nyuuzyou/wikis huggingface dataset:
Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared/Examples

== Examples of Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared ==
=== $p = 3$ ===
=== $p = 5$ ===
Category:Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared
Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared/Examples/3

== Example of Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared ==
The smallest positive integer $n$ greater than $1$ such that:
:$n^{3 - 1} \equiv 1 \pmod {3^2}$
is $8$.
== Proof ==
Only positive integers coprime to $3$ need be checked.
We have that $3^2 = 9$.
Thus:
== Sources ==
* : $64$
* : $64$
Category:Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared
Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared/Examples/5

== Example of Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared ==
The smallest positive integer $n$ greater than $1$ such that:
:$n^{5 - 1} \equiv 1 \pmod {5^2}$
is $7$.
== Proof ==
Only positive integers coprime to $5$ need be checked.
We have that $5^2 = 25$.
Thus:
== Sources ==
* : $64$
* : $64$
Category:Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared
Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared/Mistake

== Source Work ==
:
:The Dictionary
::$64$
:
:The Dictionary
::$64$
== Mistake ==
:For every prime $p$, there are values of $a$ such that $a^{p - 1} = 1$ is actually divisible by ${p^2}$. The smallest such value for $p = 3$ is $8^2 = 64$: $64 - 1$ is divisible by $3^2 = 9$.
But for any prime $p$, the positive integer $1$ fits the bill trivially:
:$1^{p - 1} - 1 = 0$
which is divisible by all integers, by Integer Divides Zero, and so of course in particular by $p^2$.
Hence the statement needs to be amended to:
:For every prime $p$, there are values of $a$ greater than $1$ such that $a^{p - 1} = 1$ is actually divisible by ${p^2}$.
== Sources ==
* : $64$
* : $64$
Category:Mistakes/Curious and Interesting Numbers
Existence of One-Sided Inverses on Natural Numbers whose Composition is Identity Mapping

== Theorem ==
Consider the set of natural numbers $\N$.
There exist mappings $f: \N \to \N$ and $g: \N \to \N$ such that:
:$g \circ f = I_\N$
where:
:$\circ$ denotes composition of mappings
:$I_\N$ denotes the identity mapping on $\N$
such that neither $f$ nor $g$ are permutations on $\N$.
== Proof ==
Let $f: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map f x = x + 1$
Let $g: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map g x = \begin {cases} x - 1 & : x > 0 \\ 0 & : x = 0 \end {cases}$
It is apparent by inspection that:
:$f$ is injective
but
:$f$ is not surjective, as there exists no $x \in \N$ such that $x + 1 = 0$
It is also apparent by inspection that:
:$g$ is surjective
but
:$g$ is not injective, as $\map g 0 = \map g 1 = 0$
Thus neither $f$ nor $g$ are bijective, and hence not permutations on $\N$.
However, it is also apparent by inspection that:
:$g \circ f = I_\N$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.17 \ \text {(b)}$
Category:Natural Numbers
Category:Inverse Mappings
Existence of Ordered Dual Basis

== Definition ==
Let $R$ be a commutative ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be an $n$-dimensional module over $R$.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $G^*$ be the algebraic dual of $G$.
Let $\sequence {a'_n}$ be the ordered dual basis of $G^*$.
This ordered dual basis $\sequence {a'_n}$ is guaranteed to exist.
== Proof ==
From Basis for $R$-Module $R$, $\set {1_R}$ is a basis of the $R$-module $R$.
Hence from Basis for Set of Linear Transformations, this ordered dual basis as defined exists.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Algebra
Existence of Ordinal with no Surjection from Set

== Theorem ==
Let $S$ be a set.
Then, there exists a non-empty ordinal $\alpha$ such that there is no surjection from $S$ to $\alpha$.
== Proof ==
By Hartogs' Lemma, let $\alpha$ be an ordinal such that there is no injection from $\alpha$ to $\powerset S$.
From Empty Mapping is Injective, it follows that $\alpha$ is non-empty.
 there is a surjection $\phi : S \to \alpha$.
Define $\psi : \alpha \to \powerset S$ as:
:$\map \psi \gamma = \set {x \in S : \map \phi x = \gamma}$
Let $\beta, \gamma \in \alpha$ such that:
:$\map \psi \beta = \map \psi \gamma$
Since $\phi$ is a surjection, there is some $x \in S$ such that:
:$\map \phi x = \beta$
from which it follows that:
:$x \in \map \psi \beta = \map \psi \gamma$
But then, by definition:
:$\map \phi x = \gamma$
so therefore, by definition of mapping:
:$\beta = \gamma$
Thus, by definition, $\psi$ is an injection from $\alpha$ to $\powerset S$.
Category:Ordinals
Category:Surjections
Existence of Orthonormal Frames

