2+1=5 nasıl mı olur az düşünün bakalım))

A altküme R ve (A , + ) bir ikili işlem olsun
a eleman A ve b elaman olmak üzere
a+b = a + b + 2 olarak ikili işlemi tanımlarsak
2 + 1 = 2 + 1 +2
2 + 1 = 5 olur .....
Not : 2 + 1 = 2 +1+2 işleminde sol taraf A dan gelen ikili işlem toplaması sağ taraf ise bildigimiz reel sayılardaki toplama ........

x=y diyelim ve sırayla aşağıdaki işlemelr uygulayalım.
xy=y^2    
xy - (x^2)=(y^2) - (x^2)  
x(y-x)=(y-x).(y+x)
x=y+x
x+1=y+x+1
x=y demistik
y gordugumuz yerlere x yazalim
x+1=x+x+1
x=2 olsun
2+1=2+2+1=5

bu ne ya inanamıyorum ya ,ispata bakınca mantıklı geliyor never helal sana

y=x kabul edilip x-y ifadeleri sadeleştirilmiş yani eşitliğin her iki yanındaki 0 lar sadeleştirilmiş.

it is not so difficult to answer such questions. If we consider the Zn group wrt addition, which most of you I think know, then suxh equalities can be done for any a, b lying in Zn.
(for the ones who dont know it is the set of equivalence classes modulo n, where n is a positive integer. if we consider the addition of these classes which is the same as the addition of numbers in Z, then the structure (Zn, +) forms a group.
but anyway: it is not worthles to point out such things because equality of two structures in mathematics does not mean equal in the real life but to be "isomorphic".
have a nice day

in addition to my previous message I also wish to say that philidor has right. simplifying 0 in the proof does not prove anything, but just a trick that does not need much thinking to understand it.