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1+1=1

\(x=y\)

\(x^2=xy\)

\(x^2 - y^2=xy - y^2\)

\((x + y)(x - y)= y(x - y)\)

\(x + y=y\)

\(1 + 1=1\)

What is the fallacy?

Note by Sharky Kesa
3 years, 5 months ago

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\(x = y\) This means \(x-y = 0\).

In step \(4\) you have cancelled \(0\) from both sides!!The world just exploded!!! Eddie The Head · 3 years ago

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@Eddie The Head What if it has? What if this is the afterlife? Sharky Kesa · 3 years ago

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it's good to see this.... see this what your proof says in terms of \(1\) \(1=1\\ { 1 }^{ 2 }=1\times 1\\ { 1 }^{ 2 }-{ 1 }^{ 2 }=1\times 1-{ 1 }^{ 2 }\\ (1+1)(1-1)=1(1-1)\\ and\quad you\quad cancelled\quad (1-1)\quad i.e.\quad 0\quad from\quad both\quad sides\\ which\quad means\quad \frac { 0 }{ 0 } \quad that\quad doesn't\quad exist\)

please give me suggestion if i am wrong Rishabh Jain · 3 years ago

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How about this?

-20=-20

4²-9x4=5²-9x5

4²-9x4+81/4=5²-9x5+81/4

(4-9/2)²=(5-9/2)²

4-9/2=5-9/2

4=5

2+2=5 Debarpan Adhikari · 3 years ago

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@Debarpan Adhikari If x^2=y^2, it doesn't necessarily mean that x=y, it means x=+-y as y can be negative also. Kushagra Sahni · 3 years ago

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@Kushagra Sahni But in the first step it is given that x=y which implies that both the variables have to have the same sign,unless they both are zero. Adarsh Kumar · 3 years ago

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@Adarsh Kumar No, you are wrong!!! What do you think sqrt. 16 is ? Of course, it is 4 but why can't it be -4 ? Kushagra Sahni · 3 years ago

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@Kushagra Sahni I get that x^2 =y^2 can have two outcomes but what I am trying to tell u is that x and y have the same signs and this can be seen in the 1st step of the falla ie.Get that. Adarsh Kumar · 3 years ago

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@Adarsh Kumar 16 is just a random example to prove that sqrt. of any real number can be both positive or negative, unless specified otherwise that x>0 or x<0 Kushagra Sahni · 3 years ago

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@Debarpan Adhikari when you take the square root of both sides, you have to add plus or minus sign, and remove extraneous solutions. Siva M. · 3 years ago

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@Debarpan Adhikari since x^1/2 is = +x therefore you can't write 4^2-9*4+81/4 = (4-9/2)^2 it must be written as

(9/2-4)^2=(5-9/2)^2 ie 9/2-4=5-9/2 => 9=9 Aakash Khandelwal · 2 years, 10 months ago

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@Debarpan Adhikari I think 4th line has defect. Abdur Rehman Zahid · 2 years, 6 months ago

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@Debarpan Adhikari Really, 2+2=5, I checked it using MS Excel: pic.twitter.com/iS9BNigN3p ;) Jurii Mariinsky · 2 years, 11 months ago

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@Debarpan Adhikari 4-9/2 is obviously a negative number it's equal to -0.5 you should have realized that the square root of a negative number do not exist imaginary only, so you cannot take the square root of the both sides :) Marchan Sy · 3 years ago

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@Debarpan Adhikari if we taking square root then there are two ans one is positive second one is negetive having magnitude same so we have to assume both cases.... Nandkishor Wankhede · 3 years ago

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This one's good,

\(-1=-1\)

\(\frac {-1}{1}=\frac {1}{-1}\)

\(\sqrt{\frac {-1}{1}}=\sqrt{\frac {1}{-1}}\)

\(\frac {\sqrt{-1}}{\sqrt{1}}=\frac {\sqrt{1}}{\sqrt{-1}}\)