== Theorem ==
Let $\struct{M, g}$ be a $n$-dimensional Riemannian manifold with or without boundary. 
Let $p \in M$ be a point.
Let $TM$ be the tangent bundle of $M$.
Let $U \subseteq M$ be an open subset.
Suppose $\tuple {X_j}$ is a smooth local frame for $TM$ over $U$.
Then for all $p \in M$ there is a smooth orthonormal frame $\tuple {E_j}$ over $U$ such that:
:$\forall k \in \N : 1 \le k \le n : \map \span {\bigvalueat {E_1} p, \ldots, \bigvalueat {E_n} p} = \map \span {\bigvalueat {X_1} p, \ldots, \bigvalueat {X_n} p}$
In particular, for every $p \in M$ there is a smooth orthonormal frame $\tuple {E_j}$ defined on some neighborhood of $p$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Definitions
Category:Orthonormal Frames
Existence of Paracompact Space which is not Compact

== Theorem ==
There exists at least one example of a paracompact topological space which is not also a compact topological space.
== Proof ==
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
From Real Number Line is Paracompact, $T$ is a paracompact space.
From Real Number Line is not Countably Compact, $T$ is not a countably compact space.
From Compact Space is Countably Compact, it follows that $T$ is not a compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Paracompact Spaces
Category:Compact Spaces
Existence of Positive Root of Positive Real Number

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
== Proof ==
=== Positive Exponent ===
=== Negative Exponent ===
== Also see ==
* Definition:Root of Number
* Existence and Uniqueness of Positive Root of Positive Real Number
Category:Existence of Positive Root of Positive Real Number
Category:Existence and Uniqueness of Positive Root of Positive Real Number
Category:Roots of Numbers
Category:Real Numbers
Existence of Positive Root of Positive Real Number/Negative Exponent

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n < 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
== Proof ==
Let $m = -n$.
Then $m > 0$.
Let $g$ be the real function defined on $\hointr 0 \to$ defined by:
:$\map g y = y^m$
Since $x \ge 0$:
:$\dfrac 1 x \ge 0$
By Existence of Positive Root of Positive Real Number: Positive Exponent there is a $y > 0$ such that:
:$\map g y = \dfrac 1 x$
It follows from the definition of power that:
Category:Existence of Positive Root of Positive Real Number
Existence of Positive Root of Positive Real Number/Positive Exponent

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
== Proof 1 ==
== Proof 2 ==
Category:Existence of Positive Root of Positive Real Number
Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 1

== Theorem ==
== Proof ==
Let $f$ be the real function defined on the unbounded closed interval $\hointr 0 \to$ defined by $\map f y = y^n$.
Consider first the case of $n > 0$.
By Strictly Positive Integer Power Function is Unbounded Above:
:$\exists q \in \R_{>0}: \map f q \ge x$
Since $x \ge 0$:
:$\map f 0 \le x$
By the Intermediate Value Theorem:
:$\exists y \in \R: 0 \le y \le q, \map f y = x$
Hence the result has been shown to hold for $n > 0$.
Category:Existence of Positive Root of Positive Real Number
Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 2

