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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, 7 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, 2 months ago

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@Eddie The Head What if it has? What if this is the afterlife? Sharky Kesa · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months 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 · 3 years ago

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@Debarpan Adhikari I think 4th line has defect. Abdur Rehman Zahid · 2 years, 8 months ago

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@Debarpan Adhikari Really, 2+2=5, I checked it using MS Excel: pic.twitter.com/iS9BNigN3p ;) Jurii Mariinsky · 3 years 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, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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@Jaydee Lucero You found it. :D Ahmad Naufal Hakim Ashari · 3 years, 2 months ago

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@Jaydee Lucero I completely agree with you Vijitendra D · 3 years, 2 months ago

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@Ahmad Naufal Hakim Ashari u have to put both + and - while removing the root sign!! Gunjas Singh · 3 years, 2 months ago

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@Gunjas Singh Wrong Sharky Kesa · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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Zero by zero is indeterminate form.. You cannot cancel zero by zero.. (x-y)=0 Steve Stanley · 3 years, 2 months 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, 2 months ago

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u r not allowed to cancel 0 from both sides...i mean (x-y) Abhishek Bakshi · 3 years, 2 months ago

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(x-y) = 0. You took 0/0 = 1, which is actually indeterminate. Ashish Siva · 1 year, 3 months ago

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in 4th line, you cannot cancel zero (x-y=0) with zero of other side. Uzair Awan · 2 years, 8 months ago

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That's not possible Pratham Sharma · 2 years, 9 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, 9 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, 9 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, 9 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, 9 months ago

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x-y is not 0. you can not cancel a 0 term Pratham Sharma · 2 years, 9 months ago

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x-y is 0.Cannot cancel a 0 term Tapan Saraph · 2 years, 9 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, 11 months ago

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well.. (x-y)=0 so,, no Ruba HuSsein · 2 years, 11 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, 11 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, 11 months ago

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This should have been framed in a form of a question. Anuj Shikarkhane · 2 years, 12 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 · 3 years 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 · 3 years ago

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dividir entre cero Oliver Garcia · 3 years ago

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x=y => x-y=0 => can't divide ( x-y ) in line 4 Nguyên Nguyễn · 3 years, 1 month 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 · 3 years, 1 month ago

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x=y Aya Mohamed · 3 years, 1 month ago

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x-y=0, in step 4 x-y can't be eliminated Radha Krishnan B · 3 years, 1 month ago

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If x=y then (x-y) =0 and hence it cannot be divided on any side. Gautam Singh · 3 years, 1 month ago

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cancellation of (x-y) on both sides is a mistake Geetha Kumar Mamidisetty · 3 years, 1 month ago

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You cannot cancel X-Y until X=Y.... Madhukar Thalore · 3 years, 1 month 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 · 3 years, 1 month 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 · 3 years, 1 month 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 · 3 years, 1 month 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 · 3 years, 1 month ago

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x+y=y is the fallacy Sonika Kumar · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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u can't divide by x-y since x -y = 0 Austin Seiberlich · 3 years, 2 months ago

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YOu cannot cannot 0 both sides. Raja Ragh · 3 years, 2 months ago

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In step 4 you have cancelled 0 from both sides which is not possible Sagnik Ghosh · 3 years, 2 months ago

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divided by zero Marjun Fernandez · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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x-y=0 Ritesh Manna · 3 years, 2 months ago

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Substitute numbers from the beginning and then you get the fault. :P Shreya R · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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

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x-y/x-y = 0/0 math error Prateek Rai · 3 years, 2 months ago

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in the 5th step Math Man · 3 years, 2 months ago

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x+y ≠ y Hans Ramírez · 3 years, 2 months ago

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The element is zero shock swallow does not participate in the multiplication operation Ibrahim Çakır · 3 years, 2 months ago

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interesting! I can't find it Nguyen Thien Vy · 3 years, 2 months ago

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@Nguyen Thien Vy x=y, therefore x-y=0. Sharky Kesa · 3 years, 2 months 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, 2 months 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, 2 months ago

