Is 2=4? Using the simple formula ( x - y )^2 = x^2 + y^2 - 2*x*y I have proved that 2 = 4?

Question

Using the simple formula ( x - y )^2 = x^2 + y^2 - 2*x*y
I have proved that 2 = 4

Here is the proof:-
-8 = -8
4-12 = 16-24 (since 4-12= -8 & 16 –24 = -8)

Now add 9 to both sides
4 – 12 + 9 = 16 – 24 + 9
2*2 - 2*2*3 + 3*3 = 4*4 – 2*4*3 + 3*3
2^2 - 2*2*3 + 3^2 = 4^2 – 2*4*3 + 3^2
this steplooks like the formula x2 + y2 - 2*x*y = ( x - y )2

( 2 – 3 )^2 = ( 4 – 3 )^2
So 2 – 3= 4 – 3
Add 3 to both sides
2 = 4

try to find the mistake in this

Answer 1

Dude the prob seems to be with the square roots i.e. from



( 2 – 3 )^2 = ( 4 – 3 )^2
So 2 – 3= 4 – 3




U (-2)^2 = 4 and also 2^2=4
its similar to the thing i showed you in ur solution.

Answer 2

A problem arises when you go from

( 2 – 3 )^2 = ( 4 – 3 )^2

to

2 - 3 = 4 - 3

because the expression on the left is negative while the expression on the right is positive (an untrue statement)

When you take a square root, the result is + or - the root.

+-(2 - 3) = +-(4 - 3)

and only pick the combinations that are true, like

+(2 - 3) = -(4 - 3)

or

-(2 - 3) = (4 - 3)

Adding 3 to both sides of either of these equations gives another true statement.

Answer 3

You cannot just eliminate the ^2 from both sides, this isn't how quadratics work.

(2-3)^2 = (4-3)^2
is really written:
(2-3)*(2-3) = (4-3)*(4-3)
Getting rid of the ^2 from both sides is not mathematically correct. This would be the same as cancelling out a *(-1) from the left and a *(1) from the right, which can't be done.

Example:

x^2 = y^2
using your method and dropping the ^2:
x = y, but this isn't always true (x = 1, y = -1 is a solution where x does not = y), because the equation is really x*x = y*y

Answer 4

You have in lines 4 and 5 from the end:
(2-3)² = (4-3)²
So 2-3 = 4-3 (error)
It should be:
(2-3)² = (4-3)²
(-1)² = (1)²
1 = 1

Powers and roots have a higher priority then Addition and Subtraction. ( Everything in side the brackets becomes subject to the power.

Answer 5

here it is.

when you went from :
2^2 - 2*2*3 + 3^2 = 4^2 – 2*4*3 + 3^2
to :
( 2 – 3 )^2 = ( 4 – 3 )^2
is your errors.

you cannot get rid of the 2*2*3 cause it is not the same on the right hand side.
Also, when dealing with exponents you cannot arbitrarily add brackets. 4^2*3^2 is not the same as (4-3)^2


Nice try sir, but incorrect.

Answer 6

A quadratic equation can have two roots.
For example,
the solution to the equation x^2 = 4 is
-2 or 2 because
(-2)^2 = 4 and (2)^2 = 4
Does that mean that
(-2)^2 = (2)^2 implies -2 = 2?
No.

Answer 7

Your proof contains:

(-1)^2 = 1^2 => -1 = 1

which is false.

Most "proofs" like this one contain one of a short list of fallacies. See the source I have listed for more.

Answer 8

infinite in trignometry=1/0
then 1=0*infinite
anything multiplying with 0 is 0 then
is 1=0 tell me in the same way is 2=4?

Answer 9

The problem arises when you try to square root. Those are the same number, but you can get two sqrt results that are not equal. You cannot do that, otherwise the equation is wrong.

Answer 10

THE MISTAKE IS THAT YOU WAISTED TIME COMING UP WITH THIS FORMULA THAT HAS NO REAL USE IN THE WORLD....WHO CARES IF 2 = 4