nyuuzyou/wikis huggingface dataset:
Positive Definite and Positive Homogeneous Map with Convex Closed Unit Ball is Norm

== Theorem ==
Let $\mathbb F$ be a subfield of $\C$. 
Let $X$ be a vector space over $\mathbb F$. 
Let $N : X \to \R_{\ge 0}$ be a positive definite and positive homogeneous function.
That is: 
:$(1): \quad$ $\map N x = 0$ $x = 0$
:$(2): \quad$ $\map N {\lambda x} = \cmod \lambda \map N x$ for each $x \in X$ and $\lambda \in \mathbb F$.
Suppose further that: 
:$B = \set {x \in X : \map N x \le 1}$ is convex.
Then: 
:$N$ is a norm on $X$. 
== Proof ==
Note that $N$ satisfies and .
So we only need to verify .
That is, we want to show that: 
:$\map N {x + y} \le \map N x + \map N y$ 
for each $x, y \in X$. 
Fix $x, y \in X$. 
If $\map N x = 0$, then $x = 0$ and we have: 
Similarly, if $\map N y = 0$, then $y = 0$ and: 
In both cases we clearly have: 
:$\map N {x + y} \le \map N x + \map N y$
So suppose that $\map N x > 0$ and $\map N y > 0$.
We then have: 
so:
:$\dfrac x {\map N x} \in B$
Similarly: 
so:
:$\dfrac y {\map N y} \in B$
Now note that we can write: 
:$1 = \dfrac {\map N x} {\map N x + \map N y} + \dfrac {\map N y} {\map N x + \map N y}$
Since $B$ is convex, we have: 
:$\dfrac {\map N x} {\map N x + \map N y} \paren {\dfrac x {\map N x} } + \dfrac {\map N y} {\map N x + \map N y} \paren {\dfrac y {\map N y} } \in B$
That is: 
That is: 
:$\map N {\dfrac {x + y} {\map N x + \map N y} } \le 1$
Using positive homogeneity, we then obtain: 
:$\dfrac {\map N {x + y} } {\map N x + \map N y} \le 1$
so:
:$\map N {x + y} \le \map N x + \map N y$
as was required.
== Sources ==
* $3.1$: Norms
Category:Normed Vector Spaces
Category:Convex Sets (Vector Spaces)
Positive Difference Relation on Reals is Transitive

== Theorem ==
Let $P \subseteq \R$ be a subset of the real numbers such that:
:$(1): \quad 1 \in P$
:$(2): \quad a, b \in P \implies a + b \in P$
:$(3): \quad$ For all $x \in \R$, exactly one of these is true:
:::$x \in P$
:::$x = 0$
:::$-x \in P$
Let $Q \subseteq \R \times \R$ be the relation on $\R$ defined as:
:$Q = \set {\tuple {a, b} \in \R: a - b \in P}$
Then $Q$ is a transitive relation.
== Proof ==
Let $a - b \in P$ and $b - c \in P$.
By condition $(2)$:
:$\paren {a - b} + \paren {b - c} \in P$
Simplifying:
:$a - c \in P$
The result follows by definition of transitive relation.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $1$
Category:Transitive Relations
Positive Elements of Ordered Ring

