By Vurdlak on April 13, 2006, with 228 Comments
Try to figure this one out! How is it possible that two is equal to one, when we all know that that isn’t true. Try to spot the mistake one of the twins made!
(38 votes) (38 votes)Comments
228 Responses
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OK. THere are two mistakes, at least one of which has already been said.
#1. if a=b, then a-b=0, and you can’t divide by zero.
#2. If 2a=a, a can only be zero, so you can’t divide by a as the last step.
As a side-note, some people have the wierdest reasons for why it doesn’t make sense. a and b are the first and second lettersof the alphabet? That’s true, but what does it have to do with anything?
In the third line the right side equals 0 because b x b is the same as ab so ab – b squared is 0
What was Annie talking about earlier!?
‘iiii…. am twelve, sixth grade… soooo… yea. It was waay over my head…’
Is she crazy? Oh yeah…I forgot I’m in higher math…lol
I’m eleven and seven months and in sixth grade, and this makes total sense to me!
I’m gonna show my teacher this! ^_^
-EXILE
Exile here! This is what I have to say!
Not sure, but this is what I think:
Problem: What went wrong?
Given: a=b
a^2=ab
a^2-b^2=ab-b^2
(a+b)(a-b)=b(a-b)
(a+b)=b
a+a=a
2xa=a
2=1
a=b we will assume a and b are 2
a^2=ab would be 2×2=2×2 (4)
a^2-b^2=ab-b^2 is 2×2-2×2=2×2-2×2 (0)
(a+b)(a-b)=b(a-b) is (2+2)x(2-2)=2x(2-2) (0)
(a+b)=b is (2+2)=2 (4=2) incorrect
(a+a)=a is (2+2)=2 (4=2) incorrect again
2xa=a is 2×2=2 (4=2) incorrect a third time
2=1 is 2=1
that can’t be right!
Any number but zero is wrong!
If we do try it with 0, however…
a^2=ab would be 0×0=0×0 (0)
a^2-b^2=ab-b^2 is 0×0-0×0=0×0-0×0 (0)
(a+b)(a-b)=b(a-b) is (0+0)x(0-0)=0x(0-0) (0)
(a+b)=b is (0+0)=0 (0) correct
(a+a)=a is (0+0)=0 (0) correct again
2xa=a is 2×0=0 (0) correct a third time
2=1 is wrong. this makes it 0=0
^_^
So technically, 2=1 is wrong, because the only thing possible is 0=0, not to mention variables aren’t neede now that we know it’s 0! Happy trails!
on the line after:
(a+b)=b
the next line written is:
a+a=a
this is impossible as the line should read:
a+b=b
which would make a=0 and b=0
lets start at the line that reads:
(a+b)=b
the next line after that reads:
a+a=a
this is impossible as the next line is supposed to read:
a+b=b
which would make a=0 and in turn making b=0…it is actually really simple cuz the one twim made the mistake…which is what your supposed to look for…duh
Huh?
THE MISTAKE THE TWINS MADE WAS WEARING THOSE CLOTHES.
seeing as you cannot solve an equasion with more then 1 unknown variable, the problem is unsolvable.. most of your explanations would be correct if A and B we’re actually 0, but there’s no way to tell that from the equation a==b. it simply states they’re of equal value
it’s realy simple, the mistake is made in the fifth line, (a+b)= a is wrong (a+b) = a+b so not b
easy problem..any tougher one?!
dude, i learned this crap in 6th grade, this is not ann illusion
i did this in A2A3 math. thats 8th grade stuff. its not that hard.
a squared is a times a not a times b
gothca
Given: a=b
a2 = ab
a2-b2 = ab-b
(a+b)(a-b)= b(a-b) remember given
therefore
(a+b)x 0 = b x 0
0 = 0
LHS = RHS
this guy sure failed math
If half door is open, it is equal to half door shut, therefore if full door is open, it means full door is shut ?!!!
(a-b)=0, therefore (a+b)(a-b)=b(a-b)=0, period.
I know something kind of like this. Three guys walk into an inn and pay ten dollars each for all three of them to stay the night. Later, the clerk realizes he overcharged them and gives five of the dollars to someone to give back to the three men. On the way, the person realizes there is no way to evenly split the five dollars, so he gives one to each of the men and keeps two for himself. So, each of the men only ends up paying nine dollars. Three times nine is twenty-seven. Add the two that the person kept and you get twenty-nine. What happened to the other dollar?
Scott, you have a very good point, but seriously… THIS IS NOTHING TO GET SO WORKED UP OVER!!! Anyway, even if they multiplied by (a-b), the math should still be constant, even if it equals zero. Don’t go all psycho. This really messed up equation is just all in good fun.
i’m in 7th grade and i’m done with algebra 1. if a=b then onthe 4th line then if you use the distributive property then it looks like (a+b)x0=bx0 aka 0=0. case closed. By the way, why the heck did these twins waste their time with this anyway?
