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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 $0x=0$?

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  • $\begingroup$ 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. $\endgroup$ – Eff Mar 28 '17 at 21:57
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    $\begingroup$ I would probably start by noting that $0+0=0$; therefore $0x=(0+0)x=0x+0x$. Subtracting $0x$ then yields $0=0x$. $\endgroup$ – Nick Peterson Mar 28 '17 at 21:57
  • $\begingroup$ @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. $\endgroup$ – user409521 Mar 28 '17 at 22:02
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    $\begingroup$ Honestly I do not understand why your first six lines have to to with anything and you make things to complicated. The crux is simply $x0 = x(0+0)$ because $0$ is the additive identity $x(0+0) = x0+x0$ by distributive property. And $x0 = x0 + x0\implies x0-x0\implies x0+x0 - x0\implies $0= x0$. $\endgroup$ – fleablood Mar 29 '17 at 7:57
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Since $0$ is the neutral element for the addition, we have that $$0x = (0 + 0)x$$ and because of distributivity we find that $$(0 + 0)x = 0x + 0x.$$ Hence we find that $$0x = 0x + 0x$$ so $0x$ also acts as the neutral element. Because of unicity of this element, we have that $0x = 0$.

$\textbf{Edit:}$ As Will Jagy commented, you could also use that $0x$ has an additive inverse, denoted by $-0x$ and applying this to both sides of the last equation immediately gives that $0 = 0x$.

$\textbf{Edit 2:}$ So what you did seems correct to me, but you basically took some extra step to find the step I used first. You could immediately take this step for the reason I gave.

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    $\begingroup$ you can also say that $0x$ has an additive inverse to subtract it from both sides of $0x=0x+0x$ $\endgroup$ – Will Jagy Mar 28 '17 at 21:59
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    $\begingroup$ @WillJagy: thanks for your comment, I will edit this. $\endgroup$ – Student Mar 28 '17 at 21:59
  • $\begingroup$ Thank you for taking time to answer and confirm what I have found, I appreciate it a lot! One last thing, is this a completely meaningless result? I'm asking because as opposed to doing induction proofs (which makes me feel like a wizard when I get them right) this proof still does not feel satisfying to me at all... $\endgroup$ – user409521 Mar 28 '17 at 22:21
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    $\begingroup$ @Phyllotactic yeah, it might feel as a quite trivial result, since you used it a lot when working with real numbers for example. For example, if you get to 'complete order' of the reals, you can also proof that $0 < 1$. The proof will give little satisfaction, since it is something which seems very trivial! Induction proof on the contrary seem very advanced, since you are proving statement which might not be obvious at all. Consider this proof and the one of $0 < 1$ as being exercises on using the right definitions at the right moment. :) $\endgroup$ – Student Mar 28 '17 at 22:30
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Starting from $ax=ax$ you can find $ax-ax=0$. Then you can take out $x$ as $x(a-a)=0$. Now, if you know that $a-a=0$, then you have finished.

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    $\begingroup$ Don't you use in this case that $-ax = (-a)x$ in the distribution step? (I only know proofs of this which use that $0x = 0$). $\endgroup$ – Student Mar 29 '17 at 15:41

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