\(\frac {i}{1}=\frac {1}{i}\)

\(i^{2}=1\)

\(-1=1.\quad \square\) Ahmad Naufal Hakim Ashari · 3 years ago

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@Ahmad Naufal Hakim Ashari Quotient rule for radicals \(\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\) only applies when \(x\) and \(y\) are non-negative, and that \(y \neq 0\). So, the third to fourth line of solution is faulty. Jaydee Lucero · 3 years ago

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@Jaydee Lucero You found it. :D Ahmad Naufal Hakim Ashari · 3 years ago

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@Jaydee Lucero I completely agree with you Vijitendra D · 3 years ago

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@Ahmad Naufal Hakim Ashari u have to put both + and - while removing the root sign!! Gunjas Singh · 3 years ago

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@Gunjas Singh Wrong Sharky Kesa · 3 years ago

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@Gunjas Singh YES, SAME CONCEPT. IF X^2=Y^2, THEN X IS NOT NECESSARILY = TO Y, IT CAN BE -Y ALSO Kushagra Sahni · 3 years ago

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If x=y , then x-y = 0 Hence we cannot cancel (x-y) in the 5th step coz a number divided by 0 is not defined in mathematics. So even if you prove that 1+1=1(which is not likely to happen).....it won't be practical in real life Vijitendra D · 3 years ago

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in 4th line do---x^2+y^2-2xy=xy-y^2

then solve it as, x^2+2y^2-3xy=0

it forms x^2-3xy+2y^2=0

x^2-xy-2xy+2y^2=0

x(x-y)-2y(x-y)=0

(x-2y)(x-y)=0

x=2y or x=y

hence x=y proved;;;;;; Yashasvini Sharma · 3 years ago

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Zero by zero is indeterminate form.. You cannot cancel zero by zero.. (x-y)=0 Steve Stanley · 3 years ago

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Mr. Sharky....you can divide an inequality only by a non zero number....here when you have assumed x=y...then x-y becomes zero....and this is the reason you are getting 2=1 Vikrant Yadav · 3 years ago

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u r not allowed to cancel 0 from both sides...i mean (x-y) Abhishek Bakshi · 3 years ago

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(x-y) = 0. You took 0/0 = 1, which is actually indeterminate. Ashish Siva · 1 year, 1 month ago

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in 4th line, you cannot cancel zero (x-y=0) with zero of other side. Uzair Awan · 2 years, 5 months ago

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That's not possible Pratham Sharma · 2 years, 7 months ago

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x=y implies x-y=0.In step 4 we have divided both sides by 0.So,it's 0/0 which is undefined. Kapil Chandak · 2 years, 7 months ago

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its wrong when x + y = y, x=y -y or x=0 so you are dead wrong man Shashank Madhusudhan · 2 years, 7 months ago

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You should watch forth step you ignored x-y ... It Means u r handling it like a real number...but if you transfer both x-y in one side you will see that u have made 0/0 undefined form...... Puneet Mehra · 2 years, 7 months ago

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\( x =y \) means

\( x -y =0\)

in step 4 you are dividing the expression by zero which is not possible . this is the fallacy Megh Choksi · 2 years, 7 months ago

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x-y is not 0. you can not cancel a 0 term Pratham Sharma · 2 years, 7 months ago

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x-y is 0.Cannot cancel a 0 term Tapan Saraph · 2 years, 7 months ago

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you have just factored 0 , there are many solutions .. just like 9 x 0 = 0 Okijo Setsu · 2 years, 9 months ago

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well.. (x-y)=0 so,, no Ruba HuSsein · 2 years, 9 months ago

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as per you second last equation, x is multiple of 2y .. so "x" never is equal to y.. Muhammad Ahsan · 2 years, 9 months ago

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since x=y as stated in the problem, dividing both sides by (x-y) will result in an undefined equation as (x-y) = 0. (x+y)/0 = Undefined so the equality does not hold. <It is equivalent to saying that 1/0=2/0 clearly undefined> Andrew Yu · 2 years, 9 months ago