Hmm…
WOW….
barring the possibility that there are little kids posting…
if you are an adult and have gone through the education system and don’t understand the simple math like the connection between a=b and a^2=a*b you shouldn’t be allowed to have kids.
Sorry about being harsh, because its fine to be stumped by this, its not ok to not know basic elementary school math
so really the solution is 2/0 = 1/0 ?
And that is true…
So the real “mistake” is saying that 2/0 = 2 and that 1/0 = 1….
And if anyone reads this far down, get a f’in life
this is not an optical illusion,they just try to confuse you and check if you at lest remmember 6grade maths.everything that’s below line 3 is not right:
(a+b)(a-b) will never equale the statement b(a-b)
always (a+b)(a-b)=a2+b2
There is a missing condition fot divison (as you all mentioned)
-> a-b /= 0 =>
-> a /= b
which it isn't, there for you can't
the mistake is in the transition from the fourth to the fifth line whereby the (a-b) on both sides is ‘cancelled’. Remembering that anything divided by 0 is undefined, divividing both sides by a-b which is really zero simply means that undefined = undefined. WOOP DE DOO. =.= , back to the drawing board.
wow some of u r in 8th grade and u cant even figure this out… and some of you have no humour, some of you think it’s a set of unrelated equations… i found the error when i was in 6th grade! and the error is that you cant divide by zero.
(a+b)=b
not true^^^
got me for a minute tho, hadda think it throughht
and comment #1… (mark)
are you saying 8th grade math is easy
cause im in 8th grade, and im taking an advanced course, and its still hard
it should come out to be 0=0
because (a+b)(a-b)=b(a-b) equals zero
and 2a=a which is the same as 0+0=0
should in fact be
0=0
so the only way the whole equation..or however they are trying to figure this out works is to ultimately say a and b equal zero. so every equation they wrote satisfys that. so 2a=a –> 2(0)=0! and (a+b)=b –> (0+0)=0!
so the only mistake they made is at the end when they assumed that a and b were 1.
so in lines 4 and 5 to say that they were dividing by zero so that doesn't work is stupid because they were writting random equations not following steps to get to the next problem duh :P
You can divide by zero in calculus though with L’Hospitals rule, ha ha
They missed out the parameters – "a=b=1" (or if you prefer "a=1", "b=12)
Need to state what a & b are ;) otherwise it is technically correct and then the last line is also wrong… you cant just assume 'a' is equal to one – needs to be stated… yes, im a uni student with nothing better to do ;)
“this is not an optical illusion,they just try to confuse you and check if you at lest remmember 6grade maths.everything that’s below line 3 is not right:
(a+b)(a-b) will never equale the statement b(a-b)
always (a+b)(a-b)=a2+b2″
First of all this must be corrected:
“always (a+b)(a-b)=a2+b2″
should read:
“always (a+b)(a-b)=a^2-b^2″
In this situation, this isn’t the error.
b(a-b) is equal to ab-b^2 and since a=b this could also read a^2-b^2.
The real error is as it has been said before, when they divide by zero by trying to cancel by dividing the (a-b) over since a=b, a-b = 0. You cannot divide by zero.
if a=b , then a and b are equal, meaning that THEY ARE THE SAME NUMBER!
therefore, the answer cant be 2=1
Don’t blame me for not getting this i’m still in primary school…(age 11) tho some of my classmates have that a – b = ? some sort of things…
Sigh… the answer, which very few people have correct, is that you cannot divide by 0. ALL of the math here is sound, EXCEPT that one of the steps is not reversible (divide by 0). That is in going from line 4 to 5. If you notice, lines 1-4 are all true as long as a = b, which was the premise. Starting at line 5, after the division by 0, everything is only correct if a = b = 0, which was not a stated condition.
(a+b)(a-b)=b(a-b) ….. isnt tht bit rong?
them dudes try devide by zero. Tards. I IS IN 8TH GRADE AND ME FIND ANTSA! NO ZERO DIVISION PPL!
It is because they divided by a-b in the 4th step and because a=b that means they divided by zero, which gets UNDEFINED not any real answer. Duh!
It is not given that a=b=1. That is the only thing that is wrong. And then there is the problem that the only solution is that a=b=0 so you are dividing with 0 which is a problem as the answer infinite.
Besides, 2a=a equals to 2a-a=0 a=0, not to 2=1.
If two dollars have the same value as one dollar, then those dollars are FAKE!
ok i’m in 7th grade and i found this easy.
the big mistake is that(a+b)(a-b) will never equal the statement b(a-b)
always (a+b)(a-b)=a2+b2
it’s a simple formula really :)
what are you talking about?
