avewright/proofwiki-math huggingface dataset:
Theorem: Rational Power of Product of Real Numbers

Let $r, s \in \R_{> 0}$ be (strictly) positive real numbers.
<onlyinclude>
Let $x \in \Q$ be a rational number.
Let $r^x$ be defined as $r$ to the power of $x$.
Then:
:$\paren {r s}^x = r^x s^x$

Proof:
Let $x = \dfrac p q$ where $p, q \in \Z$ and $q > 0$.
We have:
{{begin-eqn}}
{{eqn | l = r^x s^x
      | r = \paren {r^p}^{1 / q} \paren {s^p}^{1 / q}
      | c = 
}}
{{eqn | r = \paren {r^p s^p}^{1 / q}
      | c = 
}}
{{eqn | r = \paren {\paren {r s}^p}^{1 / q}
      | c = 
}}
{{eqn | r = \paren {r s}^{p / q}
      | c = 
}}
{{eqn | r = \paren {r s}^x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sign of Quadratic Function Between Roots

Let $a \in \R_{>0}$ be a (strictly) positive real number.
Let $\alpha$ and $\beta$, where $\alpha < \beta$, be the roots of the quadratic function:
:$\map Q x = a x^2 + b x + c$
whose discriminant $b^2 - 4 a c$ is (strictly) positive.
Then:
:$\begin {cases} \map Q x < 0 & : \text {when $\alpha < x < \beta$} \\ \map Q x > 0 & : \text {when $x < \alpha$ or $x > \beta$} \end {cases}$

Proof:
Because $b^2 - 4 a c > 0$, we have from Solution to Quadratic Equation with Real Coefficients that the roots of $\map Q x$ are real and unequal.
This demonstrates the existence of $\alpha$ and $\beta$, where by hypothesis we state that $\alpha < \beta$.
We can express $\map Q x$ as:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta}$
When $\alpha < x < \beta$ we have that:
:$x - \alpha > 0$
:$x - \beta < 0$
and so:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta} < 0$
{{qed|lemma}}
When $x < \alpha$ we have that:
:$x - \alpha < 0$
:$x - \beta < 0$
and so:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$
{{qed|lemma}}
When $x > \beta$ we have that:
:$x - \alpha > 0$
:$x - \beta > 0$
and so:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$
{{qed|lemma}}
Hence the result.
{{qed}}
Theorem: Minimum Value of Real Quadratic Function

Let $a \in \R_{>0}$ be a (strictly) positive real number.
Consider the quadratic function:
:$\map Q x = a x^2 + b x + c$
$\map Q x$ achieves a minimum at $x = -\dfrac b {2 a}$, at which point $\map Q x = c - \dfrac {b^2} {4 a}$.

Proof:
{{begin-eqn}}
{{eqn | l = \map Q x
      | r = a x^2 + b x + c
      | c = 
}}
{{eqn | r = \dfrac {4 \paren {a x}^2 + 4 a b x + 4 a c} {4 a}
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}
      | c = 
}}
{{end-eqn}}
As $\paren {2 a x + b}^2 > 0$, it follows that:
{{begin-eqn}}
{{eqn | l = a x^2 + b x + c
      | o = \ge
      | r = \dfrac {-\paren {b^2 - 4 a c} } {4 a}
      | c = 
}}
{{eqn | r = c - \dfrac {b^2} {4 a}
      | c = 
}}
{{end-eqn}}
Equality occurs when $2 a x + b = 0$, that is:
:$x = -\dfrac b {2 a}$
{{qed}}
Theorem: Number of Type Rational r plus s Root 2 is Irrational

Let $r, s \in \Q$ be rational numbers.
Then $r + s \sqrt 2$ is irrational.

Proof:
{{AimForCont}} $t = r + s \sqrt 2$ be rational.
Then:
:$\sqrt 2 = \dfrac {t - r} s$ is also rational.
This contradicts the fact that Square Root of 2 is Irrational.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Roots of Quadratic with Rational Coefficients of form r plus s Root 2

Consider the quadratic equation:
:$(1): \quad a^2 x + b x + c = 0$
where $a, b, c$ are rational.
Let $\alpha = r + s \sqrt 2$ be one of the roots of $(1)$.
Then $\beta = r - s \sqrt 2$ is the other root of $(1)$.

