Francesco Fiorani

Challenges  -

Try to find the error!﻿
44
14

No . Try again. :-)﻿

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....﻿

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

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.﻿
amir yb

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. ﻿
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Line and the result is wrong﻿

But I think it has got multiple errors...﻿
Titus L
+
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 ] ^_^﻿

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.﻿

You see how nice we all get along.﻿

Yes actually it is the answer 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

Edited to fix confusing grammar.

Just to elaborate, because  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.﻿

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). ﻿

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.﻿

You have chosen the wrong solution when took the square roots.
The other one is
x-3=pi-x
2x=pi + 3﻿

the cause for error is simple. The wrong square root is taken. -(sqrt(LHS))=+(sqrt(RHS)), which returns the original expression.   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
when square rooting, the answer is not always positive.﻿

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.﻿

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.﻿

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.﻿

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

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. ﻿

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. ﻿

Cutely a ÷0 where 0 = π-3.
(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.﻿

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