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This should have been framed in a form of a question. Anuj Shikarkhane · 2 years, 10 months ago

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since x-y=0 you can't cut x-y on both the sides of step no.4 Aakash Khandelwal · 2 years, 10 months ago

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You cancelled x-y with each other. what if values of it is 0? Cancelling of 0/0 is not 1. Btw, result is x=y=0 M.S. Saggoo · 2 years, 11 months ago

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dividir entre cero Oliver Garcia · 2 years, 11 months ago

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x=y => x-y=0 => can't divide ( x-y ) in line 4 Nguyên Nguyễn · 2 years, 11 months ago

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From (x+y)(x-y) = y(x-y) to x+y = y, you divided each side by (x-y), but since you cannot divide by 0, this only applies if x-y does not equal 0, and since you set both x and y as 1, x-y does equal 0, which makes the equation invalid. Anthony Ng · 2 years, 11 months ago

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x=y Aya Mohamed · 2 years, 11 months ago

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x-y=0, in step 4 x-y can't be eliminated Radha Krishnan B · 2 years, 11 months ago

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If x=y then (x-y) =0 and hence it cannot be divided on any side. Gautam Singh · 2 years, 11 months ago

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cancellation of (x-y) on both sides is a mistake Geetha Kumar Mamidisetty · 2 years, 11 months ago

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You cannot cancel X-Y until X=Y.... Madhukar Thalore · 2 years, 11 months ago

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because x=y so x-y= 0 => in step 4 you can change both sides to 0 Phạm Đỗ Thiên Ấn · 2 years, 12 months ago

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x-y, since x-y=0, dividing any number by 0 is undefined (0 by 0 is indeterminate). Bernard Ian Nieto · 2 years, 12 months ago

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the eqation (x+y)(x-y)=y(x-y) has two solutions ,first one is (x+y)=y and (x-y)=0.Which gives x=y,the right solution. Vishal Yadav · 2 years, 12 months ago

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Division by zero. Since x = y, x - y = 0. We cannot use the Cancellation Law of Multiplication for a factor of zero. Jaybee Penaflor · 2 years, 12 months ago

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x+y=y is the fallacy Sonika Kumar · 3 years ago

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cannot divided by 0 two sides of equation in step 4 (//because x=y so x-y=0) Phunn Boonchouchouy · 3 years ago

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Take a note from \[x=y\] The false statement is when dividing both sides by the difference of x& y where it is indeterminate with denominator of x-y=0. John Aries Sarza · 3 years ago

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x - y = 0 and any no. divided by zero is infinity so,we can't cut x-y from both sides. Mayank Mahajan · 3 years ago

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u can't divide by x-y since x -y = 0 Austin Seiberlich · 3 years ago

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YOu cannot cannot 0 both sides. Raja Ragh · 3 years ago

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In step 4 you have cancelled 0 from both sides which is not possible Sagnik Ghosh · 3 years ago

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divided by zero Marjun Fernandez · 3 years ago

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when you cancel x-y in LHS with RHS,it directly means that x not equal to y,but then you are putting x=y in next step! RamanDeep Singh · 3 years ago

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you cannot cancel x-y to both sides since x-y = 0. any number divided by 0 is undefined. Cabug John Carlo · 3 years ago

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x=y x-y=0 so (x+y)(x-y)=y(x-y) (x+y)(x-y)/(x-y)=y as (x-y)=0 (x+y)(x-y)/0 which is not exist Saranya Naha roy · 3 years ago

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x-y=0 Ritesh Manna · 3 years ago

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Substitute numbers from the beginning and then you get the fault. :P Shreya R · 3 years ago

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since x=y, x-y=0,and in the next step the division isnt possible because division by 0 is not defined in mathematics Benson Thomas · 3 years ago

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In the 4th step, (x-y) is on the both sides. dividing (x-y) by (x-y) gives 0/0, which is the indeterminate form. That is the fallacy Eshan Abbas · 3 years ago

