[tex]Method\ 1:[/tex]
[tex]\dfrac{1}{3}=\dfrac{1\cdot5}{3\cdot5}=\dfrac{5}{15}\\\\\dfrac{3}{5}=\dfrac{3\cdot3}{5\cdot3}=\dfrac{9}{15}\\\\\dfrac{3}{15} < \dfrac{9}{15}\ therefore\ \dfrac{3}{5}\ is\ bigger[/tex]
[tex]Method\ 2:[/tex]
[tex]\dfrac{1}{3} < \dfrac{1}{2}=\dfrac{1.5}{3}\\\\\dfrac{3}{5} > \dfrac{1}{2}=\dfrac{2.5}{5}\\\\therefore\ \dfrac{3}{5}\ is\ bigger[/tex]
