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
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
- 4: Proof
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of ring.
- The reader knows a definition of field.
- The reader admits the proposition that any field is an integral domain.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(R\): \(\in \{\text{ the rings }\}\)
\(F\): \(\in \{\text{ the fields }\}\)
//
Statements:
\(\forall p \in G, \forall z \in \mathbb{Z} (p^z \text{ is valid })\)
\(\land\)
\(\forall p \in R, \forall n \in \mathbb{N} (p^n \text{ is valid })\)
\(\land\)
\(\forall p \in F , \forall n \in \mathbb{N} (p^n \text{ is valid })\)
\(\land\)
\(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z} (p^z \text{ is valid })\)
\(\land\)
\(\forall p \in G, \forall z, z' \in \mathbb{Z} (p^z p^{z'} = p^{z + z'})\)
\(\land\)
\(\forall p \in R, \forall n, n' \in \mathbb{N} (p^n p^{n'} = p^{n + n'})\)
\(\land\)
\(\forall p \in F, \forall n, n' \in \mathbb{N} (p^n p^{n'} = p^{n + n'})\)
\(\land\)
\(\forall p \in F \text{ such that } p \neq 0, \forall z, z' \in \mathbb{Z} (p^z p^{z'} = p^{z + z'})\)
\(\land\)
\(\forall p \in G, \forall z, z' \in \mathbb{Z} ((p^z)^{z'} = p^{z z'})\)
\(\land\)
\(\forall p \in R, \forall n, n' \in \mathbb{N} ((p^n)^{n'} = p^{n n'})\)
\(\land\)
\(\forall p \in F, \forall n, n' \in \mathbb{N} ((p^n)^{n'} = p^{n n'})\)
\(\land\)
\(\forall p \in F \text{ such that } p \neq 0, \forall z, z' \in \mathbb{Z} ((p^z)^{z'} = p^{z z'})\)
\(\land\)
\(\forall p \in G, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(z^{n'})} = (p^z)^{z^{n' - 1}})\)
\(\land\)
\(\forall p \in R, \forall n, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(n^{n'})} = (p^n)^{n^{n' - 1}})\)
\(\land\)
\(\forall p \in F, \forall n, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(n^{n'})} = (p^n)^{n^{n' - 1}})\)
\(\land\)
\(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z}, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(z^{n'})} = (p^z)^{z^{n' - 1}})\)
\(\land\)
\(\forall p \in G, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N}, \forall n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(z^{n' n''})} = (p^{(z^{n'})})^{z^{n' n'' - n'}})\)
\(\land\)
\(\forall p \in R, \forall n, n', n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(n^{n' n''})} = (p^{(n^{n'})})^{n^{n' n'' - n'}})\)
\(\land\)
\(\forall p \in F, \forall n, n', n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(n^{n' n''})} = (p^{(n^{n'})})^{n^{n' n'' - n'}})\)
\(\land\)
\(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N}, \forall n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(z^{n' n''})} = (p^{(z^{n'})})^{z^{n' n'' - n'}})\)
//
2: Natural Language Description
For any group, \(G\), any ring, \(R\), and any field, \(F\), \(\forall p \in G, \forall z \in \mathbb{Z} (p^z \text{ is valid })\), \(\forall p \in R, \forall n \in \mathbb{N} (p^n \text{ is valid })\), \(\forall p \in F , \forall n \in \mathbb{N} (p^n \text{ is valid })\), \(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z} (p^z \text{ is valid })\), \(\forall p \in G, \forall z, z' \in \mathbb{Z} (p^z p^{z'} = p^{z + z'})\), \(\forall p \in R, \forall n, n' \in \mathbb{N} (p^n p^{n'} = p^{n + n'})\), \(\forall p \in F, \forall n, n' \in \mathbb{N} (p^n p^{n'} = p^{n + n'})\), \(\forall p \in F \text{ such that } p \neq 0, \forall z, z' \in \mathbb{Z} (p^z