Find members of this field. Let $m$ be an integer, $m \geq 2$. and let ${\mathbb Z}_m$ be the set of all positive integers less than $m$, ${\mathbb Z}_m = \{0, 1, ..., m -1\}$. If $a$ and $b$ are in ${\mathbb Z}_m$, let $a + b$ be the least positive remainder obtained by dividing the (ordinary) sum of $a$ and $b$ by $m$, and, similarly, let $ab$ be the least positive remainder obtained by dividing the (ordinary) product of $a$ and $b$ by $m$. (Example: if $m = 12$, then $3 + 11 = 2$ and $3.11 = 9$.)
a) What is $-1$ in ${\mathbb Z}_5$?
b) What is $1/3$ in ${\mathbb Z}_7$? 
I am unsure how $-1$ and $1/3$ can exist in the sets as it is meant to be made of positive integers and i know it forms a field when $m$ is a prime number, though i have n't been able to prove that yet either as i could not find justification for the following field properties.


*

*A(4) To every a there corresponds a unique scalar $-a$ such that $a + (-a) = 0$.

*B(2) Multiplication is associative $a(bg) =(ab)g$.

*B(4) To every non-zero scalar a there corresponds a unique scalar $a^{-1}$ (or $1/a$) such that $aa^{-1} = 1$.


Any help even just with the original two questions i think may give me enough to have another attempt at this proof.
Any help would be highly appreciated.
Cheers
 A: In a field $F$, if $a\in F$, then "$-a$" represents the unique element $x$ of $F$ with the property that $a+x = 0$.
In a field $F$, if $a\in F$, $a\neq 0$, then $\frac{1}{a}$ (or $a^{-1}$) represents the unique element $y$ of $F$ with the property that $ay = 1$.
So: to find $-1$ in $\mathbb{Z}_5$, you need to find the unique element $x$ of $\mathbb{Z}_5$ that, added to $1$, yields $0$ (that is, the unique solution to $x+1=0$ in $\mathbb{Z}_5$). And to find the element $\frac{1}{3}$ of $\mathbb{Z}_7$, you need to find the unique element $y$ of $\mathbb{Z}_7$ that, when multiplied by $3$, yields $1$. That is, the unique solution to $3y=1$ in $\mathbb{Z}_7$. 
The reason why in a field $F$, given $a\in F$ there is a unique $x$ such that $a+x=0$ is that the field is supposed to be an abelian group under addition. This means $+$ must be associative and commutative, that F has to have an element $0$ such that $a+0 = a$ for all $a$; and that for all $a$ there must exist a $b$ such that $a+b=0$; the fact that these are unique follow easily from these properties.
The reason why in a field $F$, given $a\in F$, $a\neq 0$, there must exist a unique $y$ such that $ay=1$ is again part of the definition: a field is required to have a multiplicative identity that is not the additive identity (an element $1\neq 0$ such that $a1 = a$ for all $a$), and every nonzero element is required to have a multiplicative inverse; uniqueness follows from these facts. In short: the reason A(4), B(2), and B(4) must hold is by definition. There is no further justification needed: in order to be able to be called a "field", those properties (among others) must be satisfied. It is a membership requirement to be able to get past the bouncer at the door of the Club of Fields.
A: $$
3\cdot 5 \text{ is the same as }1\text{ in }\mathbb{Z}_7.
$$
This should tell you how $1/3$ can exist in $\mathbb{Z}_7$.
$$
(1 + \text{something}) \text{ is the same as } 0\text{ in }\mathbb{Z}_7.
$$
This should tell you how $-1$ can exist in $\mathbb{Z}_7$.
A: Defining $\mathbb Z_m$ as you did (the set of elements... + operations...), sometimes we abuse of notation. For example, we write $-1$ but we already are refering to the least positive remainder obtained by dividing $-1$ by $m$. For example, $-1$ in $\mathbb Z_{12}$ means $11$ since 
$$
-1= 12\times 0 + 11.
$$
Actually, this happens with any integer number. $127$ is $7$ in $\mathbb Z_{12}$ since
$$
127=12\times 10 + 7.
$$
With respect to the quotients, some elements have an inverse element. In you example b), $3$ has a inverse in $\mathbb Z_7$ since 
$$
3\times 5= 15 =7\times 2+1,
$$
thus we write $3\times 5=1$ and we say that $5$ is the inverse of $3$, and sometimes we write $5=1/3.$
Note that this does not always happen. For example, $3$ has not an inverse element in $\mathbb Z_{12}$. Actually, $k$ has an inverse element in $\mathbb Z_{m}$ if and only $k$ and $m$ are coprime. One consecuence is that $\mathbb Z_m$ is a field if and only if $m$ is a prime number.