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $P$ be the set of positive elements of $R$ , that is, $P = R_{\ge 0}$.
Then:
: $(1): \quad P + P \subseteq P$
: $(2): \quad P \cap \paren {-P} = \set {0_R}$
: $(3): \quad P \circ P \subseteq P$
If $\le$ is a total ordering, that is, if $\struct {R, +, \circ, \le}$ is a totally ordered ring, then:
: $(4): \quad P \cup \paren {-P} = R$
The converse is also true:
Let $\struct {R, +, \circ}$ be a ring.
Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied.
Then there is one and only one ordering $\le$ compatible with the ring structure of $R$ such that $P = R_{\ge 0}$.
Also, if $(4)$ is also satisfied, then $\le$ is a total ordering.
== Proof ==
=== Necessary Condition ===
First, suppose that $\le$ is compatible with the ring structure of $R$.
Hence:
:$(OR1): \quad \le$ is compatible with $+$
:$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$.
$(1)$: Let $x, y \in R: 0_R \le x, 0_R \le y$.
Then $0_R + 0_R \le x + y$ by the fact that $\le$ is compatible with $+$.
Thus $0_R \le x + y$ and thus $x + y \in P$.
$(2)$: By Properties of Ordered Ring item 4, $-P = \set {x \in R: x \le 0_R}$.
Let $x \in P \cap \paren {-P}$.
Then $x \le 0_R$ and $0_R \le x$.
So from the antisymmetric nature of $\le$, $x = 0_R$.
$(3)$: This is equivalent to $0_R \le x, 0_R \le y \implies 0_R \le x \circ y$ which is one of the properties of being compatible with the ring structure of $R$.
$(4)$: Now if $\le$ is a total ordering, then $\forall x \in R$, either $x \le 0_R$ or $0_R \le x$, and the result follows.
=== Sufficient Condition ===
Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied.
By item $(OR2)$ of Properties of Ordered Ring, we have:
:$x \le y \iff 0 \le y + \paren {-x}$
so there is at most one ordering on $R$ compatible with the ring structure of $R$ such that $P = R_{\ge 0}$, namely, the one that satisfies:
:$x \le y \iff y + \left({-x}\right) \in P$
Now we need to show that $\le$ thus defined has the required properties.
;Reflexivity:
By $(2)$:
:$\forall x \in R: x \le x$ because $x + \paren {-x} = 0_R \in P$
;Antisymmetry:
Let $x \le y$ and $y \le x$.
Then $y + \paren {-x} \in P$ and $-\paren {y + \paren {-x} } = x + \paren {-y} \in P$.
Thus by $(2)$, $y + \paren {-x} = 0_R$ and thus $y = x$.
;Transitivity:
If $x \le y$ and $y \le z$, then $y + \paren {-x} \in P$ and $z + \paren {-y} \in P$.
But as $z + \paren {-x} = z + \paren {-y} + y + \paren {-x}$, we have that $z + \paren {-x} \in P$ from $(1)$.
Hence $x \le z$.
Let $x \le y$.
Then $z + x \le z + y$ since $\paren {z + y} + \paren {-\paren {x + x} } = y + \paren {-x} \in P$.
Finally, Ordering Compatible with Ring holds because of $(3)$.
If $(4)$ holds, then $\forall x, y \in R$, either $y + \paren {-x} \in P$ or $x + \paren {-y} = -\paren {y + \paren {-x} } \in P$, that is, either $x \le y$ or $y \le x$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.12$
Category:Ordered Rings
Positive Even Integers as Sum of 2 Composite Odd Integers in 2 Ways

== Theorem ==
Let $n \in \Z_{>0}$ be a positive even integer.
Let $n$ be such that it cannot be expressed as the sum of $2$ odd positive composite integers in at least $2$ different ways.
Then $n$ belongs to the set:
:$\left\{ {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 32, 34, 38, 40, 44, 46, 52, 56, 62, 68}\right\}$
== Proof ==
The sequence of odd positive composite integers begins:
:$9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, \ldots$
which is more than we need for this proof.
Generating the possible sums of these proceeds as follows:
This exhausts all possible sums of odd positive composite integers of $68$ or less.
It remains to be demonstrated that all positive even integers over $68$ can be so expressed in $2$ different ways.
Notice that one of $n - 15, n - 25, n - 35$ is divisible by $3$.
Similarly one of $n - 9, n - 15, n - 21, n - 27, n - 33$ is divisible by $5$.
For all above $n - k$ which is composite, since $k$ is composite, $\paren {n - k} + k$ is a way to express $n$ as the sum of $2$ odd positive composite integers.
Since $n > 68$, these two ways are distinct unless $n - 15$ is divisible by $15$.
In that case, $n - 45$ is also divisible by $15$, and $n - 45 \ne 15$.
Then $n = \paren {n - 15} + 15 = \paren {n - 45} + 45$ are two ways to express $n$ as the sum of $2$ odd positive composite integers.
== Historical Note ==
== Sources ==
* 
* : $68$
Category:Composite Numbers
Category:Odd Integers
Positive Even Integers as Sum of 2 Composite Odd Integers in 2 Ways/Historical Note

