250 Replies to “Math Optical Illusion”

  1. How is this an optical illusion? It’s a well known 8th grade math puzzle.

    If you don’t remember Algebra I, then none of what they’re doing makes any sense and it’s all math gibberish. If you do remember basic math, then the problem is easy to spot:

    In the step where you cancel out (a-b), you are dividing by zero, which is a mathematical no-no. (Since a=b, a-b is always zero)

  2. I think it has something to do with the fourth line. (a – b) = 0. I don’t think you can cancel zero from both sides. I guess the problem is actually the third line, then. a^2 – b^2 = 0 and so does ab-b^2. Propagating zero down the line leads to 2*0 = 0, which is true.

    … but really, I can’t see the problem in the algebra. Are you not allowed to subtract and arbitrary number from both sides? I think you are…

    grrrrrrrr (I love it…)

    . . . . . . .

  3. Now we know the answer to the age old question of what came first: the chicken or the egg.

    Clearly it was the entrepreneur (the ultimate professional) that came first. He/she introduced the question of the chicken or the egg to academia, so that academics could waste there lives in a circular reference, enabling the entreprenuer to get on with real life…

  4. Step 4 to 5 :
    You can’t divide by (a-b) because a=b. This is equivalent ti dividing by 0…

  5. Yeah, I remember this from school. In the first line you state that a equals b. You go from line 4 to 5 by dividing both sides by (a-b). If a = b, then you are in fact dividing by zero. The answer to that is undefined and causes the funny result.

  6. This makes no sense what so ever!
    Its as if each new line is a different problem with some info taken from the last line kept to the next. This IS NOT an equation being worked out, and after line 4 all the computations are totally off

    a+a != a
    a+B != b

    even if a=b, ESPECIALLY if A=B

    Lets assing arbitrary values, 5 in this case. So if A=B (or == b in my notation :P) then:

    5+5 =10 or 2a, 2b not 5

    I’m not going to tackle line 4 because I can’t remember the equasion you run it through when you have to multiply two parenthasies but I don’t think its equal.

  7. i think your mother is going downhill, you should be grateful for this website like rest of us are, its easy to give bad comments…be contructive

  8. I’m in 8th grade, and this is impossible! How in the world is A=B? and how did they get 2=1? But, A is the first letter in the alphabet, and B is the second, so maybe thats how they could have gotten it. I don’t know.

  9. The funny thing is that even if you ignore the fact the we divide by (a-b) which is zero, still at the end if 2a=a it doesn’t mean at all that 2=1.

  10. We want to solve for either a or b.

    How about we stick to the a+a=a ignoring the dividing by zero
    giving us 2a=a

    we don’t divide both sides by a but rather subtract a from both sides and get


  11. The last line says 2a = a, and therefore, a = a.

    To find out what “a” is, you’d divide the equation with “a” entirely – and when you do, you’re left with a = 0.

    Therefore, it means that the answer is a=0

  12. I think what happened was the twin on the left wrote the left side, while the twin on the right wrote the right side, so they both got different answers. Course, I’m only in 6th Grade so, whatever!

  13. Simply put, the Exterminator got it right.
    On line 5, you have a corret statement: (A + B) = B
    Forget the fact that A will solve to zero and all that bull about dividing by zero. At this time, all you know is A + B = B, therefore you reduce this equation by subtracting B from both sides leaving A + B – B = B – B, reducing we get A = 0.

  14. u guys r thinking this too hard…..the twin on the left first put (a+b) then put a+a thats not logical….i mean its common sense

  15. Ladies and gentlemen:

    If a = b as given,

    then (a-b)= 0 , correct???

    so, (fourth line becomes) (a + b) x 0 = b x 0
    then (fifth line becomes) 0 = 0

  16. I see many right answers and many wrong answers.

    I think there was an implied “for all a and b such that a = b”. From that, we can go all the way to “for all a and b such that a = b, (a + b)(a – b) = b(a – b)”. We CANNOT get to “for all a and b such that a = b, (a + b) = b” unless we know that (a – b) is never zero, when, in fact, it is always zero.