p^{z'} = p^{z + z'})\), \(\forall p \in G, \forall z, z' \in \mathbb{Z} ((p^z)^{z'} = p^{z z'})\), \(\forall p \in R, \forall n, n' \in \mathbb{N} ((p^n)^{n'} = p^{n n'})\), \(\forall p \in F, \forall n, n' \in \mathbb{N} ((p^n)^{n'} = p^{n n'})\), \(\forall p \in F \text{ such that } p \neq 0, \forall z, z' \in \mathbb{Z} ((p^z)^{z'} = p^{z z'})\), \(\forall p \in G, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(z^{n'})} = (p^z)^{z^{n' - 1}})\), \(\forall p \in R, \forall n, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(n^{n'})} = (p^n)^{n^{n' - 1}})\), \(\forall p \in F, \forall n, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(n^{n'})} = (p^n)^{n^{n' - 1}})\), \(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z}, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(z^{n'})} = (p^z)^{z^{n' - 1}})\), \(\forall p \in G, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N}, \forall n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(z^{n' n''})} = (p^{(z^{n'})})^{z^{n' n'' - n'}})\), \(\forall p \in R, \forall n, n', n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(n^{n' n''})} = (p^{(n^{n'})})^{n^{n' n'' - n'}})\), \(\forall p \in F, \forall n, n', n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(n^{n' n''})} = (p^{(n^{n'})})^{n^{n' n'' - n'}})\), \(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N}, \forall n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(z^{n' n''})} = (p^{(z^{n'})})^{z^{n' n'' - n'}})\).
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 \(\forall p \in G, \forall z \in \mathbb{Z} (p^z \text{ is valid })\).
For \(z = 0\), \(p^z = 1 \in G\).
For \(0 \lt z\), \(p^z = p ... p \in G\), where the multiplications are \(z\)-times.
For \(z \lt 0\), \(p^z = p^{-1} ... p^{-1} \in G\), where the multiplications are \(- z\)-times.
On the other hand, for a \(q \in \mathbb{Q}\), \(p^q\) may not be valid.
For example, for \(q = 1 / 2\), what does \(p^q\) mean? \(x = p^{1 / 2}\) should mean that \(x^2 = p\), but there may not be such an \(x \in G\).
And the power cannot be any element of \(G\): what does \(p^{p'}\) mean for a \(p' \in G\)?
Let us prove that \(\forall p \in R, \forall n \in \mathbb{N} (p^n \text{ is valid })\).
For \(n = 0\), \(p^0 = 1 \in R\).
For \(0 \lt n\), \(p^n = p ... p \in R\), where the multiplications are \(n\)-times.
On the other hand, for a \(z \in \mathbb{Z}\) such that \(z \lt 0\), \(p^z\) may not be valid.
That is because there may not be any \(p^{-1}\).
For a \(q \in \mathbb{Q}\), \(p^q\) may not be valid.
The reason is as before.
And the power cannot be any element of \(R\) as before.
Let us prove that \(\forall p \in F , \forall n \in \mathbb{N} (p^n \text{ is valid })\).
For \(n = 0\), \(p^n = 1 \in F\).
For \(0 \lt n\), \(p^n = p ... p \in F\), where the multiplications are \(n\)-times.
On the other hand, for any \(z \in \mathbb{Z}\) such that \(z \lt 0\) and \(p = 0\), \(p^z\) is not valid.
That is because \(p^{-1}\) does not exist.
Let us prove that \(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z} (p^z \text{ is valid })\).
For \(z = 0\), \(p^z = 1 \in F\).
For \(0 \lt z\), \(p^z = p ... p \in F\), where the multiplications are \(z\)-times.
For \(z \lt 0\), \(p^z = p^{-1} ... p^{-1} \in F\), where the multiplications are \(- z\)-times, which is possible because \(p \neq 0\).
On the other hand, for a \(q \in \mathbb{Q}\), \(p^q\) may not be valid.
