Let Rz={(a,b)|a,b is belongs to Z^+ and a-b is an odd positive integer}.

The set of integers is represented by the letter Z. An integer is any number in the infinite set,

Z={.., -3,-2,-1, 0, 1, 2 ,3, …}
Z+ is the set of all positive integers(1,2,3,….) there is no 0 here
Z- is the set of all negative integers (…,-3,-2,-1) there is no 0 here
Z^nonneg is the set of all positive integers including 0, while Z^nonpos is the set of all negative integers including 0

Source: https://calculus.nipissingu.ca/tutorials/numbers.html#:~:text=An%20integer%20is%20any%20number,%2D2%2C%20%2D1).

Soln:
Set R1 is reflexive because
Q1) For every a is belongs to Z+ . (a,a) is belongs to R . a-a = 0 is not an odd  number. So, R is not reflexive.

For every a,b is belongs to Z+ : if a,b is positive, then b-a is negative. That if( a,b) is belongs R then (b,a) is not belongs to R. So not symmetric.

Due to the same reason, as above, (a,b) and (b,a) cannot be simultaneously in R . So the condition in the definition of anti symmetry is false.

The implication holds trivially , R is anti symmetric.

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Author: zakilive