Valuation on a field and integral domain Let $R$ be an integral domain, and $K$ be its field of fractions. Let $(G,+)$ be an abelian ordered group.
Then a valuation $v$ on $K$ is a map from it to $(G,+)\cup\{\infty\}$ with the property that
(1) $v(xy)=v(x)+v(y)$
(2) $v(x+y)\ge \mbox{min}\{ v(x),v(y)\}$;
(3) $v(x)=\infty$ if and only if $x=0$.
My question is not too technical, but I did not find any reason in the books I referred.
Why do we include $\infty$ along with $G$, and define $v(0)=\infty$? Means, for what purposes, it is beneficial?
If we are playing with only non-zero elements of the field or of integral domain, then, can't we state second property as $v(x+y)\ge \mbox{min}\{ v(x), v(y)\}$ for all $x,y$ with $x+y\neq 0$?

(Ordered group: there is an ordering relation $\le $ on elements of $G$ with the property

*

*$x\le y$ and $y\le z$ implies $x\le z$;


*$x\le y$ and $y\le x$ implies $x=y$;


*for any $x,y\in G$, either $x\le y$ or $y\le x$, and it is always true that $x\le x$;


*if $x\le y$ then $x+z\le y+z$ for all $x,y,z$ in $G$.
 A: 
Why do we include $\infty$ along with $G$, and define $v(0)=\infty$? Means, for what purposes, it is beneficial?

It's best to think about what real examples look like. A standard first example of a valuation is the $p$-adic valuation $v_p:\mathbb Q\to\mathbb Z\cup\{\infty\}$ for a prime $p$. What this valuation does on integers (imprecisely speaking) is it counts the number of times the integer is divisible by $p$, so $v_2(4)=2$, $v_3(81)=4$, or generally $v_p(n)=\max\{k\in\mathbb N\mid p^k|n\}$ (at least, this makes sense when $n\neq0$). The point is that $0$ is divisible by $p^k$ for all $k$, i.e. the maximum does not exist, so the only natural definition one could make is $v_p(0)=\infty$.

If we are playing with only non-zero elements of the field or of integral domain, then, can't we state second property as $v(x+y)\ge\min\{v(x),v(y)\}$ for all $x,y$ with $x+y\neq0$?

This definition is fine and equivalent, it's just not really saying anything different. This says "if $x\neq -y$ then $v(x+y)\ge\min\{v(x),v(y)\}$", but of course this also holds when $x=-y$ so why bother excluding this case in exchange for a more complicated definition?
