesn’t depend on anything
Cool!
December 23, 2008 at 6:06 pm |
Also, from 2b=b, you can only conclude 2=1 or b=0
December 24, 2008 at 6:32 am |
We had a lot of these in school, every one of them involved dividing by zero at some point.
December 24, 2008 at 4:15 pm |
math is not misleading,, it is jus far too simple to be fooled with it can be any simplier ,,just work with the already simplified transposition framework no shortcuts and u wil never go wrong
December 25, 2008 at 6:51 pm |
COOL AND HOT
December 25, 2008 at 9:35 pm |
Chuck Norris can divide by zero.
December 25, 2008 at 10:03 pm |
Chuck Norris can make Pi rational by telling it to calm the fuck down.
December 26, 2008 at 3:51 am |
Holy Shit skullsinthestars may be the funniest mutha fucker in the whole world. That is literally the best one Ive ever heard!!!!!!!
But seriously, I think my calc prof needs to read these dumb fuckin mistakes cuz he does that stuff all the time. there is truth in the fact that more educated people overlook sinple mitakes….
December 26, 2008 at 9:52 am |
I like the one that ends with pi=3 this is brilliant!!
December 26, 2008 at 7:43 pm |
Great! This is the kind of stuff that makes math interesting
.
December 26, 2008 at 10:08 pm |
Take a look at this: ( it’s similar to the one by glubot : 2 = 1 )
x^n = x + x + … + x + x + x ( n times )
Differentiate both sides to get,
n(x^(n-1)) = 1 + 1 + … 1 + 1 + 1 ( n times )
So, we find n(x^n-1) = n !
[[ pi = 3 was superb ! ]]
December 28, 2008 at 8:16 pm |
[...] Spot the math errors! Via StumbleUpon, I came across this short text page which lists three mathematical ‘proofs’ which seem to [...] [...]
December 31, 2008 at 9:44 pm |
Fantastic, found this through stumbleupon, I was about to say you should be a teacher
then I read more of your blog!
Do you mind if I put your blog on my blogroll? I’ve only just started blogging and you share a lot of my interests
January 2, 2009 at 1:14 pm |
bgecko wrote: “Do you mind if I put your blog on my blogroll? I’ve only just started blogging and you share a lot of my interests”
Certainly I don’t mind! I’m always happy to be on another blogroll. I’m glad you’re finding the site enjoyable!
January 15, 2009 at 2:02 am |
first goes to glubot:
you mistake is a the differantial,if i remember correctly, you will be a statment will variable and not and equantion. like x/dx of x^2 = 2x
second goes to TORNADO
you have a mistake right at your first statment
x^n = x*x*x*x*x(n times)
or
1^2=1
2^2=2+2
3^2=3+3+3
etc
and
1^3=1
2^3=(2+2)+(2+2)
3^3=(3+3+3)+(3+3+3)+(3+3+3)
etc
don’t know how to write it in math way, but you get the idea
January 19, 2009 at 4:50 am |
anything wrong with this proof
0=0
sin^-1(0)=sin^-1(0)
180=0
180-90=0-90
90=-90
sin(90)=sin(-90)
1=-1
1+1=0
January 24, 2009 at 11:58 pm |
it is pettry obvious
180=0
January 26, 2009 at 10:18 pm |
@Jeffery Chanbers:
The range of the arcsine (a.k.a. “inverse sine” or “sin^-1″) function is [-pi/2, pi/2] (radian measure), which happens to be -90 degrees to 90 degrees inclusive. As a result, sin^-1(0) cannot be 180 because that is represented as either 180 or -180, which isn’t in the range at all. To summarize, your flaw is in your third step as optiranium pointed out.
Here is a strange one:
x = 1
d/dx[x] = 1
d/dx[x] = x
d/dx[x^2] = x^2
2x = x^2
x^2 – 2x = 0
x^2 – 2x + 1 = 0 + 1
x^2 – 2x + 1 = 1
(x – 1)^2 = 1
x – 1 = +-sqrt(1)
x = 0 or x = 2
1 = 0 or 1 = 2
The flaw is admittedly easy to spot. Still, it was fun. ^_^
February 3, 2009 at 3:00 pm |
Oh, how i love math. <3
February 15, 2009 at 5:42 pm |
On the last proof, are there not TWO incorrect steps? The second to last step “i = 1/i” followed by squaring both sides to obtain “i^2 = 1″ is completely false as well, and is a much more glaring error.
My reasoning is thus: (1/i)^2 = 1/(i^2) = 1/-1 = -1? So the step in red might as well not be in error, because this one is a clear violation of imaginary numbers.
Am i off base here?
February 15, 2009 at 10:09 pm |
Tyler: Ah! There is, in a sense, an ambiguity about the step from i = 1/i to the step i^2 = 1. My intention, and the intention of the proof, is that one multiplies both sides of the expression by i, which results in i^2=1. One can also derive an expression for i^2 by squaring the equation i = 1/i; if you do that, you counteract the error created by taking the wrong square root of -1/1=1/-1.
To clarify the (erroneous!) proof, I should add a step as follows:
i = 1/i
i x i = i / i
i^2 = 1
Does that make sense?
March 4, 2009 at 5:26 am |
Very interesting. I’d seen some of these before but couldn’t really explain where the faults were.
April 11, 2009 at 6:20 pm |
are you stupid or something?, I just read the first one and it’s completely wrong. If a=b, then a-b=0, so how the hell do you go from 0=0 to a+b=b? and nobody has noticed this simple problem yet?, math by its nature cannot contradict itself only stupid people practicing math can contradict themselves.
April 11, 2009 at 6:28 pm |
okay, second one is wrong too when multiplying pie + 3 and pie – 3 doesn’t give you pie sqaured -9. I didn’t pay in school and even I can tell where the problems are. the only thing I find interesting about these faulty equations is, how can someone with a limited knowledge of simple mathematical rules write these equations and have them posted and nobody tell them they’re fucked in the head.
April 11, 2009 at 6:30 pm |
and tyler hartley, you are not off-base
April 11, 2009 at 6:31 pm |
o shit, my bad I didn’t read the rest of the instructions, I didn’t know they were intentionally wrong. my bad.
April 11, 2009 at 10:02 pm |
Um, yes, step 1 in solving any math problem: READ THE INSTRUCTIONS!
April 11, 2009 at 10:09 pm |
Wow. Great example here of leaping before you look. This feels like a YouTube comment thread now. You’ll notice though i too thought i’d found a problem, i didn’t tell people they were stupid and insult them; i asked. Take a chill pill, don’t call people you’ve never met “fucked in the head”, and realize that everyone on the internet is not dumber than you.
And have a good day.
May 20, 2009 at 8:19 pm |
Hi, my name is Krzysztof.
but seriously, I think my calc prof needs to read these dumb fuckin mistakes cuz he does that stuff all the time. there is truth in the fact that more educated people overlook sinple mitakes….
December 8, 2009 at 12:13 am |
John:
The point of the first one is that you have divided by (a-b), which is, of course, a faulty step in the proof. If a did not equal b, this would be a perfectly feasible mathematical operation.
Secondly, pi + 3 times pi – 3 DOES give you pi^2 – 9. It’s called FOIL. Ever heard of it? Most people learn it in Algebra I. If you’re over the age of 15 and have never heard of it, have fun working at McDonald’s.