== Historical Note on Positive Even Integers as Sum of 2 Composite Odd Integers in 2 Ways ==
According to the footnote to the presentation of the solution to Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers in , this particular result was deduced by , who is reported to have been a student.
== Sources ==
* 
Category:Historical Notes
Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers

== Theorem ==
The positive even integers which cannot be expressed as the sum of $2$ composite odd numbers are:
:$2, 4, 6, 8, 10, 12, 14, 16, 20, 22, 26, 28, 32, 38$
== Proof ==
The smallest composite odd numbers are $9$ and $15$, so trivially $2$ to $16$ and $20$ to $22$ cannot be expressed as the sum of $2$ composite odd numbers.
We have:
It remains to investigate $26, 28$ and $32$.
This will be done by progressively subtracting smaller composite odd numbers from them, and noting that the difference is not composite.
It remains to be demonstrated that all even integers greater than $38$ can be expressed as the sum of $2$ composite odd numbers.
We note that $9 + 6 k$ is odd and a multiple of $3$.
Numbers $18$ and greater of the form $6 n$ can be expressed as:
:$\left({9 + 6 k}\right) + 9$
Numbers $34$ and greater of the form $6 n + 4$ can be expressed as:
:$\left({9 + 6 k}\right) + 25$
Numbers $44$ and greater of the form $6 n + 2$ can be expressed as:
:$\left({9 + 6 k}\right) + 35$
We have that $40$ and $42$ are of the form $6 n + 4$ and $6 n$ respectively.
Hence all even integers greater than $38$ are accounted for.
== Also see ==
* Positive Even Integers as Sum of 2 Composite Odd Integers in 2 Ways
* Largest Even Integer not expressible as Sum of 2 k Odd Composite Integers
== Sources ==
* 
* 
* : $38$
Category:Composite Numbers
Category:Odd Integers
Positive Image of Point of Continuous Real Function implies Positive Closed Interval of Domain

== Theorem ==
Let $f: \R \to \R$ be a continuous real function.
Let $a \in \R$ such that $\map f a > 0$.
Then:
:$\exists k \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \closedint {a - \delta} {a + \delta}: \map f x \ge k$
== Proof ==
Let $\map f a = l$ where $l > 0$.
As $f$ is continuous:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \size {y - x} < \delta \implies \size {\map f y - \map f x} < \epsilon$
Let $\epsilon = \dfrac l 2 = k$.
Then:
:$\exists \delta' \in \R_{>0}: \forall y \in \R: \size {y - x} < \delta' \implies \size {\map f y - \map f a} < \dfrac l 2$
Thus:
:$\forall x \in \openint {a - \delta'} {a + \delta'}: \map f x > \dfrac l 2$
Let $\delta = \dfrac {\delta'} 2$
Then $a - \delta \in \openint {a - \delta'} {a + \delta'}$ and $a + \delta \in \openint {a - \delta'} {a + \delta'}$.
Thus:
:$\closedint {a - \delta} {a + \delta} \subseteq \openint {a - \delta'} {a + \delta'}$
and hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $8$
Category:Continuous Functions
Positive Infinity is Greatest Element

== Theorem ==
Let $\left({\overline \R, \le}\right)$ be the extended real numbers with their usual ordering.
Then $+\infty$ is the greatest element of $\overline \R$.
== Proof ==
We have, by definition of the usual ordering on $\overline \R$:
:$\forall x \in \overline \R: x \le +\infty$
That is, $+\infty$ is the greatest element of $\overline \R$.
== Also see ==
* Negative Infinity is Smallest Element
Category:Order Theory
Category:Extended Real Numbers
Positive Infinity is Maximal

