nyuuzyou/wikis huggingface dataset:
Power Function Preserves Ordering in Ordered Semigroup/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$x \preceq y \implies x^n \preceq y^n$
$\map P 1$ is the case:
:$x \preceq y \implies x \preceq y$
which is trivially true.
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
We have:
and:
Hence as $\preceq$ is an ordering and hence transitive:
:$x \preceq y \implies x \circ x \preceq y \circ y$
That is:
:$x \preceq y \implies x^2 \preceq y^2$
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$x \preceq y \implies x^k \preceq y^k$
from which it is to be shown that:
:$x \preceq y \implies x^{k + 1} \preceq y^{k + 1}$
=== Induction Step ===
This is the induction step:
and:
Hence as $\preceq$ is an ordering and hence transitive:
:$x \preceq y \implies x \circ x^k \preceq y \circ y^k$
That is:
:$x \preceq y \implies x^{k + 1} \preceq y^{k + 1}$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: x \preceq y \implies x^n \preceq y^n$
Category:Power Function Preserves Ordering in Ordered Semigroup
Power Function is Completely Multiplicative

== Theorem ==
Let $K$ be a field.
Let $z \in K$.
Let $f_z: K \to K$ be the mapping defined as:
:$\forall x \in K: \map {f_z} x = x^z$
Then $f_z$ is completely multiplicative.
=== Integers ===
== Proof ==
Let $r, s \in K$.
Then:
Category:Field Theory
Category:Completely Multiplicative Functions
Power Function is Completely Multiplicative/Integers

== Theorem ==
Let $c \in \Z$ be an integer.
Let $f_c: \Z \to \Z$ be the mapping defined as:
:$\forall n \in \Z: \map {f_c} n = n^c$
Then $f_c$ is completely multiplicative.
== Proof ==
Let $r, s \in \Z$.
Then:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $29 \ \text{(a)}$
Category:Completely Multiplicative Functions
Category:Number Theory
Power Function is Convex Real Function

== Theorem ==
Let $p \ge 1$ be a real number.
Define $f : \hointr 0 \infty \to \hointr 0 \infty$ by: 
:$\map f x = x^p$
for each $x \in \hointr 0 \infty$. 
Then $f$ is a convex function.
== Proof ==
Applying Derivative of Power twice, we have that: 
:$f$ is twice differentiable on $\openint 0 \infty$
with:
:$\map {f' '} x = p \paren {p - 1} x^{p - 2}$
for each $x \in \openint 0 \infty$.
Since $p \ge 1$, we have: 
:$p \paren {p - 1} \ge 0$
and so:
:$\map {f' '} x \ge 0$
for each $x \in \openint 0 \infty$.
From Real Function with Positive Derivative is Increasing:
:$f'$ is increasing on $\openint 0 \infty$
and so from Real Function is Convex iff Derivative is Increasing:
:$f$ is convex on $\openint 0 \infty$.
It remains to assure ourselves that:
:$\map f {t x + \paren {1 - t} 0} \le t \map f x + \paren {1 - t} \map f 0 = t \map f x$ for $x \in \hointr 0 \infty$ and $t \in \closedint 0 1$. 
We have:
:$\map f {t x} = \paren {t x}^p = t^p x^p$ for $x \in \hointr 0 \infty$ and $t \in \closedint 0 1$.
Since $p \ge 1$, we have:
:$t^p \le t$ for $t \in \closedint 0 1$.
Hence we have:
:$t^p x^p \le t x^p$
Hence:
:$\map f {t x} \le t \map f x$
Hence:
:$f$ is convex.
Category:Convex Real Functions
Power Function is Monotone/Rational Number

