Question

solve this problem urgently​

Answer 1

(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.

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