4
$\begingroup$

This is one of my first proofs about fields. Please feed back and criticise in every way (including style and details).

Let $(F, +, \cdot)$ be a field. Non-trivially, $\textit{associativity}$ implies that any parentheses are meaningless. Therefore, we will not use parentheses. Therefore, we will not use $\textit{associativity}$ explicitly.

By $\textit{identity element}$, $F \ne \emptyset$. Now, let $a \in F$. It remains to prove that $0a = 0$. \begin{equation*} \begin{split} 0a &= 0a + 0 && \quad \text{by }\textit{identity element }(+ ) \\ &= 0a + a + -a && \quad \text{by }\textit{inverse element }(+ ) \\ &= 0a + 1a + -a && \quad \text{by }\textit{identity element }(\cdot) \\ &= (0 + 1)a + -a && \quad \text{by }\textit{distributivity } \\ &= (1 + 0)a + -a && \quad \text{by }\textit{commutativity }(+ ) \\ &= 1 a + -a && \quad \text{by }\textit{identity element }(+ ) \\ &= a + -a && \quad \text{by }\textit{identity element }(\cdot) \\ &= 0 && \quad \text{by }\textit{inverse element }(+ ) \end{split} \end{equation*} QED

PS: Is "Let $(F, +, \cdot)$ be a field." ok? Besides, I would not want to call $F$ a field, because $F$ is just a set. Also, what do you think about using adverbs like "Now"? How would you have said the associativity-thing?

$\endgroup$
  • 4
    $\begingroup$ Simpler: $0a=(0+0)a=0a+0a$, thus by adding $-(0a)$ on both left and right, $0a=0$. $\endgroup$ – Jean-Claude Arbaut Aug 27 '14 at 16:02
  • 2
    $\begingroup$ It all seems a bit roundabout. There is no need to use ALL of the rules of fields in the proof. Personally I would have done 0a + 0a = (0+0)a = 0a, then subtract 0a from both sides to get 0a = 0. $\endgroup$ – DavidButlerUofA Aug 27 '14 at 16:03
1
$\begingroup$

A few pointers:

  • You don't have to use "Now". You could just say "Let $a\in F$."

  • Don't say "meaningless". Rather, phrase it like so:

    Non-trivially, associativity implies that any parentheses are redundant. Hence, parenthesis will be suppressed and we will thus not explicitly employ associativity.

  • It's not wrong to say that $(F,+,\cdot)$ is a field, unless the question goes something like this: "Let $(F,+,\cdot)$ be a field, and let $0\in F$. Then show that every multiple of zero equals zero, i.e., for any $a\in F$, $0a=0$". Considering the way you phrased your proof, I don't think that this is how the question was phrased (correct me if I'm wrong).

Now, your proof is not wrong, but it's not the shortest either. Your proof could go something like this: $0a=(0+0)a=0a+0a$; hence $0a=0$. Alternatively, you could go like so: $0a=(0+0)a=0a+0a$. But $0a=0a+0$. Hence, $0a+0a=0a+0$, implying that $0a=0$. If I was a teacher, I would personally prefer the latter. However, do not take this personally; your proof is also very nice, but a tad bit longer than what I think is the conventional proof of this fact. Mathematicians are lazy; they prefer the shortest proofs (or at least that's what I think. I'm not a professional!)!

$\endgroup$
5
$\begingroup$

Your proof is fine. As the comments have pointed out, the proof can be shortened. If you still want to do it in a single chain of equalities, then you can do something like this:

\begin{equation*} \begin{split} 0a &= 0a + 0 && \quad \text{by }\textit{identity element }(+ ) \\ &= 0a + 0a + (-0a) && \quad \text{by }\textit{inverse element }(+ ) \\ &= (0 + 0)a + (-0a) && \quad \text{by }\textit{distributivity } \\ &= 0a + (-0a) && \quad \text{by }\textit{identity element }(+ ) \\ &= 0 && \quad \text{by }\textit{inverse element }(+ ) \end{split} \end{equation*}

$\endgroup$
  • $\begingroup$ This was the best proof I could find on the web. It was simple enough to explain to a 7-year-old. :) $\endgroup$ – adampasz Jan 13 '18 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.