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = a \paren {r + s \sqrt 2}^2 + b \paren {r + s \sqrt 2} + c
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a r^2 + 2 a s + br + c} + \paren {2 a + b} s \sqrt 2
      | r = 0
      | c = 
}}
{{end-eqn}}
Because $a$, $b$, $c$, $r$ and $s$ are rational, it must be that $\paren {2 a + b} s = 0$.
Hence:
{{begin-eqn}}
{{eqn | l = a \paren {r - s \sqrt 2}^2 + b \paren {r - s \sqrt 2} + c
      | r = \paren {a r^2 + 2 a s + br + c} - \paren {2 a + b} s \sqrt 2
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
and so $\beta$ is also a root of $(1)$.
{{qed}}
Theorem: Descartes's Rule of Signs

Let $\map f x$ be a polynomial equation over the real numbers:
:$a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 = 0$
where $a_j \in \R$. 
Let $s_n$ be the number of variations in sign of $\map f x$.
Let $p_n$ be the number of positive real roots of $\map f x$ (counted with multiplicity). 
Then:
:$\forall n \in \Z_{>0}: s_n - p_n$ is a nonnegative even integer.
That is:
:for every polynomial of degree $1$ or higher, the number of sign changes less than the number of positive real roots will be a nonnegative even integer.

Proof:
The proof proceeds by induction.
:For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$s_n - p_n = 2 r$ where $r \in \Z_{\ge 0}$
Theorem: Supremum is not necessarily Greatest Element

Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ admit a supremum in $S$.
Then the supremum of $T$ in $S$ is not necessarily the greatest element of $T$.

Proof:
Proof by Counterexample:
Consider the subset $T$ of the set of real numbers $\R$:
:$T := \set {x \in \R: 1 \le x < 2}$
The number $2$ cannot be the greatest element of $T$ as $2 \notin T$.
However, $2$ is the supremum of $T$ in $S$.
Indeed, by definition:
:$\forall x \in T: x < 2$
So, let $x < 2$.
Then consider $y = \dfrac x 1 + \dfrac 2 1 = \dfrac {x + 2} 2$.
We have that $x < y < 2$ by Mediant is Between.
Thus $y \in T$ but $y > x$ and so $x$ cannot be the greatest element of $T$.
By the same construction, neither can $y$ be the supremum of $T$ in $S$.
The conclusion is that there is no greatest element of $T$.
Hence the result.
{{qed}}
Theorem: Infimum is not necessarily Smallest Element

Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ admit a infimum in $S$.
Then the infimum of $T$ in $S$ is not necessarily the smallest element of $T$.

Proof:
Let $V$ be the subset of the real numbers $\R$ defined as:
:$V := \set {x \in \R: x > 0}$
From Infimum of Subset of Real Numbers: Example 3, $V$ admits an infimum:
:$\inf V = 0$
But $V$ has no smallest element, as follows.
We note that $\inf V = 0 \notin V$.
{{AimForCont}} $x \in V$ is the smallest element of $V$.
Then $\dfrac x 2$ is also in $V$ as $\dfrac x 2 > 0$.
But $\dfrac x 2 < x$ which contradicts $x$ as being the smallest element of $V$.
Hence there can be no smallest element of $V$.
{{qed}}
Theorem: Closed Interval Defined by Absolute Value

:$\set {x \in \R: \size {\xi - x} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$
where $\closedint {\xi - \delta} {\xi + \delta}$ is the closed real interval between $\xi - \delta$ and $\xi + \delta$.

Proof:
{{begin-eqn}}
{{eqn | l = \size {\xi - x}
      | o = \le
      | r = \delta
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = -\delta
      | o = \le
      | r = \xi - x \le \delta
      | c = {{Corollary|Negative of Absolute Value|2}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \delta
      | o = \ge
      | r = x - \xi \ge -\delta
      | c = Ordering of Real Numbers is Reversed by Negation
}}
{{eqn | ll= \leadstoandfrom
      | l = \xi + \delta
      | o = \ge
      | r = x \ge \xi - \delta
      | c = Real Number Ordering is Compatible with Addition
}}
{{eqn | ll= \leadstoandfrom
      | l = \xi - \delta
      | o = \le
      | r = x \le \xi + \delta
      | c = rearranging
}}
{{end-eqn}}
By definition of closed real interval:
:$\closedint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta \le x \le \xi + \delta}$
Hence the result.
{{qed}}
Theorem: Open Interval Defined by Absolute Value

:$\set {x \in \R: \size {\xi - x} < \delta} = \openint {\xi - \delta} {\xi + \delta}$
where $\openint {\xi - \delta} {\xi + \delta}$ is the open real interval between $\xi - \delta$ and $\xi + \delta$.