    Anyway, this leads to “for all a and b such that a = b and (a – b) /= 0, 2 = 1”. That statement in and of itself is perfectly true, but it’s a vacuous truth.

  17. ihope said it all… but for the ones that ask how a+b=b makes a+a=a appear is quite simple, since b=a…

    The only problem with the equation is really the fact of a – b = 0…

    This is really a good one that tricks many people… but 2 can be equal to 1 …. like 3 + 3 = 7, but that has another trick

  18. The problem is obvious, you can’t divide by zero to get from the fouth to the fifth line. It’s funny how many people keep trying to prove that a = 0, that would mean that given a=b, a=0, which essentially implies that all numbers equal zero.

  19. if a =b then a-b =0 but in step no 4 a-b has been canceled from both sides which is not possible as we cannot cancel out 0

  20. FIRST off
    it’s not a math problem
    it a bunch of equalities
    there are 4 that are correct(the first 4)
    and 4 that are incorrect (the second 4)
    if you know algebra even a little you would know that it isn’t a problem at all
    it is just confusing to the stupid
    it’s like me saying

  21. i have 1 more
    to prove door open = door closed
    solution: 1/2 door open = 1/2 door closed
    therefore 1/2 gets cancelled on both sides,
    hence door open = door closed

  22. So many answers to the same question
    With reply to Rembrance’s answer, try uneducated rather than stupid.

    All you numbskulls no nothing the answer lies here.

    a=2a-(2b or not 2b) that is the question
    or is that the answer???

    wheres your sense of fun!

  23. Let’s Begin with your Algebra I (Everything you do to one side, you must do to the other to keep the equation balanced):
    If I were to multiply both sides with a, we get:
    a*a=a*b or
    Now let’s add -b^2 to both sides (or subtract b^2 from both sides if this makes you comfortable):
    The next trick is knowing about “the difference of squares.” This just states that if I have a^2-b^2 then you can factor the problem as (a+b)(a-b). Same thing with ab-b^2. I can factor out b and get b(a-b). So now we have:
    But now we can divide the whole equation by (a-b). This is where the argument begins. Since a=b, you cannot divide the equation with (a-b) because a-b=0, and any number divided by 0 is undefined.
    BUT, if you COULD, then you would be left with:
    Now remembering that a=b then we can substitute b for a and we get:
    a+a=a or
    Now we divide the whole equation by a and we are left with:
    2 = 1.
    A great teaser for any math teacher out there (Hint, Hint)

  24. the step 4 is wrong.
    (a-b) can be cancelled both the sides unless a is not equal to b. since its given that a=b in the first step (a-b) cantt be cancelled both the sides and the result should be 0=0

  25. Yes, a-b=0, how can a number devided by zero, is infinity! The answer is not accurate! The working steps got problem. How ever, 0 divided by 0 that’s the answer is accepted by any numbers…..

  26. They tell you that a=b
    a and b are variables an can be any number and they equal.
    a-b is equal to zero and you can’t divide by zero.

  27. r u kidding me?! like im really gonna sit here all day trying to figer out this shit!!….but maybe i’ll give it a try…..oh i get it! on the 5th equation, (a+b)=b is wrong. its (a+b)=(a+b) because at the top it says a=b, so if a were to be, let’s say 2, so would b. so it would be incorrect to say (2+2)=2.

  28. Dude, the only thing weird is how they go from 2a=a, to 2=1. Don’t you subtract “a” from each side at that point?

  29. some guys really rack and crack their heads dont they

    robert, rememberance ihope are some.

    simple answer is devision be zero is not defined.

  30. All I can see is that if you multiply anything by 0 then the answer is O as seen in line 4. Besides, there should be a minus sign inbetween the two brakets on line 4.
    Anyway its no big deal, we all know the answer is wrong so we don’t need to debate.

