check this out:

if a=b --------- and a=1=b
=> a-b=0
=> a2-b2=0
=> (a+b)(a-b)=0 --------- now since we said a-b=0
=> (a+b)(a-b)=(a-b)
=> (a+b)=1
=> 2=1

But then, maybe not

NO !!!
Since you said (a-b) = 0,
then (a+b)(a-b) = (a-b) only means (a+b)x0 = 0, and (a+b) can then be anything.
Silly !!!

I know its not right but

(a+b) is 1*(a+b) ok?
so remove the parenthes and you get 1

2+(2divided by 2)=3

Dagnabbit that double posting monster!

It's even simpler than all of that. One of the statements says:
a=1=b
Therefore, both a and b are 1.
The statement that says:
(a+b)=1 must therefore be false
Because:
(1+1) does not equal 1

You're just saying this okay.

a=b <> a=1=b, since a=b
a=1
b=1

1-1=0
how do you do the square thing... 1 squared-1Squared=0
(1+1)(1-1)=0
(1+1)*1(1-1)=0----- since even you distribute it, 1 is still equal to 1
1+1=2

.



. . ---> therefore, 1=1, and 2 not equal to 1.

If wolf=hungry then rabiit dies,
if wolf=hungry-false then rabbit gets away
if wolf=cold then rabbit runs long and fast then wolf not cold and rabbit probaly ends up dead anyway.
if wolf=pissed then doorframe=in big jeopardy
if mac makes wolf angry=true then mac is in for whoopass!

Hehe,

you got confused in your conclusion or forgot to add something.if you put the numerical values in place of a and b you will see that it's not possiable to get the answer,is this something you did in school wrong and you question why or what,i am just wondering ,thats all ,please tell me friend

Nice Try. But I noticed a slight Mathematical Problem (assumeing you are in the set of Real Numbers

=> (a+b)(a-b)=(a-b)

You Dived by a-b to get

=> (a+b)=1

However, if a-b=0 then you divided by zero . . . but one of the postulates in the set of reals states that you cannot divide by 0 because it approaches infinity, which is not in the set of real numbers.

THEREFORE: your proof is inconclusive.



BK