== Theorem ==
== Proof ==
Let $S$ be the set consisting of the positive real numbers $t$ such that $t^n < x$.
Let $t = \dfrac x {1 - x}$.
Then $0 < t < 1$ and so:
:$t^n < t < x$
demonstrating that $S \ne \O$.
Let $t_0 = 1 + x$.
Then $t > t_0$ implies $t^n \ge t > x$.
So $t \notin S$.
Hence $t_0$ is an upper bound of $S$.
By the Continuum Property, $S$ has a supremum.
Let $y = \sup S$.
 $y_n < x$.
Choose $h \in \R$ such that $0 < h < 1$ and such that:
:$h < \dfrac {x - y^n} {\paren {1 + y}^n - y^n}$
Let $\dbinom n m$ denote a binomial coefficient.
By the Binomial Theorem, the coefficient of $z^m$ in the expansion of $\paren {1 + z}^n$ is
We have:
Thus we have $y + h \in S$, which contradicts the fact that $y$ is an upper bound of $S$.
 $y^n > x$.
Let $k \in \R$ be chosen such that:
:$0 < k < 1$
:$k < y$
:$k < \dfrac {y^n - x} {\paren {1 + y}^n - y^n}$
Let $t \ge y - k$.
Then:
Thus we have $y - k$ is an upper bound of $S$
This contradicts the fact that $y$ is the supremum of $S$.
It has been shown that both $y^n < x$ and $y^n > x$ lead to a contradiction.
Hence it must be the case that $y^n = x$.
Hence the result.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.37$. Theorem
Category:Existence of Positive Root of Positive Real Number
Existence of Prime-Free Sequence of Natural Numbers

== Theorem ==
Let $n$ be a natural number.
Then there exists a sequence of consecutive natural numbers of length $n$ which are all composite.
== Proof ==
Consider the number:
:$N := \paren {n + 1}!$
where $!$ denotes the factorial.
Let $i \in I$ where:
:$I = \set {i \in \N: 2 \le i \le n + 1}$
We have that $\size I = n$.
Then for all $i \in I$:
and so $N + i$ is composite.
That is:
:$N + 2, N + 3, \ldots, N + n, N + n + 1$
are all composite.
== Sources ==
* : Chapter $\text {B}.16$: The Sequence of Primes: Footnote $1$
Category:Number Theory
Category:Prime Numbers
Existence of Prime between Prime and Factorial

== Theorem ==
Let $p$ be a prime number.
Then there exists a prime number $q$ such that:
:$p < q \le p! + 1$
where $p!$ denotes the factorial of $p$.
== Proof ==
Let $N = p! + 1$.
If $N$ is prime, then the proof is complete.
Otherwise, from Positive Integer Greater than 1 has Prime Divisor, $N$ has a prime divisor $q$.
From Absolute Value of Integer is not less than Divisors:
:$q < N$
 $q \le p$.
Then by the definition of factorial:
:$q \divides p!$
But $q$ was chosen so that :$q \divides N$.
Let $p! = q m_1, N = q m_2$.
Then:
:$1 = N - p! = q \paren {m_2 - m_1}$
and so:
:$q \divides 1$
From Absolute Value of Integer is not less than Divisors:
:$q \le 1$
and so $q$ cannot be a prime number.
It follows by Proof by Contradiction that $q > p$.
== Also see ==
* Euclid's Theorem, of which this is a variant.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $4$
Category:Prime Numbers
Existence of Probability Space and Discrete Random Variable

== Theorem ==
Let $I$ be an arbitrary countable indexing set.
Let $S = \set {s_i: i \in I} \subset \R$ be a countable set of real numbers.
Let $\set {\pi_i: i \in I} \subset \R$ be a countable set of real numbers which satisfies:
:$\ds \forall i \in I: \pi_i \ge 0, \sum_{i \mathop \in I} \pi_i = 1$
Then there exists:
:a probability space $\struct {\Omega, \Sigma, \Pr}$
and:
:a discrete random variable $X$ on $\struct {\Omega, \Sigma, \Pr}$
such that the probability mass function $p_X$ of $X$ is given by:
 
== Proof ==
Take $\Omega = S$ and $\Sigma = \powerset S$ (the power set of $S$).
Then let:
:$\ds \map \Pr A = \sum_{i: s_i \mathop \in A} \pi_i$
for all $A \in \Sigma$.
Then we can define $X: \Omega \to \R$ by:
:$\forall \omega \in \Omega: \map X \omega = \omega$
This suits the conditions of the assertion well enough.
== Motivation ==
What this theorem allows us to do is ignore all the detail of sample spaces, event spaces and probability measure, and merely say:
:For each $i \in I$, let $X$ be a (discrete) random variable which takes value $s_i$ with probability $\pi_i$
and we know that such a random variable exists without having construct it every time.
== Sources ==
* : $\S 2.1$: Probability mass functions: Theorem $2 \text{A}$
Category:Probability Theory
Existence of Product of Three Distinct Primes between n and 2n