  31. Guys, stop thinking so much, the fourth line isn’t equal, distributive property.

    (a-b)(a+b) = b(a-b) do the work out

    a^2 + b^2 = ab – b^2

    the second part equals zero, the first part is a positive number

  32. the mistake there is when 2a=a,
    he used that property in equations wherein you transfer a variable to the other side of the equation,
    so when you transfer a from 2a to the other side it would be a divide by a which in Algebra is equal to 1.
    but the mistake is variables are only equavalent to numbers when they are predefined with a specific no.

  33. This is NOT and OPTICAL ILLUSION. by this logic i could take any math problem….include it in some sort of picture and call it an optical illusion. First..Just because you threw in a picture of two guys does not cover the ‘optical’ requirement in optical illusion. second….This is not an ‘Illusion’ at all. its a math puzzle, a somwhat clever math puzzle yes, but in no way is it an illusion. Unfortunatly more and more of the “Illusions” that have been appearing on this site arent even close to the definition of optical illusion.

  34. The only way for 2a to equal a is if they’re both zero. The equation is totally ligitimate, but just cause 2a=a, doesn’t mean 2=1. Also, (a+b)(a-b)=b(a-b), you’re dividing by (a-b), not 0

  35. the algebra in itself is perfect. there is only one problem: the first line:


    which later leads to a division by zero, which is undefineable, and also against the eleventh commandment:

    “Thou shalt not divide by zero.”

    the reason why this one got so many replies with the correct answers is people enjoy pointing out other people’s errors… and frankly, so do I.

    that’s all, people

  36. I’m stumped on the first line.
    Shouldn’t a = a? The supposed fact that “a = b” is just stupid.
    End of story.
    And nobody else post about “cant divide by Zero” I’ve read enough of those replies.

  37. When you divide members of the equation for a factor you have to remove the case when the factor goes zero. so when you divide for a-b you have to remove a-b = 0 from the solutions which is the first equations! so you can’t divide for a-b, because you remove all the solutions!

  38. OK! any given number divided by 0 is not defined therefore the algebraic sequence must come to a halt at that point, as anything after it will be incorrect, hence 2=1! capiche?

  39. step 1 : a=b
    step 2 : a2 = ab

    but.. if both sides r being squared…
    isnt a2 = b2 n not ab?

    so basically .. der is only 1 mistake…
    right..or not??

  40. I figured it out. It’s the sum that states “a+b=b”. This is mathematically impossible and with these rules, I can see why the twins think that “2=1”. So the mistake is in the 6th line.

  41. As an engineer and former math teacher I’m saddened to see some of the ignorance displayed here. Taking the whole thing point-by-point (using standard Maple notation) let’s talk about it…
    The first line, Let a = b, is perfectly routine and shouldn’t raise any issues.

    The second line, multiplying both sides by a, we get a^2 = a*b which is also OK.

    In the third line things get very slippery. BY subtracting b^2 from both sides to arrive at a^2–b^2= a*b–b^2 we must recall our premise. Since we are subtracting like terms because a^2=b^2=a*b, we are already in trouble since terms on both sides of our equation are zero! Everything that comes after this step is based on a false premise that IF a*0=b*0 THEN a=b. This is a somewhat subtle use of the zero property over multipication for equations. It is closely related to division by zero.

    Factoring terms (we might think) we see that (a+b)*(a-b) = b*(a-b). We’ve now collected our zero into one of our terms, (a-b). This is crucial for the operation of the trick, but we actually left good math behind at step three.

    Now the Fait au complet comes in cancelling (dividing) the common (zero) terms on both sides giving us a+b = b (which can only be true for a=b=0, not all possible values of a!)

    completing the illusion, the trickster resubstitutes a for b simplifying to 2*a = a (since a = b), and then cancels a to conclude 2 = 1!

    Those who opined that the error came in step 4 or between 4 and 5 are close, but must remember that once you have a 0=0 situation (for BOTH RHS and LHS) in ANY equation it is useless. When that happens you must work the math differently. The best way to avoid this pitfall is to use AS FEW TERMS AS POSSIBLE!