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x-y=0 => 0/0 Yuri Sacha Corrêa Lopes Da Silva · 3 years, 2 months ago

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x-y=0 João Victor · 3 years, 2 months 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, 2 months ago

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you are a master piece as you can divide 0 by 0 Uttaran Choudhurry · 3 years, 2 months ago

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u can't cancel x-y bcz it is 0. Harsh Bhavsar · 3 years, 2 months 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, 2 months 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, 2 months ago

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Hi Kesa i'm also confused with this riddle.there is an aliter proof also Abhinav Anand · 3 years, 2 months ago

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U r assuming (x-y) = o which is not the case !! Karan Nahar · 3 years, 2 months ago

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Here one can't cancel x-y from both sides Tushar Chaudhary · 3 years, 2 months ago

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Not sure that x-y=0 or not! Minh Bùi · 3 years, 2 months ago

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@Minh Bùi No, we are not assuming. If x=y, then obviously x-y=0 Kushagra Sahni · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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

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cancelling out the term (x-y) because that means you cancel zero. Darien Jonathan · 3 years, 2 months 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, 2 months ago

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so, we have fallacy in the third step itself. Arya Ukunde · 3 years, 2 months 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, 2 months ago

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You divided by 0. Robert Melville · 3 years, 2 months ago

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Division by zero is undefined Haider Ali · 3 years, 2 months ago

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x - y = 0 since 0/0 is not defined... its false Aakash Amish · 3 years, 2 months 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, 2 months 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, 2 months ago

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x-y = 0 You cannot cancel it from the equation Vivian Sudhir · 3 years, 2 months ago

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if x=y then x-y=0 we cant cancel 0 Satyam Mani · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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0 by 0 is not 1 my friend Vibhu Baibhav · 3 years, 2 months 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, 2 months ago

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you must not cancel x-y on both sides as both have value of 0 Aakash Khandelwal · 3 years, 2 months ago

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u cant cancel 0 from both the sides in step 4! x-y=0! Gunjas Singh · 3 years, 2 months 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, 2 months ago

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x squared - y squared = 0 Shivamani Patil · 3 years, 2 months ago

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why cant we divide zero by zero? Venkatesh Panganamamula · 3 years, 2 months 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, 2 months 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, 2 months 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, 2 months ago

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because of division by zero (x-y)=(1-1)=0 Mohamed Abdallah · 3 years, 2 months ago

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x = y x - y = 0 But in 5th step we cancelled x-y Vaibhav Chandan · 3 years, 2 months 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, 2 months 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, 2 months ago

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zero cannot be divided Dani Natanael · 3 years, 2 months ago

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@Dani Natanael zero cannot be divided "by" Satvik Golechha · 3 years, 2 months ago

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You got \( x + y = y \). Doesn't this imply \( x = 0 \) and consequently \( y = 0 \)? Shabarish Ch · 3 years, 2 months ago

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@Shabarish Ch Yes, but I meant for all values of \(x\) and \(y\). Sharky Kesa · 3 years, 2 months ago

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@Sharky Kesa Hmm..... Shabarish Ch · 3 years, 2 months 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, 2 months ago

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A very good one this is but you know the problem with this. Adarsh Kumar · 3 years, 2 months ago

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@Adarsh Kumar Yes I do. Sharky Kesa · 3 years, 2 months 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, 8 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, 11 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, 2 months 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, 2 months 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, 2 months 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, 6 months ago

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We can cancel only non zero common term in both sides of an equation Arun Kumar · 3 years 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 · 3 years, 1 month 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 · 3 years, 1 month 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, 7 months ago

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@Ahmed Zizo Many remarkable things are discovered as a result of clowning :) Silas Hundt Staff · 3 years, 7 months ago

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@Silas Hundt Is clowning the same as trolling, just more mathy? Alexander Sludds · 3 years, 7 months ago

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@Alexander Sludds I'm curious too, whats clowning? Yash Talekar · 3 years, 7 months ago

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