== Theorem ==
Let $a \in \R_{>0}$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f r = a^r$
where $a^r$ denotes $a$ to the power of $r$.
Then $f$ is monotone.
Further, $f$ is strictly monotone unless $a = 1$.
== Proof ==
=== Case 1: $a > 1$ ===
Let $a > 1$.
Then by Power Function on Base Greater than One is Strictly Increasing:
:$f$ is strictly increasing.
By Strictly Increasing Mapping is Increasing:
:$f$ is increasing.
=== Case 2 : $a = 1$ ===
Let $a = 1$.
Then:
:$\forall r \in \Q: a^r = 1$
and so $f$ is constant.
From Mapping is Constant iff Increasing and Decreasing:
:$f$ is both increasing and decreasing.
=== Case 3: $0 < a < 1$ ===
Let $0 < a < 1$.
Then by Power Function on Base between Zero and One is Strictly Decreasing:
:$f$ is strictly decreasing.
By Strictly Decreasing Mapping is Decreasing:
:$f$ is decreasing.
=== Conclusion ===
It has been shown that in all cases $f$ is either increasing or decreasing or both.
Thus by definition of monotone function, $f$ is monotone.
In either of the cases where $a \ne 1$, It has been shown that in all cases $f$ is either strictly increasing or strictly decreasing.
Thus by definition of strictly monotone function, $f$ is strictly monotone.
Hence the result.
Category:Powers
Power Function is Strictly Increasing on Positive Elements

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring.
Let $x, y \in R$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Let $0 < x < y$.
Then:
:$0 < \map {\circ^n} x < \map {\circ^n} y$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$0 < \map {\circ^n} x < \map {\circ^n} y$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$0 < \map {\circ^1} x < \map {\circ^1} y$
which is just:
:$0 < x < y$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$0 < \map {\circ^k} x < \map {\circ^k} y$
from which it is to be shown that:
:$0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$
=== Induction Step ===
This is the induction step:
We have:
:$0 < x < y$
:$0 < \map {\circ^k} x < \map {\circ^k} y$
By Ring Product preserves Inequalities on Positive Elements:
:$0 < \map {\circ^k} x \circ x < \map {\circ^k} y \circ y$
Hence:
:$0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$
So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that:
:$\forall n \in \N_{> 0}: 0 < \map {\circ^n} x < \map {\circ^n} y$
== Also see ==
* Odd Power Function is Strictly Increasing
Category:Ordered Rings
Category:Proofs by Induction
Power Function is Strictly Increasing over Positive Reals/Natural Exponent

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $f: \R_{>0} \to \R$ be the real function defined as:
:$\map f x = x^n$
where $x^n$ denotes $x$ to the power of $n$.
Then $f$ is strictly increasing.
== Proof ==
Proof by induction on $n$:
Let $x, y \in \R_{>0}$ be strictly positive real numbers.
For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition:
:$x < y \implies x^n < y^n$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$x < y \implies x^k < y^k$
Then we need to show:
:$x < y \implies x^{k + 1} < y^{k + 1}$
=== Induction Step ===
This is our induction step:
First:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}: x < y \implies x^n < y^n$
Category:Examples of Strictly Increasing Real Functions
Category:Powers
Power Function on Base Greater than One is Strictly Increasing/Integer

== Theorem ==
Let $a \in \R$ be a real number such that $a > 1$.
Let $f: \Z \to \R$ be the real-valued function defined as:
:$\map f k = a^k$
where $a^k$ denotes $a$ to the power of $k$.
Then $f$ is strictly decreasing.
== Proof ==
Let $a > 1$.
By Power Function on Base Greater than One is Strictly Increasing: Positive Integer, the theorem is already proven for positive integers.
It remains to be proven over the strictly negative integers.
Let $i, j$ be integers such that $i < j < 0$.
From Order of Real Numbers is Dual of Order of their Negatives:
:$0 < -j < -i$
So:
Category:Power Function on Base Greater than One is Strictly Increasing
Power Function on Base Greater than One is Strictly Increasing/Positive Integer