(a+b)(a-b)= a^2+ab-ab-b^2=a^2-b^2 so
(a+b)(a-b)=b(a-b) is true
because a^2-b^2=ab-b^2.
The part that’s not true is (a+b)=b because if
a=b, then the above cannot be true.
a=b |*a
a^2=a*b |-b^2 //correct 0=0
a^2-b^2=a*b-b^2 |faktorize //correct 0=0
(a+b)(a-b)=b(a-b)|/(a-b) //WRONG: ADDITION 2=1
qqqqqq is right!!!!!!!!! If a=b how can it be anything else. the mistake is in the second line twin on right should be copying left the whole way down visa versa.
a=b
1=1
2=2
3=3
a=b
get it your all fools. Trick question. all you accadi*#heads. pull your head out of that thing called a bum. Twins??? meant to be the same????? 1=1 2=2 3=3 and so on. copy is the key here ahhhhhh mirror, 1=1 2=2 3=3 a=b d=f s=r trick question. I is in mine 25 year from life and i did fail maths. hang on i is really from da eight grade!!!!
Sorry, love your site. some of these people claiming they are in the 8th grade are full of it!!!!!!!!! very annoying.
Everything is correct to that point where (a+b)=b. Then you see that => a = b-b = 0.
I think the mistake is..it’s supposed to be 2=a lol where did he get 1?
Okay people, lets get something straight. You can divide both sides by (a-b) as they both have it on their sides. THERE IS NO DIVISION BY 0!!! If you divide 2y=2 by 2, then y=1. That is basic algebra. Therefore (a+b)(a-b)/(a-b)=a(a-b)/(a-b) becomes 1(a+b)=1a which is (a+b)=a
This is what is called formula manipulation, and (a-b) without definition cannot equal 0 in the first place.
Since a and b were not assigned values, the mistake is assuming they both equal 1.
Really guys, everyone saying that a and b equal 1…
2a=a, divide by “a” on both sides and you get 2=1… “a” could be 1,000,000,000 or 27, and you still get 2=1
the error is when you try to divide (a-b) from both sides, because if “a” and “b” are the same (a-b)=0
yea, this is what happens when you divide by zero. idiots
Considering:
1 = 1
1^2 = 1*1
1^2-1^2 = 1*1-1^2
(1+1)(1-1) = 1(1-1) -> This is an Error.
(1+1) = 1 -> Here starts the Wrong Equation.
(1+1) = 1
2(1) = 1
2 = 1
The error is in fact, 1 Cannot be divided by 0.
(1+1) != 1 on (a+b) != a and (a+b) != b. So are 2a != b.
THere’s no error LOL XD I made a mistake.
Actually the error was on (1+1) != 1 on (a+b) != a and (a+b) != b. So are 2a != b, that’s wrong.
Lol.very simple.its like this:
a=b (no problem)
a^2=ab (no problem)
a^2-b^2=ab-b^2 (no problem)
but! a^2-b^2=0
and ab-b^2=0
sneezes look like worms
seeing from comments, people looks happy :) more easy illusions please
http://en.wikipedia.org/wiki/Mathematical_fallacy#All_numbers_equal_all_other_numbers
It all comes down to the division by 0 in the form of (a-b). Since (a-b)=(a-a)=(b-b) => (a-b)=0=0. No matter what, it’s dividing by 0, leading to an undefined number. Although, I do love seeing everybody else’s attempts to sound smart.
Im 20 now and could have answer this in middle school, its cute seeing grown folk think there smart with wrong answers
if im wrong, email me. RafCintron@Yahoo.com
1.a=b (ok, easy)
2.a^2=ab (multiply a to both sides, still equal)
3.a^2-b^2=ab-b^2 (subract b^2 from both side, fine)
heres where the problem starts
4.(a+b)(a-b)=b(a-b) IS A WHOLE DIFFERENT EQUATION
a=b so (a-b)=0 so anything time 0 equals zero
but subtract (a-b) from both sides… (a+b) does not equal b.
1+1 isnt 1, 2+2 isnt 2, and 5+5 will never be 5
hopefully my explaination was a lil easier than Rich’s up there :P lol
It is very simple….
All the people above have tried but their explanation is unsatisfactory to me
0/0 is not equal to 0… it is not defined!!
Therefore we cannot do that!!
(a+b)(a-b)=a^2+b^2=2a^2
b(a-b)=ab-bb=0
You cannot devide a number by 0. In 4th line we have (a+b)(a-b)=b(a-b). It is possible to divide the equality by a-b [so that we will have a+b=b] only if a-b is diferent from 0. But the starting point was a=b => a-b=0. So mathematically here is where the mistake is: you divide 4th line with 0 which is not allowed.
its correct .. no mistake .. you are all fool ..
