Consider the provided qustion,

If a \(\displaystyle\in{Z}_{{n}}\) then \(\displaystyle{\left|{a}\right|}=\frac{{n}}{{{\gcd{{\left({a},{n}\right)}}}}}\)

here given \(\displaystyle{Z}_{{12}}\). So, n=12

Given a = 6, b = 2

So, a + b = 6 + 2 = 8

\(\displaystyle{\left|{a}\right|}={\left|{6}\right|}=\frac{{12}}{{{\gcd{{\left({6},{12}\right)}}}}}=\frac{{12}}{{6}}={2}\)

\(\displaystyle{\left|{b}\right|}={\left|{2}\right|}=\frac{{12}}{{{\gcd{{\left({2},{12}\right)}}}}}=\frac{{12}}{{2}}={6}\)

\(\displaystyle{\left|{a}+\right|}={\left|{8}\right|}=\frac{{12}}{{{\gcd{{\left({8},{12}\right)}}}}}=\frac{{12}}{{4}}={3}\)

If a \(\displaystyle\in{Z}_{{n}}\) then \(\displaystyle{\left|{a}\right|}=\frac{{n}}{{{\gcd{{\left({a},{n}\right)}}}}}\)

here given \(\displaystyle{Z}_{{12}}\). So, n=12

Given a = 6, b = 2

So, a + b = 6 + 2 = 8

\(\displaystyle{\left|{a}\right|}={\left|{6}\right|}=\frac{{12}}{{{\gcd{{\left({6},{12}\right)}}}}}=\frac{{12}}{{6}}={2}\)

\(\displaystyle{\left|{b}\right|}={\left|{2}\right|}=\frac{{12}}{{{\gcd{{\left({2},{12}\right)}}}}}=\frac{{12}}{{2}}={6}\)

\(\displaystyle{\left|{a}+\right|}={\left|{8}\right|}=\frac{{12}}{{{\gcd{{\left({8},{12}\right)}}}}}=\frac{{12}}{{4}}={3}\)