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x-y = 0 & you cancelled 0 on both sides ( 0/0 is not 1 , but undefined). Anchit Virmani · 3 years ago

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x = y

x - y = 0

In the fourth step you divided by (x-y), or 0. And when divided by 0, it should be 0 = 0 not 1 + 1 = 1 :) Emmanuel John Baliwag · 3 years ago

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x^2 =y^2 doesnot imply x=y Sunitha Bhadragiri · 3 years ago

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(x-y) can not be divided from both sides as x=y makes it divided by zero. Tarak Das · 3 years ago

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(x+y)(x-y) = y(x-y) only proves that 0=0 x-y cannot be cancelled out as we usually do Naveen Mathew · 3 years ago

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x-y/x-y = 0/0 math error Prateek Rai · 3 years ago

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in the 5th step Math Man · 3 years ago

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x+y ≠ y Hans Ramírez · 3 years ago

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The element is zero shock swallow does not participate in the multiplication operation Ibrahim Çakır · 3 years ago

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interesting! I can't find it Nguyen Thien Vy · 3 years ago

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@Nguyen Thien Vy x=y, therefore x-y=0. Sharky Kesa · 3 years ago

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in 3rd step there is (x^2-y^2) at l.h.s. & in next step it is written as (x+y)(x-y),but it is not applicable for the equal numbers which is the condition given in 1st step.... Utkarsh Tyagi · 3 years ago

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you just cancelled two zeroes respectively from the lhs and the rhs in step 4. This cant be done Udayan Joshi · 3 years ago

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x-y=0 => 0/0 Yuri Sacha Corrêa Lopes Da Silva · 3 years ago

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x-y=0 João Victor · 3 years ago

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At the fifth step, we arrive at the result- x+y=y.Let us think at once, at what values of x and y, can this be true ?only when x=y=0.This is the only solution to x+y=y.If this solution follows , we can never get the result 1+1=1.This is the fallacy. Dheeman Kuaner · 3 years ago

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you are a master piece as you can divide 0 by 0 Uttaran Choudhurry · 3 years ago

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u can't cancel x-y bcz it is 0. Harsh Bhavsar · 3 years ago

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On 4th step (x+y)(x-y)=y(x-y), you have cancelled out x-y from both the sides. But according to the assumption made, where x=y => x-y=0. Also 0/0 is undetermined not 1 , so that step is wrong. Yash Mohan Sharma · 3 years ago

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In step 4, you divided both sides by (x - y) to arrive at x + y = y.

Meaning, there is a division by zero. James Canaveral · 3 years ago

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Hi Kesa i'm also confused with this riddle.there is an aliter proof also Abhinav Anand · 3 years ago

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U r assuming (x-y) = o which is not the case !! Karan Nahar · 3 years ago

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Here one can't cancel x-y from both sides Tushar Chaudhary · 3 years ago

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Not sure that x-y=0 or not! Minh Bùi · 3 years ago

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@Minh Bùi No, we are not assuming. If x=y, then obviously x-y=0 Kushagra Sahni · 3 years ago

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Since x=y, x-y=0... if you divide a number by zero it will be undefined, so you cannot divide the equation by x-y. Marchan Sy · 3 years ago

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there are more than one problem in your answer the first one when you divide the both side on (x-y) the rule is ( x - y != 0 ) and x = y so x - y = 0 and you can't divide the second one is when you reach to the line before the last one x + y = y so that will be x = y - y that's give x = 0 Wael Fawakhiri · 3 years ago

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in step 4 you cancelled (x-y) on both sides but as x=y that means that x-y = o and any expression divided by 0 is not defined. Abhisek Mohanty · 3 years ago

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x-y can be cancelled when you x-y is Not equal to zero means x is not equal to y Kanthi Deep · 3 years ago

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Step 1 & 2 are.. Thiliban Varadharajan · 3 years ago