Proof:
{{begin-eqn}}
{{eqn | l = \size {\xi - x}
      | o = <
      | r = \delta
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = -\delta
      | o = <
      | r = \xi - x < \delta
      | c = Negative of Absolute Value: Corollary 1
}}
{{eqn | ll= \leadstoandfrom
      | l = \delta
      | o = >
      | r = x - \xi > -\delta
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = \xi + \delta
      | o = >
      | r = x > \xi - \delta
      | c =
}}
{{end-eqn}}
But:
:$\openint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta < x < \xi + \delta}$
{{qed}}
Theorem: Set of Strictly Positive Real Numbers has no Smallest Element

Let $\R_{>0}$ denote the set of strictly positive real numbers.
Then $\R_{>0}$ has no smallest element.

Proof:
{{AimForCont}} $\R_{>0}$ has a smallest element.
Let $m$ be that smallest element.
Then we have that:
:$0 < \dfrac m 2 < m$
But as $0 < \dfrac m 2$ it follows that $\dfrac m 2 \in \R_{>0}$.
This contradicts our assertion that $m$ is the smallest element of $\R_{>0}$.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Distance from Subset of Real Numbers to Element

:$x \in S \implies \map d {x, S} = 0$

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Consider the set $T = \set {\size {x - y}: y \in S}$.
This has $0$ as a lower bound as Absolute Value is Bounded Below by Zero.
So:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} } \ge 0$
If $x \in S$ then:
:$\size {x - x} = 0 \in T$
and so:
:$\ds 0 \le \map {\inf_{y \mathop \in S} } {\map d {x, y} }$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\map d {x, y} } = 0$
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Element

:$x \in S \implies \map d {x, S} = 0$

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Element.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Supremum

Let $S$ be bounded above such that $\xi = \sup S$.
Then:
:$\map d {\xi, S} = 0$

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Let $\xi = \sup S$.
Then:
:$\forall y \in S: \size {\xi - y} = \xi - y$
So we need to show that no $h > 0$ can be a lower bound for $T = \set {\size {\xi - y}: y \in S}$.
{{AimForCont}} $\exists h > 0: \forall y \in S: \xi - y \ge h$.
But then:
:$\forall y \in S: y \le \xi - h$
and hence $\xi - h$ is an upper bound for $T$ smaller than $\xi = \sup S$.
But this contradicts the definition of supremum, that is the ''smallest'' upper bound.
So there is no such $h > 0$.
Hence by Proof by Contradiction it follows that:
:$\map d {\xi, S} = 0$.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Supremum

Let $S$ be bounded above such that $\xi = \sup S$.
Then:
:$\map d {\xi, S} = 0$

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Supremum.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Infimum

Let $S$ be bounded below such that $\xi = \inf S$.
Then:
:$\map d {\xi, S} = 0$

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Let $\xi = \inf S$.
Consider $\map d {-\xi, S'}$ where $S' = \set {-\xi: \xi \in S}$.
By Negative of Infimum is Supremum of Negatives:
:$\xi = \inf S \implies -\xi = \sup S'$
Thus from the above, $\map d {-\xi, S'} = 0$ and hence the result.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Infimum

Let $S$ be bounded below such that $\xi = \inf S$.
Then:
:$\map d {\xi, S} = 0$

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Infimum.
{{Qed}}
Theorem: Real Number at Distance Zero from Closed Real Interval is In Interval

Let $I \subseteq \R$ be a closed real interval.
Then:
:$\map d {x, I} = 0 \implies x \in I$

