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Which of the following is not an equivalence relation in Z

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Let R={(a,b):a+b is an even integer a,b∈z}R = \{(a,b):a + b\ is\ an\ even\ integer\ a,b \in z\}
For a∈Z,a+a=2aa \in Z,a + a = 2a is an even integer.
∴(a,a)∈R\therefore(a,a) \in R ∀a∈Z\forall a \in Z
∴\therefore R is reflexive.
Let(a,b)∈R⇒a+b(a,b) \in R \Rightarrow a + b is an even integer
⇒\Rightarrow R is symmetric.
Let (a,b),(b,c)R(a,b),(b,c)R
⇒a+b and b+c\Rightarrow a + b\ and\ b + c is an even integers
⇒(a+b)+(b+c)=a+c+2b\Rightarrow (a + b) + (b + c) = a + c + 2b is an even integer
⇒(a+c+2b)−2b=a+c\Rightarrow (a + c + 2b) - 2b = a + c is an even integer
⇒(a,c)R\Rightarrow (a,c)R R is transitive.
∴\therefore R is an equivalence relation.
Let R={(a,b):a−b is even integers,a,bZ}R = \{(a,b):a - b\ is\ even\ integers,a,bZ\}
For aZ,a−a=0aZ,a - a = 0 is an even integer.
∴(a,a)R\therefore(a,a)R
∴\therefore R is reflexive.
Let (a,b)∈R.⇒a−b(a,b) \in R. \Rightarrow a - b is an even integer.
⇒−(a−b)\Rightarrow - (a - b) is even integer.
⇒b−a\Rightarrow b - a is even integer.
⇒(b−a)R\Rightarrow (b - a)R
R is symmetric.
Let (a,b)(b,c)R.⇒a−b,b−c(a,b)(b,c)R. \Rightarrow a - b,b - c are even integers..
⇒(a−b)+(b−c)\Rightarrow (a - b) + (b - c) is even integer.
⇒a−c\Rightarrow a - c is even integer.
⇒(a,c)∈R\Rightarrow (a,c) \in R
⇒R is transitive.\Rightarrow R\ is\ transitive.
R is an equivalence relation.
Let R={(a,b):a<b,a,b∈Z}R = \{(a,b):a < b,a,b \in Z\}
Let aZ, a < a is false.
∴\therefore R is not reflexive.
∴\therefore R is not an equivalence relation.
Let R={(a,b):a=b,a,b∈Z}R = \{(a,b):a = b,a,b \in Z\}
It is quite easy to check that R is an equivalence relation.

Source: Grass-G / M-17 Mathematics Paper 26 Nov Second.docx

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