2024-07-28

695: Memorandum on Powers of Group, Ring, or Field Elements

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description/proof of memorandum on powers of group, ring, or field elements

Topics


About: group
About: ring
About: field

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of a memorandum on powers of group, ring, or field elements.

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:
G: ∈{ the groups }
R: ∈{ the rings }
F: ∈{ the fields }
//

Statements:
∀p∈G,∀z∈Z(pz is valid )
∧
∀p∈R,∀n∈N(pn is valid )
∧
∀p∈F,∀n∈N(pn is valid )
∧
∀p∈F such that p≠0,∀z∈Z(pz is valid )
∧
∀p∈G,∀z,z′∈Z(pzpz′=pz+z′)
∧
∀p∈R,∀n,n′∈N(pnpn′=pn+n′)
∧
∀p∈F,∀n,n′∈N(pnpn′=pn+n′)
∧
∀p∈F such that p≠0,∀z,z′∈Z(pzpz′=pz+z′)
∧
∀p∈G,∀z,z′∈Z((pz)z′=pzz′)
∧
∀p∈R,∀n,n′∈N((pn)n′=pnn′)
∧
∀p∈F,∀n,n′∈N((pn)n′=pnn′)
∧
∀p∈F such that p≠0,∀z,z′∈Z((pz)z′=pzz′)
∧
∀p∈G,∀z∈Z,∀n′∈N such that 1≤n′(p(zn′)=(pz)zn′−1)
∧
∀p∈R,∀n,n′∈N such that 1≤n′(p(nn′)=(pn)nn′−1)
∧
∀p∈F,∀n,n′∈N such that 1≤n′(p(nn′)=(pn)nn′−1)
∧
∀p∈F such that p≠0,∀z∈Z,n′∈N such that 1≤n′(p(zn′)=(pz)zn′−1)
∧
∀p∈G,∀z∈Z,∀n′∈N,∀n″∈N such that 1≤n″(p(zn′n″)=(p(zn′))zn′n″−n′)
∧
∀p∈R,∀n,n′,n″∈N such that 1≤n″(p(nn′n″)=(p(nn′))nn′n″−n′)
∧
∀p∈F,∀n,n′,n″∈N such that 1≤n″(p(nn′n″)=(p(nn′))nn′n″−n′)
∧
∀p∈F such that p≠0,∀z∈Z,∀n′∈N,∀n″∈N such that 1≤n″(p(zn′n″)=(p(zn′))zn′n″−n′)
//

2: Natural Language Description


For any group, G, any ring, R, and any field, F, ∀p∈G,∀z∈Z(pz is valid ), ∀p∈R,∀n∈N(pn is valid ), ∀p∈F,∀n∈N(pn is valid ), ∀p∈F such that p≠0,∀z∈Z(pz is valid ), ∀p∈G,∀z,z′∈Z(pzpz′=pz+z′), ∀p∈R,∀n,n′∈N(pnpn′=pn+n′), ∀p∈F,∀n,n′∈N(pnpn′=pn+n′), ∀p∈F such that p≠0,∀z,z′∈Z(pzpz′=pz+z′), ∀p∈G,∀z,z′∈Z((pz)z′=pzz′), ∀p∈R,∀n,n′∈N((pn)n′=pnn′), ∀p∈F,∀n,n′∈N((pn)n′=pnn′), ∀p∈F such that p≠0,∀z,z′∈Z((pz)z′=pzz′), ∀p∈G,∀z∈Z,∀n′∈N such that 1≤n′(p(zn′)=(pz)zn′−1), ∀p∈R,∀n,n′∈N such that 1≤n′(p(nn′)=(pn)nn′−1), ∀p∈F,∀n,n′∈N such that 1≤n′(p(nn′)=(pn)nn′−1), ∀p∈F such that p≠0,∀z∈Z,n′∈N such that 1≤n′(p(zn′)=(pz)zn′−1), ∀p∈G,∀z∈Z,∀n′∈N,∀n″∈N such that 1≤n″(p(zn′n″)=(p(zn′))zn′n″−n′), ∀p∈R,∀n,n′,n″∈N such that 1≤n″(p(nn′n″)=(p(nn′))nn′n″−n′), ∀p∈F,∀n,n′,n″∈N such that 1≤n″(p(nn′n″)=(p(nn′))nn′n″−n′), ∀p∈F such that p≠0,∀z∈Z,∀n′∈N,∀n″∈N such that 1≤n″(p(zn′n″)=(p(zn′))zn′n″−n′).

3: Note


Of course, those facts are straightforward to think carefully, but some careless confusions can happen making some confusions with powers of real numbers. So, such a memorandum will be handy.

4: Proof


Let us prove that ∀p∈G,∀z∈Z(pz is valid ).

For z=0, pz=1∈G.

For 0<z, pz=p...p∈G, where the multiplications are z-times.

For z<0, pz=p−1...p−1∈G, where the multiplications are −z-times.

On the other hand, for a q∈Q, pq may not be valid.

For example, for q=1/2, what does pq mean? x=p1/2 should mean that x2=p, but there may not be such an x∈G.

And the power cannot be any element of G: what does pp′ mean for a p′∈G?

Let us prove that ∀p∈R,∀n∈N(pn is valid ).

For n=0, p0=1∈R.

For 0<n, pn=p...p∈R, where the multiplications are n-times.

On the other hand, for a z∈Z such that z<0, pz may not be valid.

That is because there may not be any p−1.

For a q∈Q, pq may not be valid.

The reason is as before.

And the power cannot be any element of R as before.

Let us prove that ∀p∈F,∀n∈N(pn is valid ).

For n=0, pn=1∈F.

For 0<n, pn=p...p∈F, where the multiplications are n-times.

