Question

Show that for each natural number n the fraction $$\frac { 14n+3 }{ 21n+4 } $$ is in its lowest form.

Answer 1

A:

If $$n=1,\quad \dfrac { 14n+3 }{ 21n+4 } =\dfrac { 14(1)+3 }{ 21(1)+4 } =\dfrac { 14+3 }{ 21+4 } =\dfrac { 17 }{ 25 } $$

is in the lowest form

If $$n=2,\quad \dfrac { 14n+3 }{ 21n+4 } =\dfrac { 14(2)+3 }{ 21(2)+4 } =\dfrac { 28+3 }{ 42+4 } =\dfrac { 31 }{ 46 } $$

is in the lowest form

If

$$n=3,\quad \dfrac { 14n+3 }{ 21n+4 } =\dfrac { 14(3)+3 }{ 21(3)+4 } =\dfrac { 42+3 }{ 63+4 } =\dfrac { 45 }{ 67 } $$

is in the lowest form

If

$$n=4,\quad \dfrac { 14n+3 }{ 21n+4 } =\dfrac { 14(4)+3 }{ 21(4)+4 } =\dfrac { 56+3 }{ 84+4 } =\dfrac { 49 }{ 880 } $$

is in the lowest form.

Similarly by substituting natural numbers for $$n$$ we can show that $$\dfrac { 14n+3 }{ 21n+4 } $$ is in its lowest form.