(i) Let 'a' be an even positive integer.
Apply division algorithm with a and b, where b=2
a=(2×q)+r where 0≤r<2
a=2q+r where r=0 or r=1
since 'a' is an even positive integer, 2 divides 'a'.
∴r=0⇒a=2q+0=2q
Hence, a=2q when 'a' is an even positive integer.
(ii) Let 'a' be an odd positive integer.
apply division algorithm with a and b, where b=2
a=(2×q)+r where 0≤r<2
a=2q+r where r=0 or 1
Here r≠0 (∵a is not even) ⇒r=1
∴a=2q+1
Hence, a=2q+1 when 'a' is an odd positive integer.