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cancelling out the term (x-y) because that means you cancel zero. Darien Jonathan · 3 years ago

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if u remove (x-y) on both side then a condition x is not equal to y should be followed.......so it is wrong. Vivek Yadav · 3 years ago

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so, we have fallacy in the third step itself. Arya Ukunde · 3 years ago

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In the third step we subtract y^2 from both the sides. Then we get x^2 - y^2 on left hand side. But x^2 - y^2 = 0 as x = y , x^2 = y^2 , x^2 - y^2 = 0. Arya Ukunde · 3 years ago

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You divided by 0. Robert Melville · 3 years ago

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Division by zero is undefined Haider Ali · 3 years ago

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x - y = 0 since 0/0 is not defined... its false Aakash Amish · 3 years ago

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if x = y , then x-y = 0, therefore you cant use x-y to divide your algebraic expressions Jay Tio · 3 years ago

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We have seen and it is easily observed that the note posted by Sharky Kea is wrong (I mean to say 1 + 1 is not equal to 1) . The truth is when we solve a equation we actually don't know the value of the variable or variables.But in this case we already have put that x = y so by this we know that x - y = 0 so we can't just cancel x - y on both sides . In case we don't know the value or if x is equal to y or x is greater than y or something else..... In that case we can cancel x - y on both the sides as we don't know the value of it , it may be 1 or 4 or even 0 . But because can't assume that it will definitely be 0 so we can cancel x - y on both sides and then the answer will come to be x = 0 and if we substitute it , it holds true. So, 1 + 1 is not equal to 1. Utkarsh Dwivedi · 3 years ago

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x-y = 0 You cannot cancel it from the equation Vivian Sudhir · 3 years ago

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if x=y then x-y=0 we cant cancel 0 Satyam Mani · 3 years ago

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what if i say 1=1
25-24=41-40
(3x3)+(4x4)-(2x3x4)=(5x5)+(4x4)-(2x5x4)
(3-4)^2=(5-4)^2
3-4=5-4 (taking root both sides)
-1=1 Piyush Patnaik · 3 years ago

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sir you can't just remove what is spouse to be one of the solutions and say hi everybody the algebra is wrong.... Ahmed Sheweita · 3 years ago

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The error is in line 4 x-y=0 It is incorrect for divided with this. Reajul Haque · 3 years ago

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0 by 0 is not 1 my friend Vibhu Baibhav · 3 years ago

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Actually it must be: (2-1)^{2} = 2^{2}-2x2x1 +1^{2} =1^{2}-2x1x2+2^{2} =(1-2)^{2} i.e., (2-1)^{2}=(1-2)^{2} =2-1=1-2
=4=2 Hence, 2=1 Shreyans Badjatay · 3 years ago

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you must not cancel x-y on both sides as both have value of 0 Aakash Khandelwal · 3 years ago

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u cant cancel 0 from both the sides in step 4! x-y=0! Gunjas Singh · 3 years ago

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x = y i.e. x- y = 0. But we cancelled x - y from LHS and RHS in the fourth step, which means you cancelled 0!!! I was in a condition of getting a 'Heart Attack'!!!!!!!!!!!!!!! Vaibhav Chandan · 3 years ago

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x squared - y squared = 0 Shivamani Patil · 3 years ago

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why cant we divide zero by zero? Venkatesh Panganamamula · 3 years ago

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@Venkatesh Panganamamula That could assume ANY value (i.e. 0, pi, 13, 23523523523,e, you get the point where y,o,u,g,e,t,t,h,e,p,o,i,n, and t are real numbers, etc.) David Lee · 3 years ago

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I have objection ! you have said that x=y then x-y=0 then why are you cancelling x-y on both sides? Kalim Ullah · 3 years ago

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Here (X-y)=0 so we can't devide it from both side so the 4yh step is wrong Meet Patel · 3 years ago

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because of division by zero (x-y)=(1-1)=0 Mohamed Abdallah · 3 years ago