    Of my last group of AB Calculus students (who scored 3s and 4s on their AP exams) none solved this puzzle! I’ve had three students and 5 colleagues figure it out (correctly) in 23 years of engineering and math teaching. I think the students are doing quite well! They are always better at getting “out of the box.” It’s no wonder the Fields Medals only go to those 40 and younger!

  42. ok, this is where they really really messed up. there were only variables involved before and then suddenly- boom! numbers. real, rational, old-fashioned, numerical, numbers

  43. a=b
    asq(a square)=ab
    2a x 0 = a(a-a)
    2a x 0 = 1
    a x 0 = 1

    I have not had math in 20 years but whatever I remember from grade 13 indicates that when you multiply a x 0 is always equals 1.

  44. This all makes sense except for the very last line.

    a=b or 2=2

    a^2=ab or 4=4

    a^2-b^2=ab-b^2 or 16-16=16-16 or 0=0

    (a+b)(a-b)=b(a-b) or (0+0)*(0-0)=0*(0-0)

    (a+b)=b or (0+0)=0

    a+a=a or 0+0=0

    2a=a or 2*0=0

    all true to this point.

    But 2*0 does not equal 1

  45. Given that anything multiplied by zero is always equal to zero.
    You could say for instance
    5 x 0 = 2 x 0
    doesn’t mean that 5=2

  46. This doesnt even make sense because if a=b then a+b cant equal b and a plus a cant equal a makin this whole thing false

  47. I’m not so sure how this is an illusion…

    if a=b, then its safe ot say that a and b hold the same value, which is pointless, but just for arguments sake, lets say it fits the scenario, if a is equal to b, then

    (a+b) does not = b, it equals 2b
    and a + a equal 2 a, so you have 2a = 2b, reduced means a=b, nothing proven.

    Plus, in the 4th and 5th lines, im not sure what the author did, but they divided by -0 or something weird like that, because (a-b) becomes (a+b), in which case b on the outside would have become a negative b, which would then make the proof, 1=-1, and technically all numbers are reflexive to their intergers.

    It took me a while though to figure this one out, props to the author.

  48. it shouldnt be b(a-b) in the middle step. It should be (b+b)(a-b) which gives you (b+b)(a-b)

    if someone already said this, fine, i didnt read past the first 10 really stupid answers.

    both are equal to the product ab-b^2, but there is a difference between the two. that is why there is something wrong with it, it has nothing to do with dividing with zero until later on if you decide to continue with it, which you wouldnt because they would look exactly teh same and to simplify you would cancel it all out.

  49. 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?

  50. 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! ^_^


  51. 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=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 0x0=0x0 (0)
    a^2-b^2=ab-b^2 is 0x0-0x0=0x0-0x0 (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!

  52. on the line after:
    the next line written is:
    this is impossible as the line should read:
    which would make a=0 and b=0

  53. lets start at the line that reads:
    the next line after that reads:
    this is impossible as the next line is supposed to read:
    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

  54. 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

  55. gothca

    Given: a=b

    a2 = ab
    a2-b2 = ab-b
    (a+b)(a-b)= b(a-b) remember given
    (a+b)x 0 = b x 0
    0 = 0
    LHS = RHS

    this guy sure failed math

  56. 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.

  57. 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?

  58. 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.

  59. 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?

  60. 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

  61. 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

  62. 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

  63. 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.

  64. 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.

  65. (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

  66. 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

  67. 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 ;)

  68. “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.

  69. 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…

  70. 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.

  71. 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!

  72. 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.

  73. 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!

  74. 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 :)

  75. 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.

  76. 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.


    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!!!!

  77. 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.

  78. 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

  79. 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.

  80. 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.

  81. 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. [email protected]

    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

  82. 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!!

  83. 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.

  84. 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 :)

  85. the mistake is from the 4th to the 5th line.
    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.

  86. they divided by zero :)
    no, seriously though, they did, when they divided by (a-b). if a=b, then a-b=0 :D

  87. 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.

  88. 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 =)

  89. 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.)

