Exercise 0.1.5(Well defined functions)

Proof.

(a)

$f:\mathbb{Q}\to \mathbb{Z}$ “defined” by $f(a∕b)=a$ is not well defined.

To give a counterexample, $x=2∕3$ has image $2$ and $x^{\prime }=4∕6$ has image $4$, but $x=x^{\prime }$ and $f(x)\ne f(x^{\prime })$ : this contradicts the definition of a function.

Another example is $x=(-2)∕3$, where $f(x)=-2$, and $x^{\prime }=2∕(-3)$, where $f(x^{\prime })=2$. Here $x=x^{\prime }$ and $f(x)\ne f(x^{\prime })$.

To give a signification to $f$, we must say:

“let $x=a∕b$, where $\gcd (a,b)=1,b>0$. Then $f(x)=a$.”

(b)

$f:\mathbb{Q}\to \mathbb{Q}$ defined by $f(a∕b)=a^2∕b^2$ is well defined:

if $a∕b=c∕d$ then $a^2∕b^2=c^2∕d^2$.

More simply, we can define

$$f\{\begin{array}{ccc}\hfill \mathbb{Q}\hfill & \hfill \to \hfill & \mathbb{Q}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & x^2\hfill \end{array}$$

Such a definition is impossible in part (a).