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 15$.
Then between $n$ and $2 n$ there exists at least one integer which is the product of $3$ distinct prime numbers.
== Proof ==
Let $16 \le n \le 29$.
Then:
:$n < 30 < 2 n$
and we have:
:$30 = 2 \times 3 \times 5$
which is a product of $3$ distinct primes.
Hence the result holds for $n$ in that range.
Let $n \ge 30$.
Then by the Division Theorem:
:$\exists q, r \in \N: n = 6 q + r$, $0 \le r < 6$, $q \ge 5$
By Bertrand-Chebyshev Theorem, there is a prime $p$ where $5 \le q < p < 2 q$.
Hence $p$ is not $2$ or $3$, and:
This proves the result.
== Sources ==
* : No. $94$
* : $15$
Category:Prime Numbers
Category:15
Existence of Pseudo-Riemannian Adapted Orthonormal Frames

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a pseudo-Riemannian manifold.
Let $M \subseteq \tilde M$ be an embedded pseudo-Riemannian or Riemannian submanifold. 
Then for each $p \in M$, there exists a smooth orthonormal frame on a neighborhood of $p \in \tilde M$ that is adapted to $M$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Pseudo-Riemannian Geometry
Existence of Pseudocompact Space which is not Countably Compact

== Theorem ==
There exists at least one example of a pseudocompact topological space which is not also a countably compact space.
== Proof ==
Let $T$ be an infinite particular point space.
From Particular Point Space is Pseudocompact, $T$ is a pseudocompact space.
From Infinite Particular Point Space is not Weakly Countably Compact, $T$ is not a weakly countably compact space.
From Countably Compact Space is Weakly Countably Compact, $T$ is not a countably compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Pseudocompact Spaces
Category:Countably Compact Spaces
Existence of Radius of Convergence of Complex Power Series

== Theorem ==
Let $\xi \in \C$.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a (complex) power series about $\xi$.
Then there exists a radius of convergence $R \in \overline \R$ of $\map S z$.
=== Absolute Convergence ===
=== Divergence ===
== Also see ==
* Existence of Interval of Convergence of Power Series for a proof of the same result in real numbers.
== Sources ==
* : $\S 4.4$. Power Series
Category:Complex Power Series
Category:Convergence
Existence of Radius of Convergence of Complex Power Series/Absolute Convergence

== Theorem ==
Let $\xi \in \C$.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a complex power series about $\xi$.
Let $R$ be the radius of convergence of $\map S z$.
Let $\map {B_R} \xi$ denote the open $R$-ball of $\xi$.
Let $z \in \map {B_R} \xi$.
Then $\map S z$ converges absolutely.
If $R = +\infty$, we define $\map {B_R} \xi = \C$.
== Proof ==
Let $z \in \map {B_R} \xi$.
By definition of the open $R$-ball of $\xi$:
:$\cmod {z - \xi} < R$
where $\cmod z$ denotes the complex modulus of $z$.
By definition of radius of convergence, it follows that $\map S z$ converges.
Suppose $R$ is finite.
Let $\epsilon = R - \cmod {z - \xi} > 0$.
Now, let $w \in \map {B_R} \xi$ be a complex number such that $R - \cmod {w - \xi} = \dfrac \epsilon 2$.
Such a $w$ exists because, if at a loss, we can always let $w = \xi + R - \dfrac \epsilon 2$.
If $R = +\infty$, then let $w \in C$ be any complex number such that $\cmod {z - \xi} < \cmod {w - \xi}$.
Then:
From the $n$th Root Test, it follows that $\map S z$ converges absolutely.
== Also see ==
* Existence of Interval of Convergence of Power Series for a proof of the same result in real numbers.
Category:Complex Power Series
Category:Convergence
Existence of Radius of Convergence of Complex Power Series/Divergence