  90. @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)

  91. Hmmm it’s interesting but cancelling out a-b should be fine, let’s move forward

    now I never saw such cancellation where both sides are constants, so proper way to solve it is:

    hence a=0 and b=0

    Whenever you try to play with 0, 0 will play with you :)

  92. 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:)

    1. @ 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.

  93. 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.

  94. 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.

  95. 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.

  96. 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?


    1. Where a = b;
      (a+b)(a-b)= b(a-b) is true
      (a-b) = 0, therefore
      (a+b)(0) = b(0)
      0 = 0

      The problem occurs during the 5th step, where both sides are divided by (a-b), which as we know equals 0.

      Dividing by 0 can give you some interesting conclusions. Wrong conclusions, of course, but interesting ones.

    2. Dudes, i’m no math teacher, i’m no genie, but i can figure this is sooo easy….Why are u complicating so much with all those abberations with dividings and all that crap! Look at the big picture!
      Look at last 2 lines!
      2a=a is real and that gives you the value of a and b….which is zero, both, and that makes all the crap real…Except the last line…that conclusion comes from nowhere..
      So…? math experts….crap…

    3. its correct, they factories out b. if you say yours is correct multiply your working back in, its doesn’t equate

  97. We had this in math the other day and you have to see at line 4 to 5 were you divide with a-b but a-b is 0 and you can’t divide with 0

  98. I like how everyone is saying its a divide by 0 error, while in essence this is true, thats not the reason, when you divide both sides by a-b you are losing a root of the equation. Its fundamental Algebra I and II graphing and understanding. a-b=0 is one root of the equation if you graphed it out, but by dividing it out your taking away the root entirely thus making it not able to be solved with true math. Im a math major btw if anyone wants to know…

    1. Sqrt(-1)^4=1
      Sqrt(-1)=4th root of 1 = 1

      whats the mistake

      i thnk it is in 2nd line
      4th root of 1 is not only 1 but also i(iota)

      am i right or is there any other mistake.

    2. Even though I kinda drop HS, my best guess is that the mistake is the third line. On both sides the numbers have been automatically equaled to zero, from that point on.. everything else it’s a fail.

    1. A can be equal to B since they are both variables, variables can be anything. A and B could both be equal to 2 and it would still be correct.

  99. Twin A made the mistake in line 4.
    The problem is -a * -a = a squared, and so does a*a. So -a * -a = a*a. -a does not equal b.

    When a squared is broken down, Twin A ignored the -a possibility and assumed a is positive. This resulted in an equation that is one part of a simultaneous equation.

    A little work gives the simualtaneous equations
    a squared – b squared = ab – b squared, and
    b squared – a squared = ab – b squared

    This gives 0 = 2ab -2b squared, or
    2b squared = 2 ab
    so b squared = ab, and dividing by b gives
    b=a which is what we started with.

    Sorry guys, its not a division by zero problem, and for the math major, how about thinking about 2 minuses make a positive on multiplication.

    BTW, Twin B also made the same mistake on line 5

    1. Yes, it is. A=B so subtracting 2-2 is equivalent to subtracting A-B. The answer to both is zero.

  100. you cannot divide by zero.. so you cannot divide by (a-b) so you cant remove (a-b) one of the rules of basic math… cheerz

  101. I think the mistake appears in a different spot depending on the number you use.
    If a = b then we can assume any number we plug in will work fine for this equation, and i’ve worked out a way thats true.

    If we use a positive number, lets say 1, the mistake appears in line 5, 6, 7.
    Again since a and b are equal, we could use 1, 2, or 100 it wouldnt matter.

    Assume a = b *So 1 = 1*

    a² = ab SO (1² = 1*1) SO (1 = 1) true
    a²-b² = ab-b² SO (1²-1² = 1*1-1*1) SO (0=0)true
    (a+b)(a-b) = b(a-b) SO (1+1)(1-1) = 1(1-1) SO (2*0 = 1*0) SO 0=0 true
    (a+b) = b SO (1+1) = 1 SO 2 = 1 false
    a+a = a SO 1+1 = 1 SO 2 = 1 false
    2a = a SO 2*1 = 1 SO 2 = 1 false

    If a positive number is used, there seem to be 3 mistakes starting at line 5, and continuing through 6, 7 and the final solution.

