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Q.
$f :Q \rightarrow Q$ is defined by $f(x) = 15x + 7$ is
Relations and Functions - Part 2
Solution:
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \rightarrow f$ is one-one
Let $f(x) = y \Rightarrow 15 x + 7 = y \Rightarrow x = \frac{y -7}{15} \in Q $
$\forall y \, \in Q \Rightarrow f $ is onto