== Theorem ==
Let $\struct {\overline \R, \le}$ be the extended real numbers with the usual ordering.
Then $+\infty$ is a maximal element of $\overline \R$.
== Proof ==
By the definition of the usual ordering on the extended real numbers:
:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
Suppose $x \in \overline \R$ and $+\infty \le x$.
That is:
: $\tuple {+\infty, x} \in {\le}$
By the definition of union, $\tuple {+\infty, x}$ must lie in one of the three sets whose union forms $\le$.
Since ${\le_\R} \subseteq \R \times \R$ and $+\infty \notin \R$:
:$\tuple {+\infty, x} \notin {\le_\R}$
Since $+\infty \ne -\infty$ by the definition of the extended real numbers:
:$\tuple {+\infty, x} \notin \set {\tuple {-\infty, x}: x \in \overline \R}$
Therefore:
:$\tuple {+\infty, x} \in \set {\tuple {x, +\infty}: x \in \overline \R}$
and we conclude that $x = +\infty$.
That is, $+\infty$ is a maximal element of $\overline \R$.
== Also see ==
* Negative Infinity is Minimal
Category:Order Theory
Category:Extended Real Numbers
Positive Integer Greater than 1 has Prime Divisor

== Lemma ==
Every positive integer greater than $1$ has at least one divisor which is prime.
== Proof 1 ==
== Proof 2 (not using FTA) ==
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 22 \alpha$
Category:Positive Integer Greater than 1 has Prime Divisor
Category:Prime Numbers
Positive Integer Greater than 1 has Prime Divisor/Proof 2

== Lemma ==
== Proof ==
 that there are some positive integers which are not divisible by some prime.
Let $S = \set {n \in \Z: n > 1: \neg \exists p \in \Bbb P: p \divides n}$.
That is:
:$S = \set {\text {all integers not divisible by a prime} }$
Let $n \in S$ be the smallest of these.
As $S$ is bounded below by $1$, this is bound to exist, by Set of Integers Bounded Below by Integer has Smallest Element.
So:
:$\neg \exists x \in S: x < n$
Now $n$ cannot be prime itself:
:$\paren {\paren {n \in \Bbb P} \land \paren {n \divides n} \implies n \notin S} \implies n \notin \Bbb P$
So from Composite Number has Two Divisors Less Than It:
:$\exists r, s \in \Z: n = r s, 1 < r < n, 1 < s < n$
There are two possibilities:
:$(1):\quad$ Neither $r$ nor $s$ has a prime divisor
:$(2):\quad$ At least one of $r$ and $s$ has a prime divisor.
If either $r$ or $s$ has a prime divisor, then:
:$\exists p \in \Bbb P: \paren {p \divides r} \lor \paren {p \divides s} \implies p \divides n$
This contradicts our claim that $n$ is not divisible by some prime.
However, if neither $r$ nor $s$ has a prime divisor, it follows that $r, s \in S$.
But as $r, s < n$, this contradicts our choice of $n$ as the smallest element of $S$.
Therefore there can be no such $n$, therefore $S = \O$, and all positive integers greater than one are divisible by some prime.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Corollary $3$
Category:Positive Integer Greater than 1 has Prime Divisor
Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then $n$ can be expressed as the sum of $2$ or more consecutive (strictly) positive integers $n$ is not a power of $2$.
== Proof ==
=== Necessary Condition ===
Let $a$ be the smallest of $m$ consecutive (strictly) positive integers, where $m \ge 2$.
From Sum of Arithmetic Sequence, their sum is $\dfrac {m \paren {2 a + m - 1} } 2$. 
 $\dfrac {m \paren {2 a + m - 1} } 2$ is a power of $2$.
Then $m$ and $2 a + m - 1$ must also be powers of $2$.
Since $m \ge 2$, $m$ must be even.
Then $2 a + m - 1$ is odd.
Together with $2 a + m - 1$ being a power of $2$, we have $2 a + m - 1 = 1$.
However, $2 a + m - 1 > 2 \times 0 + 2 - 1 = 1$.
This is a contradiction.
Therefore any sum of $2$ or more consecutive integers cannot be a power of $2$.
=== Sufficient Condition ===
Let $n$ be an integer that is not a power of $2$.
Then $n$ contains an odd factor greater than $1$.
Let $d = 2 m + 1$ be such a factor.
Then:
:$\dfrac n d - m, \dfrac n d - m + 1, \dots, \dfrac n d + m$
is a sequence of $2 m + 1$ consecutive integers.
From Sum of Arithmetic Sequence:
It may happen that $\dfrac n d - m \le 0$.
In that case, notice that:
:$\dfrac n d - m \le m - \dfrac n d \le \dfrac n d + m$
and so $m - \dfrac n d$ is also (somewhere) in our sequence of $2 m + 1$ consecutive integers.
So all negative integers in our sequence of $2 m + 1$ consecutive integers cancel out with their positive counterparts.
Hence we can remove all numbers from $\dfrac n d - m$ to $m - \dfrac n d$.
As a result of that process, the sum remains unchanged.
We see that the number of integers eliminated is $2 \paren {m - \dfrac n d} + 1$.
Since $\dfrac n d$ is a factor of $n$, we have $\dfrac n d \ge 1$.
Hence the number of integers remaining is:
This finishes our proof.
== Examples ==
== Sources ==
* : $2$
* : $2$
Category:Number Theory
Category:2
Category:Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2
Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2/Examples/18