    If 0 is used, the ONLY mistake is the solution by both twins.
    If every number (i.e. [a=0, b=0 SO a=b SO 0=0] true) in THIS equation is 0, the only outcome POSSIBLE is 0.

    I haven’t worked a negative number yet, but according to order of operations, the outcome should be the same as with a positive number.
    I don’t feel like typing anymore lol, so if somebody wants to work that out feel free.

  102. in proving two quantities you just have to work with the left in order to arrive at the right side equation or vice versa.
    you do not have to substitute again quantities like what is made on the sixth line where a is substituted in place of b.

  103. An easy one. Line 3 is 0=0. And in the next line, 0 is factored into 0 and 0.
    There is dividing by zero a few times in this, including at the very end.

    Just because 5*0 = 3*0 doesn’t mean 5=3

  104. actually when (a+b)(a-b)=b(a-b),

    the next step should be (a+b)(a-b)-b(a-b)=0, but the (a-b) cannot be removed.

    so, (a-b)(a+b-b)=0

    (a-b)=0 or a=0

    thus, a=0, b=0

  105. Here when a=b,then a+b= b (wrong formula) also a+a=a ( wrong formula) and 2a=a (wrong formula)
    Therefore to prove

    Let a=b=1,Then

    a+b=b i.e 1+1=1 ( which is wrong,it should be a+b=2a=2b)
    a+a=a i.e 1+1=1 ( which is wrong, it should be a+a=2a=2b)but

    a^2=ab ; a^2-b^2= ab- b^2 ; (a+b)(a-b)= b(a-b); are correct formula( it holds true when a=b)

    so, 2 cannot be equal to 1

    1. Here , the last line is wrong

      => 2a = a
      => 2a – a = 0
      => a = 0

      Here , ” a ” is a variable and 2 is a constant , therefore
      we cannot divide both sides of the equation by a variable
      ( here ” a ” ) when both sides of an equation the variable
      is multiplied by constant ( here 2 and 1 ) .
      Therefore 2 is not equal to 1

      Now , if we solve the equation for ” a ” then from
      third line we get

      => a^2 – b^2 = ab – b^2
      => a^2 = ab – b^2 + b^2
      => a^2 = ab
      => a^2 – ab = 0
      => a( a – b ) = 0

      Therefore , a = 0 and a – b = 0
      => a = b

      Thus , a has got two solution , a = 0 , b

      And , If we solve the equation for “b” then from third line
      we get

      => a^2 – b^2 = ab – b^2
      => a^2 = ab
      => b = a^2/a
      => b = a

      Therefore , b has only one solution , i.e. b = a

      It shows that ” b ” is a dependent variable of independent variable ” a “

  106. to all of you ignorants out there who didn’t notice.. 3rd & 4th lines from the right side..
    you can’t take a b from b^2…

  107. in Line 4, (a+b)=a
    according to Additive Identity Property this is only possible when b=0 and if is b=0 then a also become 0. If we replace the value of and be in any line then it well satisfy the equation except last.
    And as Me =D at 8th comment said. we cannot divide the any term with zero. and in the equation the 5th and 8th step result after dividing by 0 which is error.

  108. It’s absolutely wrong math operation. You can’t do step 5 (from step4 to step5). Div 0/0=(a-b)/(a-b), due this indefinite operation or forbidden math operation. 0/0 can be anything sample:
    So this is not math…:D

  109. hello…
    if u can prove 1=2, try to prove by taking two different quantities. when u assume a=b itself in the first line, it indicates u already accept 1=2. u r just substituting ‘a’ as 1and ‘b’ as2. i think your approach is wrong. sorry if my words hurt u…

  110. the mistake that the twin made is they added a and b at the fifth step… it should be a-b… that the right step… right?
    so, the final answer would be a=1
    thus 1 always equal as 1…

  111. That’s easy… if ab=b then it must mean that the product of the numbers is the same as the numbers themselves, therefore its 1. :)

