A Proof that 0 = 1 (Can You Spot the Mistake?)
Tricky Elementary School Problem, Can you solve it in 20 seconds? https://www.youtube.com/watch?v=jPxfk2ioWls
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A Proof that 0 = 1. I do not know what to believe any more... except that there is obviously a mistake! :)
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patrickJMT
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Iceman ___
You don't have to. You do √(1/2)(1/2)

Morad Morad
nice try

Girl More
ok that's also a 6...

Mehraj Chhetri
whenever we remove square root we have to remember that we will get 2 values simultaneously
one positive value and one negetive value...
x=√1 means x=+1 or x=1
in the above problem
(49/2)²=(59/2)²
(49/2)=√(59/2)²
49/2=(59/2) or 49/2=59/2
4=4 or 4=5 but 4 !=9 so 4=4 is correct
even 49/2=1/2 and 59/2=1/2
that means
(49/2)=(59/2)
when we open roots we will get two values one is positive something and one is negetive something

Black Widower XIX
Lol dude sq root (x^2) = x

Discover The World
Here is the mistake
(49/2)^2 = (59/2)^2
must be
49/2=59/2

Exterminator9000
If a=b, and b=c, that does not necessarily mean a=c...
Here's what I mean:
A=5
B=sqrt25
C=5
In that case, (where sqrt = square root of), A=/=C, since the properties of exponents and square roots are a bit different than those of usual addition, subtraction, multiplication, and division.

Noone At all
a=49/2= 1/2
b=59/2=1/2
Yes a^2=b^2 (they are both 1/4). But a != b. a^2=b^2 concludes that a=b OR a=  b, not that a=b

MlgEpicBanana69
how the heck is 49/2 = *5*9/2 its one number more. :P

Jordan
Step 8 is incorrect

Pi
If you squareroot, or in general ^(1/2n) n being a natural number, you need to use the absolute value.
So (49/2)² = (59/2)² becomes 49/2 = 59/2 and if you compute this it gets 0.5 = 0.5. :P
We can say we end up with 2 solutions: ±(49/2) = ∓(59/2)

Miloš Lošmi
U cant root both side

brett knoss
My daddy taught me that 0 x 0 =Fag! Daddy drank.

Bogdan Colesiu
You can't square root like that.

Mahesh Reddy
u r rong

john johnson
The step where you have square the factors will have to be invalid(they become positive), because that would mean that 20=20.

ConnorConnor
0:51 didja know that 6² = 36 aswell?

vineet nair
the square thing u did after factorising was wrong. (ab)*2 is (a+b)(ab) not (ab)(ab)

//Latest News & Videos // اخر الاخبار والفيديوهات//
close but (49/2)" "2 = (59/2)'' '' 2 does not mean 49/2= 59/2

Lusti Tilk
boring'

Zion J
It took me 3 years to find the mistake (I watched the video every six months), but I FINALLY found it! It's the step in which you take the square root of both sides. Just because x^2 = y^2 does not mean that x = y (although both x and y need to equal something wild; in the video they certainly do).

Derek Xiong
Just the magic about squared

Nikita Nopants
Where did 9 over 2 come from? If you want to write the number 9 as a fraction it's 9 over 1.

thomi dhia
i dont know if it can be factored

Dzenis Abdukic
am I the only one who realizes that he forgot brackets in the 4th line... or am I just stupit

Rendy Setiawan
1st statement is true
20 = 20
2nd statement is true
16  36 = 2545
20 = 20
3rd statement is true
4^2  4.9 = 5^2  5.9
16  36 = 2545
20 = 20
4th statement is the beginning of disaster ...
4^2  4.9 + 81/4 = 5^2  5.9 + 81/4
4^2  4.9 = 5^2  5.9
16  36 = 2545
20 = 20
But, if we calculate directly
4^2  4.9 + 81/4 = 1/4
and also
5^2  5.9 + 81/4 = 1/4
Contradiction, right ?
Why ?
Since we try to force the 4th statement is equivalent with 3rd statement, so we can prrof this wrong with a simple logic, here we go :
In general ways,
Let a,b,c,d are Real numbers and assume a  b = c  d (true), s.t
(a  b = c  d) \u003c\u003e (a  b + e = c  d + e) \u003e [(a  b) = a  b + e ] and [(c d) = c  d + e] IS WRONG
Proof :
a  b = c  d \u003e a b + e = c  d + e (TRUE \u003e TRUE : TRUE)
a b + e = c  d + e \u003e a  b = c d (TRUE \u003e TRUE : TRUE)
so,
a  b = c  d \u003c\u003e a b + e = c  d + e (ALSO TRUE)
[(a  b) = a  b + e ] (FALSE)
[(c d) = c  d + e] (FALSE)
so,
[(a  b) = a  b + e ] and [(c d) = c  d + e] (ALSO FALSE)
Therefore,
(a  b = c  d) \u003c\u003e (a  b + e = c  d + e) \u003e [(a  b) = a  b + e ] and [(c d) = c  d + e]
(TRUE \u003c\u003e TRUE) \u003e (False and False)
TRUE \u003e False
False
In case, we can briefly prove this :
4^2  4.9 = 5^2  5.9 \u003c\u003e [4^2  4.9 + 81/4 = 5^2  5.9 + 81/4] \u003e [4^2  4.9 = 4^2  4.9 + 81/4] AND [5^2  5.9 =5^2  5.9 + 81/4 ]
[20 = 20 \u003c\u003e 1/4 = 1/4] \u003e [20 = 1/4] AND [20 = 1/4]
[TRUE \u003c\u003e TRUE] \u003e [FALSE] AND [FALSE]
TRUE \u003e [FALSE]
FALSE...
DISASTER OF LOGIC THINKING

