No +Marcel Wagner . Try again. :-)

### Francesco Fiorani

Challenges -Try to find the error!

44

14

51 comments

Marcel Wagner

+

1

2

1

2

1

+Francesco Fiorani at the first Moment i thought it was pi at the legt side and the right sie xD

Taking the square root at the end leaves you with two solutions (+/-). The one shown is the wrong one, the other one will lead to exactly what we started with.

You know what, it's impossible to solve this formula, because #9 really doesn't exist. That's the problem we have in this World, we have made it imperfect. Number 9 is a clown & a joker number.

This is a good one, I've read it like five times and still haven't noticed anything wrong.

But to answer an imperfect answer to your formula, 3.14141414....

(Pi+3)(pi-3) is not pi^2 - 9 ?

+Peter Van Biesen Nice try but that's a correct step, ask wolfram if you dont believe http://www.wolframalpha.com/input/?i=%28Pi%2B3%29%28pi-3%29

Pie-3 is 0 and it can't be multiplied on both sides

You see, tell your collective audience that you will always get different answers....#9 is an illusion number that never existed....it is us, imperfect human that brought about this Clown number, the Misleading number into existence. E= MC2N2D4 Absolute Relativity....not a theory E=MC2

+Ondřej Hruška should have used brain before posting ... :-) if the outcome is impossible, there's a multiplication by zero somewhere ...

It's like saying 2=-2 because both are roots of 4.

In the 7th line it should be (+/~~) (3-x) = (+/~~) (pi-x). Then, using opposite possibilities for the (+/-) on both sides of the equation returns the first line as a value for a x.

Im with +Pranathi Krishnam ...

Translate

Line and the result is wrong

But I think it has got multiple errors...

Titus L

+

1

2

1

2

1

(3-x)²=(π-x)²

→ 3-x = x-π 【not 3-x=3-π】

∵ 3<x<π

[ ∵ 3< (3+π)/2 <π ] ^_^

→ 3-x = x-π 【not 3-x=3-π】

∵ 3<x<π

[ ∵ 3< (3+π)/2 <π ] ^_^

Kevin Battisti

+

3

4

3

4

3

The error occurs when you remove the powers of 2. The reason for this is that [abs(x)]^2=x^2, but abs(x) might not equal x.

Example (-2)^2 = 2^2, but -2 doesn't equal 2.

Example (-2)^2 = 2^2, but -2 doesn't equal 2.

Ondřej Hruška

+

2

3

2

3

2

k guys here I got us a quality answer ;)

http://math.stackexchange.com/questions/788247/strange-riddle-with-pi#788250

http://math.stackexchange.com/questions/788247/strange-riddle-with-pi#788250

You see how nice we all get along.

Yes actually it is the answer +Alexander N. Benner is right by taking the case given we get the question itself...:)

the step removing power is wrong.

because,

1. 3.14 >x > 3.07 (because of 1st step)

2. so when we removing power,

(3-x)^2 becomes x-3, and (pi-x)^2 becomes pi -x

3. so, x-3 = pi -x

4. 2x = pi +3

5. x = pi+3 / 2

because,

1. 3.14 >x > 3.07 (because of 1st step)

2. so when we removing power,

(3-x)^2 becomes x-3, and (pi-x)^2 becomes pi -x

3. so, x-3 = pi -x

4. 2x = pi +3

5. x = pi+3 / 2

Mike Aben

+

1

2

1

2

1

Edited to fix confusing grammar.

Just to elaborate, because +Kevin Battisti is correct.

sqrt(x^2) does not equal x. As such,

sqrt(x^2) = abs(x).

So that means the second last line should read

abs(3 - x) = abs(pi - x)

This results in two possibilities.

3 - x = pi - x

OR (this being the key word)

3 - x = - (pi - x)

It is the second possibility that applies here as solving it just gets

x = (pi + 3) / 2 which is just how x was defined in the first place.

Just to elaborate, because +Kevin Battisti is correct.

sqrt(x^2) does not equal x. As such,

sqrt(x^2) = abs(x).

So that means the second last line should read

abs(3 - x) = abs(pi - x)

This results in two possibilities.

3 - x = pi - x

OR (this being the key word)

3 - x = - (pi - x)

It is the second possibility that applies here as solving it just gets

x = (pi + 3) / 2 which is just how x was defined in the first place.

+Mike Aben clearly you didn't understand the math that I presented or the point I was trying to make. In a mathematical proof of contradiction, all I need is one example of a contradiction (which I provided).

+Kevin Battisti Sorry, when I was saying this is correct, the "this" was what you said. I was providing the elaboration for others.

I understood what you said completely.

I understood what you said completely.

Hervé Arki

+

1

2

1

2

1

You have chosen the wrong solution when took the square roots.

The other one is

x-3=pi-x

2x=pi + 3

The other one is

x-3=pi-x

2x=pi + 3

Teresa Leclercq

+

2

3

2

3

2

the cause for error is simple. The wrong square root is taken. -(sqrt(LHS))=+(sqrt(RHS)), which returns the original expression. +Alexander N. Benner is correct

it's not allowed to

**add**x² on both sides of an equation.Only multiply or divide.Good one, it took me a little while. The problem is the removing of the squares

yeah. i agree with +Andre Wemans

when square rooting, the answer is not always positive.

when square rooting, the answer is not always positive.

9-6x+x^2 is (x-3)^2, also.

So abs(x-3)=abs(3-x)=abs(pi-x).

I guess it doesn't matter. They both have the same value.

So there is no proof at all that pi=3, but x is halfway between 3 and pi.

On the 3rd step shouldn't it be 2x-pi-3

+Pranathi Krishnam how pi-3 is zero??????????

Nm didn't see what was going on there

i think the square root should have been replaced with the step of moving the (pi-x)^2 to the LHS and difference of 2 squares.

+Dustin Pour poda thendii

I think it is like this he drew out a conclusion that pie=3 then pie-3=0 so he multiplied wid 0 on both sides....

sqrt(x^2)=|x|

if you ask wolfram alpha if (3-x)^2-(pi-x)^2=(3-x)-(pi-x), its comes back with false, so the mistake must be in the power reduction.

+Andre Wemans actually i am pretty sure u can add and subtract to both sides of an equation, think about completing the square.

oops not +Andre Wemans , but +Wouter Tomme

All you can infer from x^2 = y^2 is that EITHER x= y OR x=- y. The last step is fallacious.

I've probably just been sucked in by trolls.

I've probably just been sucked in by trolls.

Sorry 2nd to last.

X = (pi + 3)/2 > 3

So where you have:

(3 - x)^2 = (pi - x)^2

The LHS: 3 - x is actually negative.

So by taking it's square root you need to take into account an imaginary part.

So where you have:

(3 - x)^2 = (pi - x)^2

The LHS: 3 - x is actually negative.

So by taking it's square root you need to take into account an imaginary part.

I think its correct.

Cutely a ÷0 where 0 = π-3.

(The other solution is the original.)

(The other solution is the original.)

Where in the proof did we establish that 2πx = 9?

It just sort of came out of nowhere on the left side of step 5.

Then we used this baseless assumption to switch out the 9 on the right side of step 5 with 2πx.

It just sort of came out of nowhere on the left side of step 5.

Then we used this baseless assumption to switch out the 9 on the right side of step 5 with 2πx.

+Francesco Fiorani if we write the seventh step (x-3)^2 = (π-x)^2, equality is preserved and returns us to the first step.

Add a comment...