== Theorem ==
Let $a \in \R$ be a real number such that $a > 1$.
Let $f: \Z_{\ge 0} \to \R$ be the real-valued function defined as:
:$\map f n = a^n$
where $a^n$ denotes $a$ to the power of $n$.
Then $f$ is strictly increasing.
== Proof ==
Fix $n \in \Z_{\ge 0}$.
From Ordering of Reciprocals:
:$0 < \dfrac 1 a < 1$
From Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer:
:$\paren {\dfrac 1 a}^{n + 1} < \paren {\dfrac 1 a}^n$
From Real Number to Negative Power: Positive Integer:
:$\dfrac 1 {a^{n + 1} } < \dfrac 1 {a^n}$
From Ordering of Reciprocals:
:$a^n < a^{n + 1}$
Hence the result.
Category:Power Function on Base Greater than One is Strictly Increasing
Power Function on Base Greater than One is Strictly Increasing/Rational Number

== Theorem ==
Let $a \in \R$ be a real number such that $a > 1$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f q = a^q$
where $a^q$ denotes $a$ to the power of $q$.
Then $f$ is strictly increasing.
== Proof ==
Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z$ are integers and $s, u \in \Z_{>0}$ are strictly positive integers.
Let $\dfrac r s < \dfrac t u$.
From Ordering of Reciprocals:
:$0 < \dfrac 1 a < 1$
So:
Hence the result.
Category:Power Function on Base Greater than One is Strictly Increasing
Power Function on Base Greater than One is Strictly Increasing/Real Number

== Theorem ==
Let $a \in \R$ be a real number such that $a > 1$.
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = a^x$
where $a^x$ denotes $a$ to the power of $x$.
Then $f$ is strictly increasing.
== Proof ==
Let $x, y \in \R$ be such that $x < y$.
Let $\delta = \dfrac {y - x} 2$.
From Rational Sequence Decreasing to Real Number, there is some rational sequence $\sequence {x_n}$ that decreases to $x$.
From Rational Sequence Increasing to Real Number, there is some rational sequence $\sequence {y_n}$ that increases to $y$.
From Convergent Real Sequence is Bounded:
:$\exists N_1 \in \N: n \ge N_1 \implies x - \delta < x_n < x + \delta$
Since $\sequence {x_n}$ is decreasing:
:$n \ge N_1 \implies x \le x_n < x + \delta$
From Convergent Real Sequence is Bounded:
:$\exists N_2 \in \N : n \ge N_2 \implies y - \delta < y_n < y + \delta$
Since $\sequence {y_n}$ is increasing:
:$n \ge N_2 \implies y - \delta < y_n \le y$
Let $N = \max \set {N_1, N_2}$.
Then, for $n \ge N$:
 
From Power Function on Strictly Positive Base is Continuous and Sequential Continuity is Equivalent to Continuity in the Reals:
:$x_n \to x \implies a^{x_n} \to a^x$
:$y_n \to y \implies a^{y_n} \to a^y$
Also, from Power Function on Base Greater than One is Strictly Increasing: Rational Number:
:$\sequence {a^{x_n} }$ decreases to $a^x$
and:
:$\sequence {a^{y_n} }$ increases to $a^y$.
So, for $n \ge N$:
Hence the result.
Category:Power Function on Base Greater than One is Strictly Increasing
Category:Real Analysis
Power Function on Base between Zero and One Tends to One as Power Tends to Zero/Rational Number

== Theorem ==
Let $a \in \R_{> 0}$ be a strictly positive real number such that $0 < a < 1$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f r = a^r$
where $a^r$ denotes $a$ to the power of $r$.
Then:
:$\ds \lim_{r \mathop \to 0} \map f r = 1$
== Proof ==
From Ordering of Reciprocals:
:$0 < a < 1 \implies 1 < \dfrac 1 a$
So:
Category:Powers
Power Function on Base between Zero and One is Strictly Decreasing/Integer

== Theorem ==
Let $a \in \R$ be a real number such that $0 < a < 1$.
Let $f: \Z \to \R$ be the real-valued function defined as:
:$\map f k = a^k$
where $a^k$ denotes $a$ to the power of $k$.
Then $f$ is strictly decreasing.
== Proof ==
Let $0 < a < 1$.
By Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer, the theorem is already proven for positive integers.
It remains to be proven over the negative integers.
Let $i, j$ be integers such that $i < j < 0$.
From Order of Real Numbers is Dual of Order of their Negatives:
:$0 < -j < -i$
So:
Hence the result.
Category:Power Function on Base between Zero and One is Strictly Decreasing
Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer

