# Prove Well-defined functions

This problem is from my textbook, chapter of Isomorphism and Invariant.

(a) $f:\mathbb{Q}\to \mathbb{Q}$ defined by $f(\frac a b)=\frac {a+b} b$

(b) $f:\mathbb{Z}_4\to \mathbb{Z}_8$ defined by $f([a]_4)=[3a]_8$

(c) $f:\mathbb{Z}_6\to \mathbb{Z}_7$ defined by $f([a]_6)=[a+k]_7$ for any integer $k$

By the definition, well-defined means that a mapping $f:S\to T$ is well defined if $f(a)=f(b)$

whenever $a=b$ in $S$.

So, in order to prove the above, do I assume that $a=b$ ? or pick any arbitary numbers in domains and see if I can counter examples? How do I prove if it is well defined?

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First, you should try a few examples to see if it seems well-defined. If not, then you are done. If it seems to be well-defined, you need to prove this. –  Tobias Kildetoft Dec 8 '13 at 20:31
Ok, for example, (b), $f([2]_4)=[6]_8$. And can I say this case is well defined since $[6]_4=[2]_4$? is my calculation correct? –  Wes Dec 8 '13 at 20:34
@Wes:But you need to know what is the equivalence relation between fractions. –  user99680 Dec 8 '13 at 20:37

For example (a):

$$\frac ab=\frac mn\iff a=\frac{bm}n\;,\;\;or\;\;m=\frac{an}b\implies$$

$$\frac{a+b}b=\frac{\frac{bm}n+b}{b}=b\frac{m+n}{bn}=\frac{m+n}n$$

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The standard equivalence relation $R$ defined on $\mathbb Q$ is $$[\frac {a}{b}]=[\frac {c}{d}]$$ iff $ad=bc$. Try showing that if $[\frac {a}{b}]=[\frac{c}{d}]$ , then $[f (\frac {a}{b})]=[f (\frac {c}{d})]$.

So, for example, for 1): Assume [$\frac {a}{b}$]=$[\frac {c}{d}]$. Then $f(\frac {a}{b})=\frac {a+b}{b}$ . Is it equivalent to $f(\frac{c}{d})=\frac{c+d}{d}$ (under $R$)?

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