== Examples of Use of Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2 ==
$18$ is not a power of $2$.
In fact, it has $3$ and $9$ as its odd factors.
Hence by the construction, we can express $18$ as:
Category:Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2
Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$2$
== Mistake ==
:An integer is the sum of a sequence of consecutive integers it is not a power of $2$.
The property stated holds only for positive integers.
Otherwise, trivially, we would have:
:$2 = -1 + 0 + 1 + 2$
What it ought to say is:
:A positive integer is the sum of two or more consecutive (strictly) positive integers it is not a power of $2$.
This mistake has been partially corrected in .
:A positive integer is the sum of two or more consecutive integers it is not a power of $2$.
== Sources ==
* : $2$
Category:Mistakes/Curious and Interesting Numbers
Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$2$
== Mistake ==
:A positive integer is the sum of two or more consecutive integers it is not a power of $2$.
This should actually say:
:A positive integer is the sum of two or more consecutive positive integers it is not a power of $2$.
It is understood that "positive integer" is taken in its strictly positive sense.
With the statement as it stands, all integers, whether positive or negative, are trivially the sum of a sequence of integers, for example:
:$2 = \paren {-1} + 0 + 1 + 2$
:$-4 = \paren {-4} + \paren {-3} + \paren {-2} + \paren {-1} + 0 + 1 + 2 + 3$
This mistake is in fact a partial correction to its statement in the $1986$ edition :
:An integer is the sum of a sequence of consecutive integers it is not a power of $2$.
== Sources ==
* : $2$
Category:Mistakes/Curious and Interesting Numbers
Positive Integer is Well-Defined

== Theorem ==
"Positive" as applied to an integer is well-defined.
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
Thus, what we are trying to prove is:
:$\eqclass {a, b} {} = \eqclass {c, d} {} \land b < a \implies d < c$
By definition:
:$\eqclass {a, b} {} = \eqclass {c, d} {} \iff a + d = b + c$
So:
Category:Integers
Positive Integers Expressible by Sum of Integers whose Reciprocals Sum to 1

== Theorem ==
Every positive integer over $77$ can be expressed as the sum of positive integers whose reciprocals add up to $1$.
The full sequence of numbers that cannot be expressed as such is:
:$2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 33, 34, 35, 36, 39, 40, 41, 42, 44, 46, 47, 48, 49, 51, 56, 58, 63, 68, 70, 72, 77$
== Proof ==
== Examples ==
== Sources ==
* 
* : $77$
* : $77$
Category:Reciprocals
Category:Positive Integers Expressible by Sum of Integers whose Reciprocals Sum to 1
Positive Integers Not Expressible as Sum of Distinct Non-Pythagorean Primes