ronit mann
The answer is completely simple guys
See
20=20
Later he adds 81/4=20
If we see this as
0=0
Then 0*1=0*2
1=2
That's how it is

MC cashMax
The equation on the seventh row from the top, if you remove the exponent and do the math you will have a positive number on one side and negivitve on the other side of the equation but the absolute value of both numbers are equal, making the equation false. I believe you wrongly factored the fith row because the fith row is correct, however you disguised the mistake by exploiting the root problem.
The problem being, the square root of every number is both the square root of the number and the square root of the number as a negative number.Example: Btw 25^(1/2) is another means of describing the square root of 25, I just didn't have a root symbol
25^(1/2) = 525^(1/2) = (5)
5 x 5 = 255 x 5 = 25

Chandan Yogesh
the mistake was that u added 81

Priyank Oza
x^2=y^2 does not implies that x=y // we can also have x=y!! which is so in our case

halim taouilt
(44.5)^0.5 faux écriture

فائز زائف
hhhhhhhhhhhhh 2*4*9/2 and 2*5*9/2

Andy Dong
Kinda expected something more interesting than that

Gabriel Pawłowski
well this is not a proof

Laila Alalmaniya
you had to solve the binomic formulas, not just delete the "square"

Som Nath
We can write20=20*1 then 1 equals  20/20 which is little simple

sundar pal
X squre =y squre does not mean x = y

Hoozi Tomas
everything's ok until you assassinated maths with your way of solving an equasion of sort a^2=b^2

Navjot Allachauria
7th step is wrong step. You can't neglect the negative values whenever u take the sqrt.
(49/2)^2=(59/2)^2
(49/2)=+ or (59/2)
in this case
(49/2)=(59/2)
49/2=5+9/2
9=9.

lingqi Wen
The problem is on the completing the square between 2:25 to 2:49

Andrew Qi
I think (I don't know) 4 (9/2)^2 is not equal to 5(9/2)^2.

Anthony Kioupidis
The mistake has done when you erased the squares:
(49/2 )^2= (59/2)^2 =\u003e [ (89)/2]^2 = [(109)/2]^2 =\u003e ( 89)^2 = (109)^2 =\u003e (1)^2 = 1^2 =\u003e 1=1

David Masabo
first mistake is that u didn't group 4×9 and 5×9 so by system of operations the equation is wrong. finally, just because 1^2=1^2 doesn't mean that 1=1. example x^2=y^2 then x doesn't have to be equal to y. x^2=y^2, x^2=y^2 so in this case y can actually have 2 values and so can x so right answer should be x^2=+y^2

Aniket Ghosh
How is the square of a number negative?

Aniket Ghosh
Shouldn't the square root of these numbers be a lateral number?

Dominic Cingoranelli
Not to mention the issue of the squares, if you assume the sides are equal then of course it'll work. The thing with proofs, is you are changing one side (With clever forms of one, etc.) so that you get the other side.

X Magistrty X
Its easy n^2 = (n)^2
But square rooting both side give us n = n thats why it gives us 0 = 1
if you just solve without square rooting it gives.....
(1/2)^2 = (1/2)^2
L.H.S = R.H.S Hence proved

Siddharth Rodrigues
you can't cancel the squares on both sides
well in first look u got me

Franz
you can't square root both sides like that, if you evaluate (49/2)^2 and (59/2)^2 they both equal to a quarter.

Connor Sponsler
The 6th step has no basis is any real mathematics.

Giovanni Violo
ha sbagliato a "comporre" il quadrato di binomio di (49/2) e quindi di (59/2)

Μαρία Μουστοπούλου
ΠΟΎ ΕΊΝΑΙ ΤΟ ΑΠΌΛΥΤΟ ΡΕ ΓΕΛΟΊΕ ????

Rohan Albal
the real answer? you cant divide by zero !