== Theorem ==
Let $\xi \in \C$.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a (complex) power series about $\xi$.
Let $R$ be the radius of convergence of $\map S z$.
Let $\map { {B_R}^-} \xi$ denote the closed $R$-ball of $\xi$.
Let $z \notin \map { {B_R}^-} \xi$.
Then $\map S z$ is divergent.
== Proof ==
Let $z \notin \map { {B_R}^-} \xi$.
Then by definition of the closed $R$-ball of $\xi$:
:$\cmod {z - \xi} > R$
where $\cmod z$ denotes the complex modulus of $z$.
By definition of radius of convergence, there exists $w \in \C$ such that:
:$\cmod {w - \xi} < \cmod {z - \xi}$
and $S \paren w$ is divergent.
Then:
From the $n$th Root Test, it follows that $\map S z$ is divergent.
== Also see ==
* Existence of Interval of Convergence of Power Series for a proof of the same result in real numbers.
Category:Complex Power Series
Category:Convergence
Existence of Rational Powers of Irrational Numbers/Proof 1

== Theorem ==
== Proof ==
We have that:
: $\sqrt 2$ is irrational.
: $2$ is trivially rational, as $2 = \dfrac 2 1$.
Consider the number $q = \sqrt 2^{\sqrt 2}$, which is irrational by the Gelfond-Schneider Theorem.
Thus:
: $q^{\sqrt 2} = \left({\sqrt 2^{\sqrt 2}}\right)^{\sqrt 2} = \sqrt 2 ^{\left({\sqrt 2}\right) \left({\sqrt 2}\right)} = \sqrt 2^2 = 2$
is rational.
So $a = q = \sqrt 2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrationals.
Category:Existence of Rational Powers of Irrational Numbers
Existence of Rational Powers of Irrational Numbers/Proof 2

== Theorem ==
== Proof ==
Given that $2$ is rational and $\sqrt 2$ is irrational, consider the number $q = \sqrt 2^{\sqrt 2}$.
We consider the two cases.
:$(1): \quad$ If $q$ is rational then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired irrational numbers.
:$(2): \quad$ If $q$ is irrational then $q^{\sqrt 2} = \paren {\sqrt 2^{\sqrt 2} }^{\sqrt 2} = \sqrt 2^{\paren {\sqrt 2} \paren {\sqrt 2} } = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt 2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrational numbers.
== Sources ==
* 
* : $\S 1.2.5$: An aside: proof by contradiction: Theorem $1.26$
* 
Category:Existence of Rational Powers of Irrational Numbers
Existence of Rational Powers of Irrational Numbers/Proof 3

== Theorem ==
== Proof ==
Consider the equation:
:$\paren {\sqrt 2}^{\log_{\sqrt 2} 3} = 3$
We have that $\sqrt 2$ is irrational and $3$ is (trivially) rational.
It remains to be proved $\log_{\sqrt 2} 3$ is irrational.
We have:
From Irrationality of Logarithm, $\log_2 3$ is irrational.
Therefore $2 \, \log_2 3$ is irrational.
Thus there exist two irrational numbers $a = \sqrt 2$ and $b = \log_{\sqrt 2} 3$ such that $a^b$ is rational.
== Also see ==
* Positive Rational Number as Power of Number with Power of Itself for an interesting related result.
== Sources ==
* 
Category:Existence of Rational Powers of Irrational Numbers
Existence of Real Logarithm