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x = y x - y = 0 But in 5th step we cancelled x-y Vaibhav Chandan · 3 years ago

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IN STEP 4, YOU CANCELLED X-Y FROM BOTH SIDES AND X-Y WILL BE 0. SO ACTUALLY YOU DIVIDED BY 0 WHICH IS NOT DEFINED. SO, YOU CAN'T PROCEED TO STEP 5. GOOD IF YOU KNOW THIS, IT MEANS THAT YOU ARE GOOD AT MATHS BECAUSE YOU KNOW THE FUNDAMENTALS OF IT. THESE KIND OF QUESTIONS STRENGTHEN YOUR CONCEPT IN MATHEMATICS. I WOULD BE OBLIGED IF YOU POST MORE QUESTIONS OF THIS KIND. I LOVE MATHS AND WANT TO STRENGTHEN EACH OF MY CONCEPTS:) Kushagra Sahni · 3 years ago

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if x=y then you should not cancell x-y on both sides of equality Sai Krishna Chary · 3 years ago

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zero cannot be divided Dani Natanael · 3 years ago

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@Dani Natanael zero cannot be divided "by" Satvik Golechha · 3 years ago

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You got \( x + y = y \). Doesn't this imply \( x = 0 \) and consequently \( y = 0 \)? Shabarish Ch · 3 years ago

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@Shabarish Ch Yes, but I meant for all values of \(x\) and \(y\). Sharky Kesa · 3 years ago

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@Sharky Kesa Hmm..... Shabarish Ch · 3 years ago

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@Shabarish Ch y can be any value in this case, since only x has to be 0. ((According to x + y = y)) Zaid Baig · 3 years ago

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A very good one this is but you know the problem with this. Adarsh Kumar · 3 years ago

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@Adarsh Kumar Yes I do. Sharky Kesa · 3 years ago

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If \(x=y\)then \(x-y=0\) so in the 4th step you are factoring as \((x+y)(0)=y(0)\) and in the the 5th step you are dividing both sides by \(0\) and taking the result as \(1\) which is not allowed Abdur Rehman Zahid · 2 years, 6 months ago

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Why x=y if x=y then at last what is the necessity that x+y=x Subrat Panigrahi · 2 years, 9 months ago

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here, given that x=y and it means x-y=0.... and in 4th step we cannot cancel out (x-y) from both side... so this is totally wrong Preyans Raval · 3 years ago

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since x=y,x-y=0 in 4th step u have divided by x-y=0 which is not allowed Maninder Kaur · 3 years ago

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(x + y )(x - y) = y(x - y) can yield to (x+y) = y only when x not equal to y. Chenchu Krishna · 3 years ago

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Comment deleted Dec 23, 2014

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@Shashi Kant Sharma No, we are supposing \(x = y\). If they were equal this could be possible. Sharky Kesa · 3 years, 4 months ago

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We can cancel only non zero common term in both sides of an equation Arun Kumar · 2 years, 10 months ago

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@Sharky Kesa Hi This note is similar to my question 'Is it True?'.. check out https://brilliant.org/community-problem/is-it-true/?group=enKxDb2d6cg4&ref_id=201435 Ritu Roy · 2 years, 11 months ago

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@Ritu Roy Well, technically, the problem is similar to my note. I posted this over 5 months ago, Eddie had reshared about 3 weeks ago and it became a hit. Before, it wasn't. Sharky Kesa · 2 years, 11 months ago

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This talk is wrong, because it can not be divided by zero to zero, or equal to, this clowning Ahmed Zizo · 3 years, 5 months ago

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@Ahmed Zizo Many remarkable things are discovered as a result of clowning :) Silas Hundt Staff · 3 years, 5 months ago

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@Silas Hundt Is clowning the same as trolling, just more mathy? Alexander Sludds · 3 years, 5 months ago

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@Alexander Sludds I'm curious too, whats clowning? Yash Talekar · 3 years, 5 months ago

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