What relations define a mapping?

Question

Which of the following relations $f: \mathbb{Q} \rightarrow \mathbb{Q}$ define a mapping? In each case, supply a reason why $f$ is or is not a mapping. Also p and q are integers and q cannot equal zero.

(a) $f(p/q) = \frac{p+1}{p-2}$

I don't believe this is a mappong because 2 in the domain does not map to anything?

(b) $f(p/q) = \frac{3p}{3q}$

(c) $f(p/q) = \frac{p+q}{q^2}$

(d) $f(p/q) = \frac{3p^2}{7q^2} - \frac{p}{q}$

The rest of these three functions seems like mappings to me because all elements of the domain map to a unique element in the range. Is this an appropriate answer?

Answer 1

You have to check that different representations of a rational $p/q$ map to the same value.

(a) is not a mapping since $f(1/2)= -2$, but $f(4/8) = 5/2$.

(b) is the identity mapping $f(x) = x$.

(c) is not a mapping since $f(1/2) = 3/4$, but $f(2/4) = 3/8$.

(d) is is the mapping $f(x) = \frac{3}{7}x^2 - x$

Answer 2

Suppose that $x=\frac{1}{2}$. Clearly $x=\frac{2}{4}$ also.

Now for the first question, $f(\frac{1}{2})=-2$ but $f(\frac{2}{4})$ is undefined.

Now for (b), $f(x)=\frac{3p}{3q}=\frac{p}{q}$ which is the same for all $x \in \Bbb Q$.

For (c), $f(\frac{1}{2})=\frac{3}{4}$ but $f(\frac{2}{4})=\frac{6}{16}=\frac{3}{8}$

You do part (d). In general mappings of $\Bbb Q$ need to be handled carefully as it is easy to create functions that are not well-defined.