967: For Field, Positive-Natural-Number-th Root of 0 Is 0

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description/proof of that for field, positive-natural-number-th root of 0 is 0

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Topics

About: field

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of field.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$n$: $\in \mathbb{N}\setminus \{0\}$
$R_{0,n}$: $=\{\alpha \in F|{\alpha }^n=0\}$
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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 $r\in F$, $r0=0$: $r(1+0)=r1=r$, because $0$ is the additive identity and $1$ is the multiplicative identity, but $r(1+0)=r1+r0$, by the distributability, $=r+r0$, because $1$ is the multiplicative identity; so, $r+r0=r$, and $r0=-r+r+r0=-r+r=0$.

Especially, $0^2=00=0$.

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 $1<n$ hereafter.

Let us suppose that $\alpha \ne 0$. 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, $\alpha =0$.

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References

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