By Vurdlak on April 13, 2006, with 240 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!
(47 votes) (47 votes)Comments
240 Responses
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The mistake is in the 4th line :
(a+b)(a-b)= b(a-b) – This is false.
It should be:
(a+b)(a-b)= 2b(a-b)
And there you go..
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.
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…
its correct, they factories out b. if you say yours is correct multiply your working back in, its doesn’t equate
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
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…
Sqrt(-1)^4=1
Sqrt(-1)=4th root of 1 = 1
-1=1^2
-1=1
0=2
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.
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.
A cannot be equal to B
i get it 2 people equal as a whole!!<3
that is a fallacious solution. you cannot divide by 0 to attain a real whole no. since a=b and a-b=0.
Whoa, those answers are amazing o.0 I didn’t understand a word of them : D !!
a+a=a is fake. a+a=b
2 twins make 1 person because they think the same
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
a-b not equal to zero
that was magnificent..cant belive it but at long last its true..
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
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.
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.
Looking to the end result only, 2a=a doesn’t mean 2=1 it only means a=0, as 2×0=1×0
i = sqrt(-1) = (-1)^(1/2) = (-1)^(2/4) = [(-1)^2]^(1/4) = [1]^(1/4) = 1
(A+B)=B, Thats when they effed up!
Your equation is false:
(a-b)=0
though you devided by (a-b) (from 4th to 5th line)
It’s forbidden to divide by 0!
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
*LOL*
we should not cancel out (a-b) on both side of the equation… that is why the equation of 2=1 is arithmetically wrong…
answer is simple..
a=b
=>a-b=0
hence
a-b/a-b is not 1 means
0/0 is not equal to 1…. it is undefined
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
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)
and
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
2B or not 2B – that is the question!
dummy a-b is zero u cant divide by zero
since a=b then a-b=0 so we can not divide by zero this is the mistake
how can a=b? the mistake is a=b coz is not equal to b
you can not divide any thing by zero [a-a ]. That is meaningless in maths.
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…
a+b=b
Subtract b from both sides and u get
a=0
The end.
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.
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:
0/0=1=2=3=5=99999=anything
So this is not math…:D
a + b =/= b!!! simple as that!!! a=b, so a+b cant = b
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…
So freaking eazy, on step 5 you can’t take out (a-b) becuse a-b=0
toooo easy
IDIOTTTTTTTT
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…
As you know,
because a=b(a-b=0), you cannot divide both sides of the equation by (a-b).
You can not multiply by a-b
(a+b)(a-b)=2b(a-b) not b(a-b)
what are you on about?
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. :)
a to the power of 2 is aa not ab
the mistake is in the second line
Lol what its written a=b
then a^2=ab
As a=b so a-b=0
Therefore in 4th step,
i.e (a+b)(a-b)=b(a-b)
(a+b)(0)=b(0)
0=0
The Problem lies in the fourth line
(a+b)(a-b)=a^2-b^2
so if the both values are same then …
(a+a)(a-a)=a^2-a^2=0
therefore in the fourth line let us consider the value of a=b=4
(4+4)(4-4)=8*0=0
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
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!
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
the problems is in the 2nd line…
a^2 ≠ ab
a x a ≠ a x b
From a+a=a onwards…
a+a=a
=>a=a-a
=>a=o…
or,
a+a=a
=>2a=a
=>2a/a=0
=>2=0
therefore, 2 is not equal to 1.. either a=0 or 2=0
or a=2…:P
all this can be only
true if a and b = 0
all other is mess and so 2 isnt = 1 and will never be! :)
here in third line
a^2-b^2=ab-b^2
here,
a^2+b^2=b^2+b^2(since,a=b)
or,b^2+b^2=b^2+b^2(since,a=b so a^2=b^2)
or,2b^2=2b^2
therefore o=o
YOU CANT DIVIDE BY 0 AHOLES
b(a-b) = b(0) = 0
a = b
make a = 1
make b = 2
so 1 = 2
by saying its so doesn’t make it so
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.
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
a=b so (a-b)=0 and you can’t divide by 0
Can’t help but to comment on this.
The problem is in the 4th line.
a^2 – b^2 is NOT equal to (a+b)(a-b)
OK. I guess I’m wrong. The 4th line is correct after computing for it.
(a+b)(a-b)
a(a+b)-b(a+b)
a^2+ab-ba-b^2
a^2-b^2
Now I have to look again.
I guess these people answered the mystery, that you can’t divide by 0.
a+b=b means a=0.
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.
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!!!
A-b does not = 0
Hence 1=2
If that were true than 3=0
And 6=0
And 10= 0
And so on
By the way im 12
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