1+1=1

$x = y$ This means $x-y = 0$.

In step $4$ you have cancelled $0$ from both sides!!The world just exploded!!!

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

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

This one's good,

$-1=-1$

$\frac {-1}{1}=\frac {1}{-1}$

$\sqrt{\frac {-1}{1}}=\sqrt{\frac {1}{-1}}$

$\frac {\sqrt{-1}}{\sqrt{1}}=\frac {\sqrt{1}}{\sqrt{-1}}$

$\frac {i}{1}=\frac {1}{i}$

$i^{2}=1$

$-1=1.\quad \square$

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

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

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

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

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;;;;;;

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

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

zero cannot be divided

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

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:)

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

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

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

I have objection ! you have said that x=y then x-y=0 then why are you cancelling x-y on both sides?

why cant we divide zero by zero?

x squared - y squared = 0

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

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

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

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

0 by 0 is not 1 my friend

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

sir you can't just remove what is spouse to be one of the solutions and say hi everybody the algebra is wrong....

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

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

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

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.

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

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

Division by zero is undefined

You divided by 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.

so, we have fallacy in the third step itself.

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

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

Step 1 & 2 are..

x-y can be cancelled when you x-y is Not equal to zero means x is not equal to y

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.

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

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.

Not sure that x-y=0 or not!

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

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

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

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

Meaning, there is a division by zero.

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.

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

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

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.

x-y=0

x-y=0 => 0/0

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

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....

interesting! I can't find it

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

x+y ≠ y

in the 5th step

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

(x+y)(x-y) = y(x-y) only proves that 0=0 x-y cannot be cancelled out as we usually do

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

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

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 :)

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

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

since x=y, x-y=0,and in the next step the division isnt possible because division by 0 is not defined in mathematics

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

x-y=0

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

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

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!

divided by zero

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

YOu cannot cannot 0 both sides.

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

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

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.

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

x+y=y is the fallacy

Division by zero. Since x = y, x - y = 0. We cannot use the Cancellation Law of Multiplication for a factor of zero.

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.

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

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

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

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

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

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

x=y

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.

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

dividir entre cero

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

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

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

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>

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

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

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

x-y is 0.Cannot cancel a 0 term

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

$x =y$ means

$x -y =0$

in step 4 you are dividing the expression by zero which is not possible . this is the fallacy

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......

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

x=y implies x-y=0.In step 4 we have divided both sides by 0.So,it's 0/0 which is undefined.

That's not possible

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

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

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

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

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

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

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

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

@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

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