== Theorem ==
Let $a \in \R$ be a real number such that $0 < a < 1$.
Let $f: \Z_{\ge 0} \to \R$ be the real-valued function defined as:
:$\map f n = a^n$
where $a^n$ denotes $a$ to the power of $n$.
Then $f$ is strictly decreasing.
== Proof ==
Proof by induction on $n$:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a^{n + 1} < a^n$
$\map P 0$ is the case:
=== Basis for the Induction ===
$\map P 1$ is true, since:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$a^{k + 1} < a^k$
Then we need to show:
:$a^{k + 2} < a^{k + 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: a^{n + 1} < a^n$
Hence the result, by definition of strictly decreasing.
Category:Power Function on Base between Zero and One is Strictly Decreasing
Power Function on Base between Zero and One is Strictly Decreasing/Rational Number

== Theorem ==
Let $a \in \R$ be a real number such that $0 < a < 1$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f q = a^q$
where $a^q$ denotes $a$ to the power of $q$.
Then $f$ is strictly decreasing.
== Proof ==
Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z, s, u \in \Z_{>0}$.
Let $\dfrac r s < \dfrac t u$.
Then:
Category:Power Function on Base between Zero and One is Strictly Decreasing
Power Function on Base between Zero and One is Strictly Decreasing/Real Number

== Theorem ==
Let $a \in \R$ be a real number such that $0 \lt a \lt 1$.
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = a^x$
where $a^x$ denotes $a$ to the power of $x$.
Then $f$ is strictly decreasing.
== Proof ==
Let $x, y \in \R$ be such that $x < y$.
Since $0 < a < 1$, we have that:
:$\dfrac 1 a > 1$
Then we have that:
The result follows.
Category:Power Function on Base between Zero and One is Strictly Decreasing
Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number

== Theorem ==
Let $a \in \R_{> 0}$ be a strictly positive real number such that $a > 1$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f r = a^r$
where $a^r$ denotes $a$ to the power of $r$.
Then:
:$\ds \lim_{r \mathop \to 0} \map f r = 1$
== Proof ==
We start by treating the right-sided limit.
Let $0 < r < 1$.
=== Lemma ===
From the lemma:
:$1 < a^r < 1 + a r$
Also:
So from the Squeeze Theorem for Functions:
:$\ds \lim_{r \mathop \to 0^+} a^r = 1$
We now treat the left-sided limit.
Let $-1 < r < 0$.
Also:
So from the Squeeze Theorem for Functions:
:$\ds \lim_{r \mathop \to 0^-} a^r = 1$
From Limit iff Limits from Left and Right:
:$\ds \lim_{r \mathop \to 0} a^r = 1$
Hence the result.
Category:Powers
Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma

== Theorem ==
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \Q_{> 0}$ be a strictly positive rational number such that $r < 1$.
Then:
:$1 < a^r < 1 + a r$
== Proof ==
Define a real function $g_r: \R_{> 0} \to \R$ as:
:$\map {g_r} a = 1 + a r - a^r$
Then differentiating $a$ gives:
:$D_a \map {g_r} a = r \paren {1 - a^{r - 1} }$
We show now that the derivative of $g_r$ is positive for all $a > 1$:
So $D_a \map {g_r} a$ is positive for all $a > 1$.
Whence, by Derivative of Monotone Function, $g_r$ is increasing for all $a > 1$.
Now:
:$\map {g_r} 1 = r > 0$
So $\map {g_r} a$ is positive for all $a > 1$.
That is:
:$0 < 1 + a r - a^r$
Adding $a^r$ to both sides of the above yields:
:$a^r < 1 + a r$
Finally:
So, for $0 < r < 1$:
:$1 < a^r < 1 + a r$
Category:Powers
Power Function on Complex Numbers is Epimorphism