For example, for \(q = 1 / 2\), what does \(p^q\) mean? \(x = p^{1 / 2}\) should mean that \(x^2 = p\), but there may not be such an \(x \in F\). Note that when \(F = \mathbb{R}\), \(p^q\) exists, but that is not for general field. For example, when \(F = \mathbb{Q}\), \(2^{1 / 2} \in F\) does not exist.
And the power cannot be any element of \(F\) as before. Note that when \(F = \mathbb{R}\), the power can be any element of \(F\), but that is not because \(\mathbb{R}\) is a field.
Let us prove that \(\forall p \in G, \forall z, z' \in \mathbb{Z} (p^z p^{z'} = p^{z + z'})\).
When \(z = 0\), \(p^z p^{z'} = 1 p^{z'} = p^{z'} = p^{z + z'}\).
When \(z' = 0\), \(p^z p^{z'} = p^{z} 1 = p^{z} = p^{z + z'}\).
When \(0 \lt z, z'\), \(p^z p^{z'} = p ... p p ... p = p^{z + z'}\).
When \(0 \lt z\) and \(z' \lt 0\), \(p^z p^{z'} = p ... p p^{-1} ... p^{-1} = p^{z + z'}\).
When \(z \lt 0\) and \(0 \lt z'\), \(p^z p^{z'} = p^{-1} ... p^{-1} p ... p = p^{z + z'}\).
When \(z, z' \lt 0\), \(p^z p^{z'} = p^{-1} ... p^{-1} p^{-1} ... p^{-1} = p^{z + z'}\).
Let us prove that \(\forall p \in R, \forall n, n' \in \mathbb{N} (p^n p^{n'} = p^{n + n'})\).
When \(n = 0\), \(p^n p^{n'} = 1 p^{n'} = p^{n'} = p^{n + n'}\).
When \(n' = 0\), \(p^n p^{n'} = p^{n} 1 = p^{n} = p^{n + n'}\).
When \(0 \lt n, n'\), \(p^n p^{n'} = p ... p p ... p = p^{n + n'}\).
A proof of \(\forall p \in F, \forall n, n' \in \mathbb{N} (p^n p^{n'} = p^{n + n'})\) is the same with the ring case.
A proof of \(\forall p \in F \text{ such that } p \neq 0, \forall z, z' \in \mathbb{Z} (p^z p^{z'} = p^{z + z'})\) is the same with the group case.
Let us prove that \(\forall p \in G, \forall z, z' \in \mathbb{Z} ((p^z)^{z'} = p^{z z'})\).
When \(z' = 0\), \((p^z)^{z'} = 1 = p^{z z'}\).
When \(0 \lt z'\), \((p^z)^{z'} = p^z ... p^z = p^{z + ... + z} = p^{z z'}\).
When \(z' \lt 0\), \((p^z)^{z'} = (p^z)^{-1} ... (p^z)^{-1}\); when \(z = 0\), \(= 1^{-1} ... 1^{-1} = 1 = p^{z z'}\); when \(0 \lt z\), \(= (p ... p)^{-1} ... (p ... p)^{-1} = (p^{-1} ... p^{-1}) ... (p^{-1} ... p^{-1}) = p^{z z'}\); when \(z \lt 0\), \(= (p^{-1} ... p^{-1})^{-1} ... (p^{-1} ... p^{-1})^{-1} = (p ... p) ... (p ... p) = p^{z z'}\).
Let us prove that \(\forall p \in R, \forall n, n' \in \mathbb{N} ((p^n)^{n'} = p^{n n'})\).
When \(n' = 0\), \((p^n)^{n'} = 1 = p^{n n'}\).
When \(0 \lt n'\), \((p^n)^{n'} = p^n ... p^n = p^{n + ... + n} = p^{n n'}\).
A proof of \(\forall p \in F, \forall n, n' \in \mathbb{N} ((p^n)^{n'} = p^{n n'})\) is the same with the ring case.
A proof of \(\forall p \in F \text{ such that } p \neq 0, \forall z, z' \in \mathbb{Z} ((p^z)^{z'} = p^{z z'})\) is the same with the group case. Note that \((p^z)^{z'}\) makes sense because \(p^z \neq 0\): any field is an integral domain.
