# 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
6 years, 11 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

-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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

No, you are wrong!!! What do you think sqrt. 16 is ? Of course, it is 4 but why can't it be -4 ?

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 4 months ago

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 :)

- 6 years, 6 months ago

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

- 6 years, 5 months ago

I think 4th line has defect.

- 6 years ago

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....

- 6 years, 6 months ago

This one's good,

$-1=-1$

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

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

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

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

$i^{2}=1$

$-1=1.\quad \square$

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

You found it. :D

- 6 years, 6 months ago

I completely agree with you

- 6 years, 6 months ago

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

- 6 years, 6 months ago

Wrong

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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;;;;;;

- 6 years, 6 months ago

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

- 6 years, 6 months ago

Yes I do.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

Yes, but I meant for all values of $x$ and $y$.

- 6 years, 6 months ago

Hmm.....

- 6 years, 6 months ago

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

- 6 years, 6 months ago

zero cannot be divided

- 6 years, 6 months ago

zero cannot be divided "by"

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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:)

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

why cant we divide zero by zero?

- 6 years, 6 months ago

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.)

- 6 years, 6 months ago

x squared - y squared = 0

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

0 by 0 is not 1 my friend

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

Division by zero is undefined

- 6 years, 6 months ago

You divided by 0.

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

so, we have fallacy in the third step itself.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

Step 1 & 2 are..

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

Not sure that x-y=0 or not!

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

Meaning, there is a division by zero.

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

x-y=0

- 6 years, 6 months ago

x-y=0 => 0/0

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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....

- 6 years, 6 months ago

interesting! I can't find it

- 6 years, 6 months ago

x=y, therefore x-y=0.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

x+y ≠ y

- 6 years, 6 months ago

in the 5th step

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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 :)

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

x-y=0

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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!

- 6 years, 6 months ago

divided by zero

- 6 years, 6 months ago

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

- 6 years, 6 months ago

YOu cannot cannot 0 both sides.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

x+y=y is the fallacy

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 5 months ago

x=y

- 6 years, 5 months ago

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.

- 6 years, 5 months ago

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

- 6 years, 5 months ago

dividir entre cero

- 6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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>

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

x-y is 0.Cannot cancel a 0 term

- 6 years, 1 month ago

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

- 6 years, 1 month ago

$x =y$ means

$x -y =0$

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

- 6 years, 1 month ago

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......

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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

- 6 years, 1 month ago

That's not possible

- 6 years, 1 month ago

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

- 6 years ago

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

- 4 years, 8 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 6 months ago

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

- 6 years, 3 months ago

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

- 6 years ago

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

- 6 years, 11 months ago

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

Staff - 6 years, 11 months ago

Is clowning the same as trolling, just more mathy?

- 6 years, 11 months ago

I'm curious too, whats clowning?

- 6 years, 11 months ago

@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

- 6 years, 5 months ago

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.

- 6 years, 5 months ago

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

- 6 years, 5 months ago