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $f_n: \C_{\ne 0} \to \C_{\ne 0}$ be the mapping from the set of complex numbers less zero to itself defined as:
:$\forall z \in \C_{\ne 0}: \map {f_n} z = z^n$
Then $f_n: \struct {\C_{\ne 0}, \times} \to \struct {\C_{\ne 0}, \times}$ is a group epimorphism.
The kernel of $f_n$ is the set of complex $n$th roots of unity.
== Proof ==
From Non-Zero Complex Numbers under Multiplication form Group, $\struct {\C_{\ne 0}, \times}$ is a group.
Therefore $\struct {\C_{\ne 0}, \times}$ is closed by .
Let $w, z \in \C_{\ne 0}$.
Thus $f_n$ is a group homomorphism.
Now suppose $w = r \paren {\cos \alpha + i \sin \alpha}$, expressing $w$ in polar form.
Then $w = \map {f_n} z$ where:
:$z = r^{1/n} \paren {\cos \dfrac \alpha n + i \sin \dfrac \alpha n}$
and so:
:$\forall w: w \in \map {f_n} {\C_{\ne 0} }$
That is, $f_n$ is a surjection.
Being a group homomorphism which is also a surjection, by definition $f_n$ is then a group epimorphism.
The kernel of $f_n$ is the set:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
which follows from Complex Roots of Unity in Exponential Form.
== Examples ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Example $12.2$
Category:Complex Numbers
Category:Group Epimorphisms
Category:Power Function on Complex Numbers is Epimorphism
Power Function on Complex Numbers is Epimorphism/Examples/Cube

== Example of Use of Power Function on Complex Numbers is Epimorphism ==
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $f_3: \C_{\ne 0} \to \C_{\ne 0}$ be the mapping from the set of complex numbers less zero to itself defined as:
:$\forall z \in \C_{\ne 0}: \map {f_n} z = z^3$
Then $f_3: \struct {\C_{\ne 0}, \times} \to \struct {\C_{\ne 0}, \times}$ is a group epimorphism.
The kernel $U_3$ of $f_3$ is the set of complex $3$rd roots of unity:
:$U_3 = \set {1, \omega, \omega^2}$
where:
Hence for all $ z \in \C_{\ne 0}$, the coset $z U_3$ is the set:
:$z U_3 = \set {z, z \omega, z \omega^2}$
and multiplication on $\C_{\ne 0} / U_3$ of all such cosets satisfies:
:$\set {z_1, z_1 \omega, z_1 \omega^2} \times \set {z_2, z_2 \omega, z_2 \omega^2} = \set {z_1 z_2, z_1 z_2 \omega, z_1 z_2 \omega^2}$
Hence the associated isomorphism $g: \C_{\ne 0} / U_3 \to \C_{\ne 0}$ takes the equivalence class:
:$\set {z, z \omega, z \omega^2}$
into the cube:
:$z^3 = \paren {z \omega}^3 = \paren {z \omega^2}^3$
of any one of its elements.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Example $12.2$
Category:Power Function on Complex Numbers is Epimorphism
Power Function on Strictly Positive Base is Continuous/Rational Power

== Theorem ==
Let $a \in \R_{>0}$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f x = a^x$
where $a^x$ denotes $a$ to the power of $x$.
Then $f$ is continuous.
== Proof ==
Category:Powers
Power Function on Strictly Positive Base is Continuous/Real Power

== Theorem ==
Let $a \in \R_{>0}$.
Let $f : \R \to \R$ be the real function defined as:
:$\map f x = a^x$
where $a^x$ denotes $a$ to the power of $x$.
Then $f$ is continuous.
== Proof ==
By definition, $a^x$ is the unique continuous extension of $a^r$, for rational $r$.
By definition, continuous extensions are continuous.
Hence the result.
== Also see ==
* Power Function to Rational Power permits Unique Continuous Extension, where such a unique continuous extension is shown to exist.
Category:Powers
Power Function on Strictly Positive Base is Convex