I just want to give an answer for all the guys that thought that this equation only has it’s solution for a=0.Well that’s false,because there is a mistake there where we divide by (a-b),that means dividing to 0,which is impossible from mathematical point of view speaking :)
the mistake is from the 4th to the 5th line.
(a+b)(a-b)=0
because (a-b)=0 and everything multiplied with zero logically becomes zero.
and don’t be so snobby here. everybody has their individual talent and only cause you figured this one out doesn’t mean you can judge weather people can have kids or not. you can be stupid in many ways, ergo you can be really good at maths and still be stupid in real life.
they divided by zero :)
no, seriously though, they did, when they divided by (a-b). if a=b, then a-b=0 :D
IDK!! math is for stupid people anyways…
Different ways of looking at the equation.
Last line 2 = 1 is wrong however.
as pointed out earlier since a=b, a-b = 0
so (a+b)(a-b) = b(a-b) => (a+b)0 = b*0 still holds true.
now, the tricky part, 2a = a
the correct way of solving this is not cancelling out ‘a’ since there is not other variable left
it is indeed as follows.
2a = a
a + a = a
a = a – a
a = 0
since a = b, b = 0
So the solution is correct but the last line is where the mistake lies.
Simple. If a=b, then a-b=0.
And we all know that we can’t divide by zero coz we could create an awful black hole.
That’s the mistake =)
Technically the last step is correct because you can divide both sides of the equation by a. The fourth step is also correct by factoring. Also, @Anonymous (three posts down), I wish my class learned factoring in sixth grade. (I’m in seventh now, and we still haven’t learned it.)
@Rafael, Rich was correct; you weren’t. LOL.
It’s summer right now next year i’ll be in 5th grade…I’M CONFUESED!!! All i knoe is that B=2…
@lalalalalalalaa no it isn’t…if u don’t know math it limits the things u could do in ur life in the future… (by a lot)
Hmmm it’s interesting but cancelling out a-b should be fine, let’s move forward
b+a=a
a+a=a
2a=a,
now I never saw such cancellation where both sides are constants, so proper way to solve it is:
2a-a=0
hence a=0 and b=0
Whenever you try to play with 0, 0 will play with you :)
i found it, at the left side suppose to be:
(a+b)(a-b)=b(a-b)
(2a)(0)=b(0)
0=0
first we have “(a+b)(a-b)=b(a-b)” and after it we have “a+b=b”, but we cant divide both parts on “(a-b)”, because “a=b => (a-b)=0″ =)
children learn this, when they are 10 years old:)
I`m 10 and I still have not learned it.
i did not learn it til i was 12 and i was not held back
Yep, divide- by-zero error
Try this
Sqrt(-1)^4=1
Sqrt(-1)=4th root of 1 = 1
-1=1^2
-1=1
0=2
1=0
@Matthew Wales
That’s interesting. I don’t know what causes that though.
@ Matthew Wales:
You can convert your imagination to reality. Once it is real, it cant be your imagination. Same thing with sqrt(-1), an imaginary number i.
this is NOT an OPTICAL illusion. You’ve got to be sharp and remember mathematics, I didn’t come here to THINK, I came here to LOOK.
I am 28, my brain doesn’t work this way, but I have won awards for artwork and have played drums for several bands…I would consider myself successful but I’m totally lost and I am far from 10…I guess America is trained to think this way.
The problem ends when both sides of the equal sign are multiplied by (a – b). The solution is 0 = 0. Reason: if a = b, then (a – b) equals 0. At that point 0 = 0. You can’t then go on to divide both sides by (a – b) because (a – b)= 0 and 0 in the denominator is UNDEFINED.
easy enough….
this does not work because (a+b)(a-b) does not equal a squared b squared
Let’s suppose that ‘a’ exists, and ‘b’ exists… and that their existences are wholly defined by each other; They each exist entirely as the other does not. ‘a’ and ‘b’ would constitute a ‘universe’, so to speak.
The EXISTENCE of ‘a’ IS ‘b’, but ‘a’ is NOT ‘b’. In this case, the possibility for 2=1 exists.
The entities are undefined except to the extent that they can be related to and defined by one another in some way.
Does 2 = 1? Yep, it sure does… but 2 WHAT be one? 2 absences perhaps? If so, then how is the absence defined? Is ’1′ absent? Is ’1′ BEING absence? Is there an ‘outside’ to this system?
WOW HOW LONG DID IT REALLY TAKE YOU TO COME UP WITH THIS THEORY?I GIVE PROPS FOR COMING UP WIT SUCH ELABORATE CONCLUSION BUT I DONT REALLY SEE HOW THIS IS EVEN POSSIBLE SEEING HOW COMPLICATED THIS I BELIVE THAT THATS PROBABLY WHY THE PEAPLE IN THE FRIST DONT BOTHER WITH THIS ANYWAYS.