# Proof of $0x=0$

I don't feel like I proved much here, all my life I took for granted that $0x=0$ was obvious but I really never even thought of questioning why that is so. I want to get ahead and learn the basics of Analysis before taking a course on it so I am looking to get basic facts right before moving on to the good stuff.

Thus, to see if I understand the concept of a field $-$I will call it $F-$ it was suggested to me that I try producing a convincing proof, based on what I know, of the presumably obvious fact that $$0x=0$$

I start by noting that there exists a "$0$ element" in $F$ such that for all $x \in F$, $0+x=x$.

\begin{align} (0+x)x&=0x+xx\\ (0+x)x-xx & = 0x \\ 0x+xx-xx & = 0x\\ \end{align}

I want to get rid of both $xx$ terms using axioms I know. The simplest for me is to use the distributive law $x(a+b)=xa+xb$ along with the property that there is an "inverse" element in $F$ for all elements of $F$ such that their sum is $x+(-x)=0$.

I get the following result: \begin{align} 0x+((x)(x))-((x)(x))&=0x\\ 0x+((x)(x))+((x)(-x))&=0x\\ 0x+x(x+(-x))&=0x\\ x(0+0)&=0x\\ x(0+0)-0x&=0\\ x((0+0)-0)&=0\\ \therefore \; 0x&=0 \tag*{\blacksquare} \\ \end{align}

I don't know why but this result does not feel satisfying to me at all and anything I tried so far either ended up this way or lead me to a circular reasoning tailspin. Is this acceptable and/or what would be a better way to prove that $0x=0$?

• I think this question has been answered a few times on the site. See here where I just searched for it, and see an example here. – Eff Mar 28 '17 at 21:57
• I would probably start by noting that $0+0=0$; therefore $0x=(0+0)x=0x+0x$. Subtracting $0x$ then yields $0=0x$. – Nick Peterson Mar 28 '17 at 21:57
• @Eff: sorry about that I did not mean to produce a duplicate, I was wondering whether my particular reasoning based my limited knowledge was correct. – user409521 Mar 28 '17 at 22:02