description/proof of that for field, positive-natural-number-th root of 0 is 0
Topics
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of field.
Target Context
- The reader will have a description and a proof of the proposition that for any field, any positive-natural-number-th root of 0 is 0.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(R_{0, n}\): \(= \{\alpha \in F \vert \alpha^n = 0\}\)
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Statements:
\(R_{0, n} = \{0\}\)
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2: Proof
Whole Strategy: Step 1: see that \(0 \in R_{0, n}\); Step 2: see that for each \(\alpha \in R_{0, n}\), \(\alpha = 0\).
Step 1:
Let us see that \(0^n = 0\).
For each \(r \in F\), \(r 0 = 0\): \(r (1 + 0) = r 1 = r\), because \(0\) is the additive identity and \(1\) is the multiplicative identity, but \(r (1 + 0) = r 1 + r 0\), by the distributability, \(= r + r 0\), because \(1\) is the multiplicative identity; so, \(r + r 0 = r\), and \(r 0 = - r + r + r 0 = - r + r = 0\).
Especially, \(0^2 = 0 0 = 0\).
Supposing \(0^m = 0\), \(0^{m + 1} = 0^m 0 = 0^2 = 0\).
So, by the induction principle, \(0^n\).
Step 2:
Let \(\alpha \in R_{0, n}\) be any.
\(\alpha^n = 0\).
When \(n = 1\), \(\alpha^1 = \alpha = 0\).
Let us suppose that \(1 \lt n\) hereafter.
Let us suppose that \(\alpha \neq 0\). Then, there would be a multiplicative inverse, \(\alpha^{-1} \in F\). \(\alpha^{n - 1} = \alpha^n \alpha^{-1} = 0 \alpha^{-1} = 0\). When \(n - 1 = 1\), \(\alpha^{n - 1} = \alpha = 0\), a contradiction. Otherwise, \(\alpha^{n - 2} = 0\), likewise. When \(n - 2 = 1\), \(\alpha = 0\), a contradiction. Otherwise, ..., etc. Anyway, eventually, \(\alpha = 0\), a contradiction.
So, \(\alpha = 0\).