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Because $I$ is an interval, if $x \notin I$ then $x$ is either an upper bound or a lower bound for $I$.
Suppose $x$ is an upper bound for $I$.
Let $B$ be the supremum of $I$.
Then because $I$ is closed, $B \in I$.
So:
{{begin-eqn}}
{{eqn | q = \forall y \in I
      | l = \size {x - y}
      | r = x - y
      | c = 
}}
{{eqn | r = x - B + B - y
      | c = 
}}
{{eqn | r = x - B + \size {B - y}
      | c = 
}}
{{end-eqn}}
Now from Infimum Plus Constant:
:$\inf_{y \mathop \in S} \size {x - y} = x - B + \inf_{y \mathop \in S} \size {B - y}$
But we also have:
:$x - B \ge 0$
:$\map d {B, S} \ge 0$
:$\map d {x, S} = 0$
So it follows that $x = B$ and so $x \in I$.
A similar argument applies if $x$ is a lower bound for $I$.
{{Qed}}
Theorem: Existence of Real Number at Distance Zero from Open Real Interval not in Interval

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:
{{begin-eqn}}
{{eqn | q = \exists a, b \in \R
      | l = I
      | r = \openint a b
}}
{{eqn | q = \exists a \in \R
      | l = I
      | r = \openint a \to
}}
{{eqn | q = \exists b \in \R
      | l = I
      | r = \openint \gets b
}}
{{end-eqn}}
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 $b$.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Element/Proof 1

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Then:
{{:Distance from Subset of Real Numbers to Element}}

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Consider the set $T = \set {\size {x - y}: y \in S}$.
This has $0$ as a lower bound as Absolute Value is Bounded Below by Zero.
So:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} } \ge 0$
If $x \in S$ then:
:$\size {x - x} = 0 \in T$
and so:
:$\ds 0 \le \map {\inf_{y \mathop \in S} } {\map d {x, y} }$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\map d {x, y} } = 0$
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Element/Proof 2

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Then:
{{:Distance from Subset of Real Numbers to Element}}

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Element.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Supremum/Proof 1

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
{{:Distance from Subset of Real Numbers to Supremum}}

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Let $\xi = \sup S$.
Then:
:$\forall y \in S: \size {\xi - y} = \xi - y$
So we need to show that no $h > 0$ can be a lower bound for $T = \set {\size {\xi - y}: y \in S}$.
{{AimForCont}} $\exists h > 0: \forall y \in S: \xi - y \ge h$.
But then:
:$\forall y \in S: y \le \xi - h$
and hence $\xi - h$ is an upper bound for $T$ smaller than $\xi = \sup S$.
But this contradicts the definition of supremum, that is the ''smallest'' upper bound.
So there is no such $h > 0$.
Hence by Proof by Contradiction it follows that:
:$\map d {\xi, S} = 0$.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Supremum/Proof 2

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
{{:Distance from Subset of Real Numbers to Supremum}}

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Supremum.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Infimum/Proof 1

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
{{:Distance from Subset of Real Numbers to Infimum}}

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Let $\xi = \inf S$.
Consider $\map d {-\xi, S'}$ where $S' = \set {-\xi: \xi \in S}$.
By Negative of Infimum is Supremum of Negatives:
:$\xi = \inf S \implies -\xi = \sup S'$
Thus from the above, $\map d {-\xi, S'} = 0$ and hence the result.
{{Qed}}
Theorem: Distance from Subset of Real Numbers to Infimum/Proof 2

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
{{:Distance from Subset of Real Numbers to Infimum}}

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Infimum.
{{Qed}}
Theorem: Infimum of Set of Reciprocals of Positive Integers

Let $S$ be the subset of the set of real numbers defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
Then:
:$\inf S = 0$
where $\inf S$ denotes the infimum of $S$.

Proof:
We have that $\Z_{>0}$ contains only (strictly) positive integers.
So it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $S$ contains only (strictly) positive real numbers.
Hence $0$ is a lower bound for $S$.
{{AimForCont}} that $0$ is not the infimum of $S$.
Then $\exists h \in \R_{>0}$ such that $h$ is a lower bound for $S$.
Then:
:$\forall n \in \Z_{>0}: \dfrac 1 n \ge h$
Hence from Reciprocal Function is Strictly Decreasing:
:$\forall n \in \Z_{>0}: n \le \dfrac 1 h$
where $\dfrac 1 h \in \R$.
But from the Axiom of Archimedes:
:$\exists n \in \Z_{>0}: n > \dfrac 1 h$
This is a contradiction.
It follows by Proof by Contradiction that no such $h > 0$ exists.
Hence $0$ is the infimum of $S$.
{{qed}}
Theorem: Set of Numbers of form n - 1 over n is Bounded Above

Let $S$ be the subset of the set of real numbers $\R$ defined as:
:$S = \set {\dfrac {n - 1} n: n \in \Z_{>0} }$
$S$ is bounded above with supremum $1$.
$S$ has no greatest element.