  112. The Problem lies in the fourth line
    so if the both values are same then …
    therefore in the fourth line let us consider the value of a=b=4
    and b(a-b) = 4(4-4) = 4×0 = 0
    and also in the next step we see (a+b)=b which is not possible as we know a=b and b(a-b) = b*0
    and therefore (a-b)\(a-b) = 0/0 which is infinite

    SO easy solved in 10 seconds Lol and nobody here could solve this freaking thing :P

  113. I agree with Akshay.

    These steps simply manipulate the zero with multiplication & Division. Any equation should stop using them (x & /)if both sides become zero.In this equation, it had happened in the 3rd step itself.

    Consider this:
    5×0=0 & 100×0=0 (substitute 5 or 100 as you like)
    Can I say then, 5=100?
    That way, anything can be equal to anything else!

  114. Hello, everyone.. The condition for a^2-b^2 = (a+b)(a-b), is when a-b is not equal to zero. ie a should not be equal to b…

    The Complete formula is

    a^2-b^2 = (a+b)(a-b)
    when a-b != 0

  115. From a+a=a onwards…

    therefore, 2 is not equal to 1.. either a=0 or 2=0
    or a=2…:P

  116. here in third line
    or,b^2+b^2=b^2+b^2(since,a=b so a^2=b^2)
    therefore o=o

  117. The mistake is at the end.
    The equation only makes sense if a=0=b and you can’t divide by 0. Otherwise you get 2=1 which is obviously wrong.

  118. you cannot divide both sides by a-b because since they are the same you will get zero and you can’t divide by zero because you’ll get all these wacky answers

  119. OK. I guess I’m wrong. The 4th line is correct after computing for it.
    Now I have to look again.

  120. Going from the fourth to the fifth line we are dividing by (a-b). However, a = b, thus (a – b) = 0 and hence we are dividing by zero which is not possible. Thus the error.

  121. uh im in grade five and have no clue what that mathy stuff even is, but one guy is grabbing the exclamation mark ! this is a optical illusion website not a nerdy correction area!!!

  122. There is no problem with the equation.
    A equal B
    2nd line: xA on both sides
    3rd line: minus b squared
    4th line: correct factorizing
    5th line: minus (a-b)
    6th line: switched the b with a since a equal b
    7th line: divide by a on both sides
    8th line: 2=1 is the final anser therefor the first line is the mistake of the anser shod b a=b=0

  123. Here , the last line is wrong

    => 2a = a
    => 2a – a = 0
    => a = 0

    Here , ” a ” is a variable and 2 is a constant , therefore we cannot divide both sides of the equation by a variable ( here ” a ” ) when both sides of an equation the variable is multiplied by constants ( here 2 and 1 )

    Therefore 2 is not equal to 1

    Now , if we solve the equation for ” a ” then from
    third line we get

    => a^2 – b^2 = ab – b^2
    => a^2 = ab – b^2 + b^2
    => a^2 = ab
    => a^2 – ab = 0
    => a( a – b ) = 0

    Therefore , a = 0 and a – b = 0
    => a = b

    Thus , a has got two solution , a = 0 , b

    And , If we solve the equation for “b” then from third line
    we get

    => a^2 – b^2 = ab – b^2
    => a^2 = ab
    => b = a^2/a
    => b = a

    Therefore , b has only one solution , i.e. b = a

    It shows that ” b ” is a dependent variable of independent variable ” a “

  124. in the first line it is already stated that a=b.Therefore,a-b=0.Now, in the 4th line it is(a+b)(a-b)=b(a-b).and then in the fifth line it is shown as (a+b)=b by cancelling (a-b) from both sides which is wrong as 0/0 is indeterminate and it cant be cancelled out.There is the mistake.

  125. The only thing is that you cant take further calculation after step (a+b)(a-b)=b(a-b) as here a and b are equal so a-b = 0 and multiplication both side by 0 (a-b) will make both side 0, and as u know 0 = 0.

Leave a Reply

Your email address will not be published. Required fields are marked *