== Theorem ==
The following positive integers cannot be expressed as the sum of distinct non-pythagorean primes:
:$1, 2, 4, 5, 6, 8, 9, 12, 13, 15, 16, 17, 20, 24, 25, 27, 28, 32, 35, 36, 39, 48, 51, 55$
All positive integers greater than $55$ can be so expressed.
== Proof ==
== Sources ==
* : $55$
* : $55$
Category:Prime Numbers
Positive Integers not Expressible as Sum of Fewer than 19 Fourth Powers

== Theorem ==
The following positive integers are the only ones which cannot be expressed as the sum of fewer than $19$ fourth powers:
=== $79$ as Sum of $19$ Fourth Powers ===
=== $159$ as Sum of $19$ Fourth Powers ===
=== $319$ as Sum of $19$ Fourth Powers ===
=== $399$ as Sum of $19$ Fourth Powers ===
== Proof ==
== Also see ==
* Hilbert-Waring Theorem
Category:Fourth Powers
Category:Hilbert-Waring Theorem
Positive Integers under Scaled Euclidean Metric is not Complete Metric Space

== Theorem ==
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the scaled Euclidean metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Then $\struct {\Z_{>0}, \delta}$ is not a complete metric space.
== Proof ==
Consider the sequence $\sequence {x_n}$ in $\Z_{>0}$ defined as:
:$\forall n \in \N: x_n = n$
From Cauchy Sequence in Positive Integers under Scaled Euclidean Metric:
:$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$.
But $\sequence {x_n}$ is not convergent to any $m \in \Z_{>0}$.
Hence the result, by definition of complete metric space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $10$
Category:Scaled Euclidean Metric
Category:Complete Metric Spaces
Positive Integers under Usual Metric is Complete Metric Space

== Theorem ==
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Then $\struct {\Z_{>0}, d}$ is a complete metric space.
== Proof ==
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.
From Cauchy Sequence in Positive Integers under Usual Metric is eventually Constant:
:$\sequence {x_n}$ is a convergent sequence to some $n \in \Z_{>0}$.
Hence the result by definition of complete metric space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $10$
Category:Euclidean Metric
Category:Complete Metric Spaces
Positive Integers which are Euler Phi Value for 17 Integers

== Theorem ==
There are $17$ positive integers which have an Euler $\phi$ value of the following:
:$72, 96, 120, \ldots$
== Proof ==
From Numbers with Euler Phi Value of 72, it is seen that $17$ positive integers have an Euler $\phi$ value of $72$:
:$73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252, 270$
== Sources ==
* : $72$
Category:Euler Phi Function
Positive Integers which are not Divisor Sum Values

== Theorem ==
The following positive integers are not the values of the divisor sum function $\map {\sigma_1} n$ for any $n$:
:$2, 5, 9, 10, 11, 16, 17, 19, 21, 22, 23, 25, 26, 27, 29, 33, 34, 35, 37, 41, 43, 45, 46, 47, 49, 50, 51, 52, 53, 55, 58, 59, 61, 64, 65, \ldots$
== Proof ==
Category:Divisor Sum Function
Positive Integers whose Square Root equals Sum of Digits

== Theorem ==
The following positive integers have a square root that equals the sum of their digits:
:$0, 1, 81$
and there are no more.
== Proof ==
By considering the square roots, we are looking for positive integers with a square for which its sum of digits is equal to the original number.
It is easy to verify the result up to $36$.
We prove that for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.
Let $n$ be a $d$-digit number.
Then $n < 10^d$.
Thus $n^2 < 10^{2 d}$.
Therefore $n^2$ has at most $2 d$ digits.
Hence the sum of digits of $n^2$ is at most $18 d$.
If $n$ is a $2$-digit number, we have $n > 36 = 18 d$.
Now suppose $d \ge 3$.
Then:
Therefore for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.
== Sources ==
* : $81$
Category:Square Roots