Proof:
We have that:
:$\dfrac {n - 1} n = 1 - \dfrac 1 n$
As $n > 0$ it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $\dfrac 1 n > 0$.
Thus $1 - \dfrac 1 n < 1$ and so $S$ is bounded above by $1$.
Next it is to be shown that $1$ is the supremum of $S$.
Suppose $x$ is the supremum of $S$ such that $x < 1$.
Then:
:$1 - x = \epsilon$
where $\epsilon \in \R_{>0}$.
By the Axiom of Archimedes we have that:
:$\exists m \in \Z_{>0}: n > \dfrac 1 \epsilon$
and so from Reciprocal Function is Strictly Decreasing:
:$\exists m \in \Z_{>0}: \dfrac 1 n < \epsilon$
Thus:
:$1 - \dfrac 1 m > 1 - \epsilon = x$
and so $x$ is not the supremum of $S$ after all.
Thus $\sup S$ cannot be less than $1$.
It follows that:
:$\sup S = 1$
Next it is noted that:
:$\forall n \in \Z_{>0}: \dfrac 1 n > 0$
and so:
:$\forall x \in S: x < 1$
Thus as $\sup S \notin S$ it follows from Greatest Element is Supremum that $S$ has no greatest element.
{{qed}}
Theorem: Set of Rational Numbers Strictly between Zero and One has no Greatest or Least Element

Let $S \subseteq \Q$ be the subset of the set of rational numbers defined as:
:$S = \set {r \in \Q: 0 < r < 1}$
Then $S$ has no greatest or smallest element.
However, $S$ has a supremum $1$ and an infimum $0$.

Proof:
We have that:
:$\forall r \in S: 0 < r$
and:
:$\forall r \in S: r < 1$
Hence $0$ and $1$ are lower and upper bounds of $S$ respectively.
Let $s \in S$.
Then $s \in \Q: 0 < s < 1$.
{{AimForCont}} $s$ is the greatest element of $S$.
But then we have:
:$0 < s < \dfrac {s + 1} 2 < 1$
and so $\dfrac {s + 1} 2 \in S$ but $s < \dfrac {s + 1} 2$.
This contradicts our assertion that $s$ is the greatest element of $S$.
{{AimForCont}} $s$ is the smallest element of $S$.
But then we have:
:$0 < \dfrac s 2 < s < 1$
and so $\dfrac s 2 \in S$ but $\dfrac s 2 < s$.
This contradicts our assertion that $s$ is the smallest element of $S$.
So $S$ cannot have either a greatest or smallest element.
Let $H$ be such that $0 < H < 1$.
Suppose $H$ is a smaller upper bound of $S$ than $1$.
Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $H < r < 1$.
But then $r$ is a rational number between $0$ and $1$ which is greater than $H$.
Hence $H$ cannot be an upper bound of $S$.
Thus, by definition, $1$ is the supremum of $S$
Suppose $H$ is a greater lower bound of $S$ than $0$.
Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $0 < r < H$.
But then $r$ is a rational number between $0$ and $1$ which is smaller than $H$.
Hence $H$ cannot be a lower bound of $S$.
Thus, by definition, $0$ is the infimum of $S$.
{{Qed}}
Theorem: Between two Real Numbers exists Irrational Number

Let $a, b \in \R$ be real numbers where $a < b$. 
Then there exists an irrational number $\xi \in \R \setminus \Q$ such that:
:$a < \xi < b$

Proof:
From Number of Type Rational r plus s Root 2 is Irrational we have that a real number of the form $r \sqrt 2$, where $r \ne 0$ is rational, is irrational.
From Between two Real Numbers exists Rational Number there exists a rational number $r$ such that:
:$\dfrac a {\sqrt 2} < r < \dfrac b {\sqrt 2}$
and so:
:$a < r \sqrt 2 < b$
giving us $\xi = r \sqrt 2$ as the irrational number we seek.
If $a < 0$ and $b > 0$, it may be that the $r$ that we find happens to be zero
Then we locate a second rational number $s$ such that:
:$0 < s < \dfrac b {\sqrt 2}$
which leads us to:
:$a < 0 < s \sqrt 2 < b$
and so in this case $\xi = s \sqrt 2$ is the irrational number we seek.
{{qed}}