== Theorem ==
Let $b, y \in \R$ such that $b > 1$ and $y > 0$.
Then there exists a unique real $x \in \R$ such that $b^x = y$.
This $x$ is called the logarithm of $y$ to the base $b$.
Also see the definition of a (general) logarithm.
== Proof ==
We start by establishing a lemma:
=== Lemma 1 ===
Let $t \in \R$ be such that $t > 1$, and let $n \in \N$ be such that $n > \dfrac {b - 1} {t - 1}$.
Then:
:$b^{\frac 1 n} < t$
Because $b > 1$, we have that $b^{\frac 1 n} > 1$.
From Sum of Geometric Sequence and this observation, we find that:
:$\ds \frac {\paren {b^{\frac 1 n} }^n - 1} {b^{\frac 1 n} - 1} = \sum_{k \mathop = 0}^{n - 1} \paren {b^{\frac 1 n} }^k > n$
Rewriting this inequality, we obtain:
:$b - 1 > n \paren {b^{\frac 1 n} - 1}$
As $n > \dfrac {b - 1} {t - 1}$, this means that:
:$t - 1 > b^{\frac 1 n} - 1$
We conclude that:
:$b^{\frac 1 n} < t$
Now let $w \in \R$ be any number satisfying $b^w < y$.
Then $t = b^{-w} y > 1$, so we can apply Lemma 1.
It follows that for $n > \dfrac {b - 1} {t - 1}$:
:$b^{\frac 1 n} < b^{-w} y$
and thus:
:$b^{w + \frac 1 n} < y$
Similarly, when $b^w > y$ and $n > \dfrac {b - 1} {t - 1}$, we have:
:$b^{w - \frac 1 n} > y$
Next, let $W = \set {w \in \R: b^w < y}$.
Let $x = \sup W$.
The following lemma helps us to verify existence of $x$.
=== Lemma 2 ===
For any real number $z > 0$, we have $n \in \N$ such that:
:$b^{-n} < z < b^n$
Observe that, as $z > 0$, $b^{-n} < z$ whenever $b^n > \dfrac 1 z$.
Therefore, it suffices to show that there is an $n \in \N$ such that $b^n > z$ exists.
By Sum of Geometric Sequence, we obtain for any $n \in \N$:
:$\ds \paren {b - 1} \sum_{k \mathop = 0}^{n - 1} b^k = b^n - 1$
As $b > 1$, we obtain:
:$n \paren {b - 1} < b^n - 1$
This implies that any $n$ with $n \paren {b - 1} > z - 1$ has the desired property.
Lemma 2 shows that $x$ is the supremum of a non-empty set with an upper bound, and therefore it exists.
We will now prove that $b^x = y$.
Suppose $b^x < y$.
We have seen that there exists an $n \in \N$ such that $b^{x + \frac 1 n} < y$.
That is:
:$x + \dfrac 1 n \in W$
But this is impossible as $x = \sup W$.
Thus:
:$b^x \ge y$
Now suppose $b^x > y$.
We have seen that there exists an $n \in \N$ such that $b^{x - \frac 1 n} > y$.
That is:
:$x - \dfrac 1 n \notin W$
As $x - \dfrac 1 n$ is not the supremum of $W$, we find a $w \in W$ such that:
:$x - \dfrac 1 n < w < x$
But then we have, as $b > 1$:
:$b^{x - \frac 1 n} - b^w = b^{x - \frac 1 n} \paren {1 - b^{w - x + \frac 1 n} } < 0$
This contradicts our initial choice of $x - \dfrac 1 n$ outside $W$.
We conclude that $b^x = y$.
Lastly, suppose that $b^x = y = b^z$.
Then:
:$b^{z - x} = \dfrac {b^z} {b^x} = 1 = \dfrac {b^x} {b^z} = b^{x - z}$
It is concluded that $x = z$.
Category:Logarithms
Existence of Real Number at Distance Zero from Open Real Interval not in Interval

== Theorem ==
Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ denote the distance between $x$ and $S$.
Let $I \subseteq \R$ be an open real interval such that $I \ne \O$ and $I \ne \R$.
Then:
:$\exists x \notin I: \map d {x, I} = 0$
== Proof ==
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
As $I \ne \O$ and $I \ne \R$ it follows that one of the following applies:
It follows by the definition of open real interval that $I$ has either an infimum $a$, or a supremum $b$, or both.
 
Thus the required value of $x$ is either $a$ or $b$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (5) \ \text {(iii)}$
Category:Distance Function
Existence of Real Polynomial with no Real Root

== Theorem ==
There exist polynomials in real numbers $\R$ which have no roots in $\R$.
== Proof ==
Proof by Counterexample
Take the quadratic equation:
:$(1): \quad x^2 + 1 = 0$
From the Quadratic Formula, the solution to $(1)$ is:
But there is no real number $x$ such that:
:$x^2 = -1$
Hence the result.
== Sources ==
* : $1$: Complex Numbers: The Complex Number System
Category:Polynomial Theory
Category:Analysis
Existence of Regular Space which is not Tychonoff

== Theorem ==
There exists at least one example of a topological space which is a regular space, but is not also a Tychonoff space.
== Proof ==
Let $T$ be a Tychonoff corkscrew.
From Tychonoff Corkscrew is Regular, $T$ is a regular space.
From Tychonoff Corkscrew is not Completely Regular, $T$ is not a Tychonoff space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Separation Axioms
Category:Regular Spaces
Category:Tychonoff Spaces