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real number.
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = a^x$
where $a^x$ denotes $a$ to the power of $x$.
Then $f$ is convex.
== Proof ==
Let $x, y \in \R$.
Note that, from Power of Positive Real Number is Positive: Real Number:
:$\forall t \in \R: a^t > 0$.
So:
Hence $a^x$ is midpoint-convex.
Further, from Power Function on Strictly Positive Base is Continuous: Real Power, $a^x$ is continuous.
Thus, from Continuous Midpoint-Convex Function is Convex, $a^x$ is convex.
Category:Real Analysis
Category:Powers
Power Function tends to One as Power tends to Zero/Rational Number

== Theorem ==
Let $a \in \R_{> 0}$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f q = a^q$
where $a^q$ denotes $a$ to the power of $q$.
Then:
:$\ds \lim_{x \mathop \to 0} \map f x = 1$
== Proof ==
=== Case 1: $a > 1$ ===
If $a > 1$, then:
:$\ds \lim_{x \mathop \to 0} \map f x = 1$
from Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number.
=== Case 2: $a = 1$ ===
If $a = 1$, then:
=== Case 3: $0 < a < 1$ ===
If $0 < a < 1$, then:
:$\ds \lim_{x \mathop \to 0} \map f x = 1$
from Power Function on Base between Zero and One Tends to One as Power Tends to Zero/Rational Number.
Hence the result.
Category:Powers
Power Function to Rational Power permits Unique Continuous Extension

== Theorem ==
Let $a \in \R_{> 0}$.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$\map f q = a^q$
where $a^q$ denotes $a$ to the power of $q$.
Then there exists a unique continuous extension of $f$ to $\R$.
== Proof ==
Consider $I_k := \openint {-k} k$ for arbitrary $k \in \N$.
Let $I_k' = I_k \cap \Q$.
Note that, for all $x, y \in I_k'$:
where $M = \map \max {a^{-k}, a^k}$.
Fix $\epsilon \in \R_{> 0}$.
From Power Function tends to One as Power tends to Zero: Rational Number:
where $M$ is defined as above.
Therefore, $\exists \delta \in \R_{> 0}$ such that:
That is, $f$ is uniformly continuous on $I_k'$.
From Rationals are Everywhere Dense in Topological Space of Reals, $I_k'$ is everywhere dense in $I_k$.
From Continuous Extension from Dense Subset, there exists a unique continuous extension of $f$ to $I_k$.
Call this function $f_k$.
Define:
:$\ds F = \bigcup \set {f_k : k \in \N}$
Note that $\sequence {I_k}$ is an exhausting sequence of sets in $\R$.
By the Union of Functions Theorem, $F$ defines a function $\R \to \R$
From the Pasting Lemma, $F$ is continuous on $\R$.
Hence the result.
Category:Powers
Power Function with Cancellable Element Preserves Strict Ordering in Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y \in S$ be such that:
:$(1): \quad x \prec y$
:$(2): \quad$ either $X$ or $y$ (or both) is cancellable for $\circ$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then:
:$x^n \prec y^n$
where $x^n$ is the $n$th power of $x$.
== Proof ==
, suppose $x$ is cancellable for $\circ$.
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$x \prec y \implies x^n \prec y^n$
$\map P 1$ is the case:
:$x \prec y \implies x \prec y$
which is trivially true.
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
We have:
and:
Hence:
:$x \prec y \implies x \circ x \prec y \circ y$
That is:
:$x \prec y \implies x^2 \prec y^2$
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$x \preceq y \implies x^k \prec y^k$
from which it is to be shown that:
:$x \preceq y \implies x^{k + 1} \prec y^{k + 1}$
=== Induction Step ===
This is the induction step:
and:
Hence:
:$x \prec y \implies x \circ x^k \prec y \circ y^k$
That is:
:$x \prec y \implies x^{k + 1} \prec y^{k + 1}$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: x \prec y \implies x^n \prec y^n$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.4$
Category:Ordered Semigroups