Let $A \subseteq \mathbb{R}, B \subset \mathbb{R}^{+} \cup \{0\}$, and $A=\{{x\in\mathbb{Q}\mid -12\leq x \leq4}\}.\;$ Find $f(A)$, where $f(x)= |\frac x2|$.
How to start? $\mathbb{Q}$ is the set of rational numbers, so in $A$ will be $1/2,1/3,1/4$... to $4$ and in the negative to $-12$? There will be a lot of numbers to write! Please help.
Is it $$A=\{−12,−11,−10,−9,−8,−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4\}$$ or no?
By definition, the image of $f$ is $$\{|x/2|:x\in\mathbb{Q}\cap[-12,4]\}.$$ I think it will be easier to see what this reduces to if you notice that \begin{align} \{|x/2|:x\in\mathbb{Q}\cap[-12,4]\}&= \{x/2:x\in\mathbb{Q}\cap[0,12]\} \end{align} This is because $f(x)=\left|x/2\right| \geq 0, \forall x \in \mathbb Q\cap [-12, 4]$, so we can just assume we're working on a set of non-negative numbers, and then the absolute value bars aren't needed.
I claim that $$\{x/2:x\in\mathbb{Q}\cap[0,12]\}= \mathbb{Q}\cap[0,6].$$ To show these sets are equal, observe that the left-hand-side is clearly contained in the RHS because if we take a rational number in $[0,12]$ and divide it by $2$, it is now in $[0,6]$ and still rational. Conversely, given any rational number $q\in[0,6]$, $2q$ is a rational number in $[0,12]$ and $q=2q/2$. Therefore the image of $f$ is the set of rational numbers between $0$ and $6$. Hence, $$f(A)=\mathbb{Q}\cap[0,6]$$