On the other hand, for any z∈Z such that z<0 and p=0, pz is not valid.

That is because p−1 does not exist.

Let us prove that ∀p∈F such that p≠0,∀z∈Z(pz is valid ).

For z=0, pz=1∈F.

For 0<z, pz=p...p∈F, where the multiplications are z-times.

For z<0, pz=p−1...p−1∈F, where the multiplications are −z-times, which is possible because p≠0.

On the other hand, for a q∈Q, pq may not be valid.

For example, for q=1/2, what does pq mean? x=p1/2 should mean that x2=p, but there may not be such an x∈F. Note that when F=R, pq exists, but that is not for general field. For example, when F=Q, 21/2∈F does not exist.

And the power cannot be any element of F as before. Note that when F=R, the power can be any element of F, but that is not because R is a field.

Let us prove that ∀p∈G,∀z,z′∈Z(pzpz′=pz+z′).

When z=0, pzpz′=1pz′=pz′=pz+z′.

When z′=0, pzpz′=pz1=pz=pz+z′.

When 0<z,z′, pzpz′=p...pp...p=pz+z′.

When 0<z and z′<0, pzpz′=p...pp−1...p−1=pz+z′.

When z<0 and 0<z′, pzpz′=p−1...p−1p...p=pz+z′.

When z,z′<0, pzpz′=p−1...p−1p−1...p−1=pz+z′.

Let us prove that ∀p∈R,∀n,n′∈N(pnpn′=pn+n′).

When n=0, pnpn′=1pn′=pn′=pn+n′.

When n′=0, pnpn′=pn1=pn=pn+n′.

When 0<n,n′, pnpn′=p...pp...p=pn+n′.

A proof of ∀p∈F,∀n,n′∈N(pnpn′=pn+n′) is the same with the ring case.

A proof of ∀p∈F such that p≠0,∀z,z′∈Z(pzpz′=pz+z′) is the same with the group case.

Let us prove that ∀p∈G,∀z,z′∈Z((pz)z′=pzz′).

When z′=0, (pz)z′=1=pzz′.

When 0<z′, (pz)z′=pz...pz=pz+...+z=pzz′.

When z′<0, (pz)z′=(pz)−1...(pz)−1; when z=0, =1−1...1−1=1=pzz′; when 0<z, =(p...p)−1...(p...p)−1=(p−1...p−1)...(p−1...p−1)=pzz′; when z<0, =(p−1...p−1)−1...(p−1...p−1)−1=(p...p)...(p...p)=pzz′.

Let us prove that ∀p∈R,∀n,n′∈N((pn)n′=pnn′).

When n′=0, (pn)n′=1=pnn′.

When 0<n′, (pn)n′=pn...pn=pn+...+n=pnn′.

A proof of ∀p∈F,∀n,n′∈N((pn)n′=pnn′) is the same with the ring case.

A proof of ∀p∈F such that p≠0,∀z,z′∈Z((pz)z′=pzz′) is the same with the group case. Note that (pz)z′ makes sense because pz≠0: any field is an integral domain.

Let us prove that ∀p∈G,∀z∈Z,∀n′∈N such that 1≤n′(p(zn′)=(pz)zn′−1).

It makes sense, because while Z is a ring, zn′∈Z, p(zn′)∈G; pz∈G, zn′−1∈Z, and (pz)zn′−1∈G.

p(zn′)=p(zzn′−1)=(pz)zn′−1.

Of course, p(zn′)≠(pz)n′ in general.

Let us prove that ∀p∈R,∀n,n′∈N such that 1≤n′(p(nn′)=(pn)nn′−1).

It makes sense, because as nn′∈N, p(nn′)∈R; pn∈R, nn′−1∈N, and (pn)nn′−1∈R.

p(nn′)=p(nnn′−1)=(pn)nn′−1.

A proof of ∀p∈F,∀n,n′∈N such that 1≤n′(p(nn′)=(pn)nn′−1) is the same with the ring case.

A proof of ∀p∈F such that p≠0,∀z∈Z,n′∈N such that 1≤n′(p(zn′)=(pz)zn′−1) is the same with the group case. Note that (pz)zn′−1 makes sense, because pz≠0: any field is an integral domain.

Let us prove that ∀p∈G,∀z∈Z,∀n′∈N,∀n″∈N such that 1≤n″(p(zn′n″)=(p(zn′))zn′n″−n′).

It makes sense, because n′n″∈N, zn′n″∈Z, p(zn′n″)∈G; zn′∈Z, p(zn′)∈G, n′n″−n′=n′(n″−1)∈N, zn′n″−n′∈Z, and (p(zn′))zn′n″−n′∈G.

p(zn′n″)=p(zn′+n′n″−n′)=p(zn′zn′n″−n′)=(p(zn′))zn′n″−n′.

Let us prove that ∀p∈R,∀n,n′,n″∈N such that 1≤n″(p(nn′n″)=(p(nn′))nn′n″−n′).

It makes sense, because n′n″∈N, nn′n″∈N, p(nn′n″)∈R; nn′∈N, p(nn′)∈R, n′n″−n′=n′(n″−1)∈N, nn′n″−n′∈N, and (p(nn′))nn′n″−n′∈R.

p(nn′n″)=p(nn′+n′n″−n′)=p(nn′nn′n″−n′)=(p(nn′))nn′n″−n′.

A proof of ∀p∈F,∀n,n′,n″∈N such that 1≤n″(p(nn′n″)=(p(nn′))nn′n″−n′) is the same with the ring case.

A proof of ∀p∈F such that p≠0,∀z∈Z,∀n′∈N,∀n″∈N such that 1≤n″(p(zn′n″)=(p(zn′))zn′n″−n′) is the same with the group case.

References
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