no it would not be equivalent to find a bijection becuase Q contains N so a trivial function would be . You need to find a function . Also bijection means both injection and surjection. so u don't need to single out injective.

for example, to prove that is denumerable. notice you can break into classes depending on what the numerator and denominator adds up to like class{2} = 1/1, class{3} = 1/2, 2/1 class{4} = 3/1, 1/3 etc.... and each of these classes is finite...