Let us prove that \(\forall p \in G, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(z^{n'})} = (p^z)^{z^{n' - 1}})\).
It makes sense, because while \(\mathbb{Z}\) is a ring, \(z^{n'} \in \mathbb{Z}\), \(p^{(z^{n'})} \in G\); \(p^z \in G\), \(z^{n' - 1} \in \mathbb{Z}\), and \((p^z)^{z^{n' - 1}} \in G\).
\(p^{(z^{n'})} = p^{(z z^{n' - 1})} = (p^z)^{z^{n' - 1}}\).
Of course, \(p^{(z^{n'})} \neq (p^z)^{n'}\) in general.
Let us prove that \(\forall p \in R, \forall n, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(n^{n'})} = (p^n)^{n^{n' - 1}})\).
It makes sense, because as \(n^{n'} \in \mathbb{N}\), \(p^{(n^{n'})} \in R\); \(p^n \in R\), \(n^{n' - 1} \in \mathbb{N}\), and \((p^n)^{n^{n' - 1}} \in R\).
\(p^{(n^{n'})} = p^{(n n^{n' - 1})} = (p^n)^{n^{n' - 1}}\).
A proof of \(\forall p \in F, \forall n, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(n^{n'})} = (p^n)^{n^{n' - 1}})\) is the same with the ring case.
A proof of \(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z}, n' \in \mathbb{N} \text{ such that } 1 \le n' (p^{(z^{n'})} = (p^z)^{z^{n' - 1}})\) is the same with the group case. Note that \((p^z)^{z^{n' - 1}}\) makes sense, because \(p^z \neq 0\): any field is an integral domain.
Let us prove that \(\forall p \in G, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N}, \forall n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(z^{n' n''})} = (p^{(z^{n'})})^{z^{n' n'' - n'}})\).
It makes sense, because \(n' n'' \in \mathbb{N}\), \(z^{n' n''} \in \mathbb{Z}\), \(p^{(z^{n' n''})} \in G\); \(z^{n'} \in \mathbb{Z}\), \(p^{(z^{n'})} \in G\), \(n' n'' - n' = n' (n'' - 1)\in \mathbb{N}\), \(z^{n' n'' - n'} \in \mathbb{Z}\), and \((p^{(z^{n'})})^{z^{n' n'' - n'}} \in G\).
\(p^{(z^{n' n''})} = p^{(z^{n' + n' n'' - n'})} = p^{(z^{n'} z^{n' n'' - n'})} = (p^{(z^{n'})})^{z^{n' n'' - n'}}\).
Let us prove that \(\forall p \in R, \forall n, n', n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(n^{n' n''})} = (p^{(n^{n'})})^{n^{n' n'' - n'}})\).
It makes sense, because \(n' n'' \in \mathbb{N}\), \(n^{n' n''} \in \mathbb{N}\), \(p^{(n^{n' n''})} \in R\); \(n^{n'} \in \mathbb{N}\), \(p^{(n^{n'})} \in R\), \(n' n'' - n' = n' (n'' - 1) \in \mathbb{N}\), \(n^{n' n'' - n'} \in \mathbb{N}\), and \((p^{(n^{n'})})^{n^{n' n'' - n'}} \in R\).
\(p^{(n^{n' n''})} = p^{(n^{n' + n' n'' - n'})} = p^{(n^{n'} n^{n' n'' - n'})} = (p^{(n^{n'})})^{n^{n' n'' - n'}}\).
A proof of \(\forall p \in F, \forall n, n', n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(n^{n' n''})} = (p^{(n^{n'})})^{n^{n' n'' - n'}})\) is the same with the ring case.
A proof of \(\forall p \in F \text{ such that } p \neq 0, \forall z \in \mathbb{Z}, \forall n' \in \mathbb{N}, \forall n'' \in \mathbb{N} \text{ such that } 1 \le n'' (p^{(z^{n' n''})} = (p^{(z^{n'})})^{z^{n' n'' - n'}})\) is the same with the group case.
