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

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......

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>

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.

@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

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.

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.

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....

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.

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.

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

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.

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.

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

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.

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'!!!!!!!!!!!!!!!

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

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 :)

@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
–
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.

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:)

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## Comments

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TopNewest(x-y) = 0. You took 0/0 = 1, which is actually indeterminate.

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in 4th line, you cannot cancel zero (x-y=0) with zero of other side.

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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

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That's not possible

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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.

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its wrong when x + y = y, x=y -y or x=0 so you are dead wrong man

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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......

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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

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x-y is not 0. you can not cancel a 0 term

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x-y is 0.Cannot cancel a 0 term

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Why x=y if x=y then at last what is the necessity that x+y=x

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you have just factored 0 , there are many solutions .. just like 9 x 0 = 0

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well.. (x-y)=0 so,, no

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as per you second last equation, x is multiple of 2y .. so "x" never is equal to y..

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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>

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This should have been framed in a form of a question.

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since x-y=0 you can't cut x-y on both the sides of step no.4

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We can cancel only non zero common term in both sides of an equation

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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

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dividir entre cero

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x=y => x-y=0 => can't divide ( x-y ) in line 4

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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.Log in to reply

x=y

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x-y=0, in step 4 x-y can't be eliminated

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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

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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.

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If x=y then (x-y) =0 and hence it cannot be divided on any side.

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cancellation of (x-y) on both sides is a mistake

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You cannot cancel X-Y until X=Y....

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because x=y so x-y= 0 => in step 4 you can change both sides to 0

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x-y, since x-y=0, dividing any number by 0 is undefined (0 by 0 is indeterminate).

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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.

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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.

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x+y=y is the fallacy

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cannot divided by 0 two sides of equation in step 4 (//because x=y so x-y=0)

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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.

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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

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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.

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u can't divide by x-y since x -y = 0

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YOu cannot cannot 0 both sides.

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In step 4 you have cancelled 0 from both sides which is not possible

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divided by zero

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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!

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you cannot cancel x-y to both sides since x-y = 0. any number divided by 0 is undefined.

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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

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x-y=0

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Substitute numbers from the beginning and then you get the fault. :P

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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

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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;;;;;;

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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

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since x=y,x-y=0 in 4th step u have divided by x-y=0 which is not allowed

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Zero by zero is indeterminate form.. You cannot cancel zero by zero.. (x-y)=0

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x-y = 0 & you cancelled 0 on both sides ( 0/0 is not 1 , but undefined).

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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 :)

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x^2 =y^2 doesnot imply x=y

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(x-y) can not be divided from both sides as x=y makes it divided by zero.

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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

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x-y/x-y = 0/0 math error

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in the 5th step

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x+y ≠ y

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The element is zero shock swallow does not participate in the multiplication operation

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interesting! I can't find it

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x=y, therefore x-y=0.

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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....

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you just cancelled two zeroes respectively from the lhs and the rhs in step 4. This cant be done

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x-y=0 => 0/0

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x-y=0

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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.

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(x + y )(x - y) = y(x - y) can yield to (x+y) = y only when x not equal to y.

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you are a master piece as you can divide 0 by 0

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u can't cancel x-y bcz it is 0.

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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.

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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.

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Hi Kesa i'm also confused with this riddle.there is an aliter proof also

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U r assuming (x-y) = o which is not the case !!

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Here one can't cancel x-y from both sides

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No, we are not assuming. If x=y, then obviously x-y=0

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Not sure that x-y=0 or not!

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No, we are not assuming. If x=y, then obviously x-y=0

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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.

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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

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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.

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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

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Step 1 & 2 are..

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cancelling out the term (x-y) because that means you cancel zero.

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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.

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so, we have fallacy in the third step itself.

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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.

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You divided by 0.

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Division by zero is undefined

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x - y = 0 since 0/0 is not defined... its false

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if x = y , then x-y = 0, therefore you cant use x-y to divide your algebraic expressions

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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.

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x-y = 0 You cannot cancel it from the equation

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if x=y then x-y=0 we cant cancel 0

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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

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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....

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The error is in line 4 x-y=0 It is incorrect for divided with this.

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0 by 0 is not 1 my friend

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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

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you must not cancel x-y on both sides as both have value of 0

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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

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u cant cancel 0 from both the sides in step 4! x-y=0!

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This one's good,

\(-1=-1\)

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

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

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

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

\(i^{2}=1\)

\(-1=1.\quad \square\)

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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.

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I completely agree with you

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You found it. :D

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u have to put both + and - while removing the root sign!!

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Wrong

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YES, SAME CONCEPT. IF X^2=Y^2, THEN X IS NOT NECESSARILY = TO Y, IT CAN BE -Y ALSO

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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'!!!!!!!!!!!!!!!

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x squared - y squared = 0

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why cant we divide zero by zero?

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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.)

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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?

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

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because of division by zero (x-y)=(1-1)=0

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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

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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

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I think 4th line has defect.

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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

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Really, 2+2=5, I checked it using MS Excel: pic.twitter.com/iS9BNigN3p ;)

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4-9/2 is obviously a negative number

it's equal to -0.5you should have realized that the square root of a negative number do not existimaginary only, so you cannot take the square root of the both sides :)Log in to reply

If x^2=y^2, it doesn't necessarily mean that x=y, it means x=+-y as y can be negative also.

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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.

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when you take the square root of both sides, you have to add plus or minus sign, and remove extraneous solutions.

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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....

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x = y x - y = 0 But in 5th step we cancelled x-y

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u r not allowed to cancel 0 from both sides...i mean (x-y)

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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:)

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if x=y then you should not cancell x-y on both sides of equality

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zero cannot be divided

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zero cannot be divided "by"

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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

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You got \( x + y = y \). Doesn't this imply \( x = 0 \) and consequently \( y = 0 \)?

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ycan be any value in this case, since only x has to be 0. ((According tox + y = y))Log in to reply

Yes, but I meant for all values of \(x\) and \(y\).

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Hmm.....

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A very good one this is but you know the problem with this.

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Yes I do.

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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!!!

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What if it has? What if this is the afterlife?

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This talk is wrong, because it can not be divided by zero to zero, or equal to, this clowning

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Many remarkable things are discovered as a result of clowning :)

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Is clowning the same as trolling, just more mathy?

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