With all due respect, no teacher of math is going to agree with your silly math errors. The laws of mathematics do not change with country or culture. You need to study a little harder and focus on more serious endeavors.

Your statement that "some agree ("are agree" is incorrect English) and some disagree (more bad English)" is false. The ones that agree are simply not sufficiently skilled in simple algebra to know better. This is High School level math in the US.

Let we take a value,
20=20
applying -ve sin on both sides,
-20=-20
or,
16-36=25-45
adding (9\2)^2 on both sides we get,
4^2 (9\2)^2-36 =5^2 (9\2)^2-45
[qoute]
which is formula of (a-b)^2, in both sides as we see. You cann't ignore this.

Teachers? I knew that the USA has some lousy teachers, but I didn't know that such a problem existed in other nations.

I already demonstrated that the alleged "proof" that 2 2=5 is mathematically flawed.

Although it is true that the square of 0.5 equals the square of -0.5 (which is part of the alleged "proof"), it is not true that one can merely eliminate both squarings and get two quantities of equal value.

As the (a-b)^2 is equal to (b-a)^2 and also the value of (0.5)^2 is equal to that of (-0.5)^2 and also squre root of a^2 is equal to 'a'.
On the basis of these rules
my proof is correct.

M I right or M I right.? TOingg tOingg tOingg

If you're willing to accept in any way, shape, or form. that 2 2=5
Then I propose this, I'll sell you four items. Then you can pay me for five. When you say "Hey why am I getting charged, For an additional item?" I'll write this equation out for you. 2 2 = 5
That's about as mastermind as this nonsense is going to get. Any Takers?

Not really. However there are false proofs of untrue mathematical statements, based on false assumptions or faulty reasoning.

For example, below is a 'proof' that 1=0 (http://answers.yahoo.com/question/index?qid=1006051531250)
If you can prove that 1=0, then you can prove that 2+2=5, or indeed that practically anything equals anything else. Can you spot the fallacy?

Let a & b each be equal to 1.
Since a ^ b are equal,
b^2 = ab (eq.1)
Since a equals itself, it is obvious that
a^2 = a^2 (eq.2)
Subtract equation 1 from equation 2.
(a^2) - (b^2) = (a^2)-ab (eq. 3)
We can factor both sides of the equation:
(a^2)-ab equals a(a-b).
Likewise, (a^2)-(b^2) equals (a + b)(a - b)
(Nothing fishy is going on here. This statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3, we get
(a+b)(a-b) = a (a-b)
So far, so good. Now divide both sides of the equation by (a-b) and we get
a + b = a
b = 0
But we set b to 1 at the very beginning of this proof, so this means that
1 = 0

Read more: Is it possible to prove that 2+2=5? | Answerbag http://www.answerbag.com/q_view/471599#ixzz2Qe9fpgyf