hcju/proofwikiqa huggingface dataset:
:300px Let $AB$ be a minor straight line. Let $CD$ be commensurable in length with $AB$. It is to be demonstrated that $CD$ is a minor straight line. Let $BE$ be the annex of $CD$. Therefore by definition of minor straight line: : $AE$ and $EB$ are incommensurable in square : $AE^2 + EB^2$ is rational : $2 \cdot AE \cdot EB$ is a medial rectangle. From {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$BE : DF = AB : CD$ From {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$AE : CF = AB : CD$ and so from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = BE : DF$ We have that $AE$ and $EB$ are incommensurable in square. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$CF$ and $FD$ are incommensurable in square. We have that: :$AE : CF = BE : DF$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ So from {{EuclidPropLink|book = VI|prop = 22|title = Similar Figures on Proportional Straight Lines}}: :$AE^2 : EB^2 = CF^2 : FD^2$ Therefore from {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$ But $EB^2$ is commensurable with $FD^2$. So from: :{{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: we have that: :$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$. But by definition $AE^2 + EB^2$ is rational. Therefore by {{EuclidDefLink|X|4|Rational Area}}: :$CF^2 + FD^2$ is rational. We have that: :$AE : EB = CF : FD$ Therefore: :$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$ Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$. Therefore $CF \cdot FD$ is medial. Thus $CF$ and $FD$ are such that: : $CF$ and $FD$ are incommensurable in square : $CF^2 + FD^2$ is rational : $2 \cdot CF \cdot FD$ is a medial rectangle. Hence the result. {{qed}} {{Euclid Note|105|X}}
:300px Let $AB$ be a straight line which produces with a rational area a medial whole. Let $CD$ be commensurable in length with $AB$. It is to be demonstrated that $CD$ is a straight line which produces with a rational area a medial whole. Let $BE$ be the annex of $CD$. Therefore by definition of a straight line which produces with a rational area a medial whole: : $AE$ and $EB$ are incommensurable in square : $AE^2 + EB^2$ is medial : $2 \cdot AE \cdot EB$ is a rational rectangle. From {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$BE : DF = AB : CD$ From {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$AE : CF = AB : CD$ and so from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = BE : DF$ We have that $AE$ and $EB$ are incommensurable in square. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$CF$ and $FD$ are incommensurable in square. We have that: :$AE : CF = BE : DF$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ So from {{EuclidPropLink|book = VI|prop = 22|title = Similar Figures on Proportional Straight Lines}}: :$AE^2 : EB^2 = CF^2 : FD^2$ Therefore from {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$ But $EB^2$ is commensurable with $FD^2$. So from: :{{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: we have that: :$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$. But by definition $AE^2 + EB^2$ is medial. Therefore by $CF^2 + FD^2$ is medial. We have that: :$AE : EB = CF : FD$ Therefore: :$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$ Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$. Therefore $CF \cdot FD$ is rational. Thus $CF$ and $FD$ are such that: : $CF$ and $FD$ are incommensurable in square : $CF^2 + FD^2$ is medial : $2 \cdot CF \cdot FD$ is a rational rectangle. Hence the result. {{qed}} {{Euclid Note|106|X}}
:300px Let $AB$ be a straight line which produces with a medial area a medial whole. Let $CD$ be commensurable in length with $AB$. It is to be demonstrated that $CD$ is a straight line which produces with a medial area a medial whole. Let $BE$ be the annex of $CD$. Therefore by definition of a straight line which produces with a rational area a medial whole: : $AE$ and $EB$ are incommensurable in square : $AE^2 + EB^2$ is medial : $2 \cdot AE \cdot EB$ is a medial rectangle. From {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$BE : DF = AB : CD$ From {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$AE : CF = AB : CD$ and so from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = BE : DF$ We have that $AE$ and $EB$ are incommensurable in square. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$CF$ and $FD$ are incommensurable in square. We have that: :$AE : CF = BE : DF$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ So from {{EuclidPropLink|book = VI|prop = 22|title = Similar Figures on Proportional Straight Lines}}: :$AE^2 : EB^2 = CF^2 : FD^2$ Therefore from {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$ But $EB^2$ is commensurable with $FD^2$. So from: :{{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: we have that: :$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$. But by definition $AE^2 + EB^2$ is medial. Therefore by $CF^2 + FD^2$ is medial. We have that: :$AE : EB = CF : FD$ Therefore: :$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$ Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$. Therefore $CF \cdot FD$ is medial. Thus $CF$ and $FD$ are such that: : $CF$ and $FD$ are incommensurable in square : $CF^2 + FD^2$ is medial : $2 \cdot CF \cdot FD$ is a medial rectangle. Hence the result. {{qed}} {{Euclid Note|107|X}}
:400px Let $BC$ be a rational area. Let the medial area $BD$ be subtracted from $BC$. It needs to be demonstrated that the "side" of the remainder $EC$ is either an apotome or a minor straight line. Let $FG$ be a rational straight line. Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth. Let the area $GK$ equal to $BC$ be subtracted from $GH$. Then the remainder $EC$ is equal to $LH$. We have that $BC$ is rational and $BD$ is medial. We also have: :$BC = GH$ and: :$BD = GK$ Therefore $GH$ is rational and $GK$ is medial. But $GH$ and $GK$ are applied to a rational straight line $FG$. Therefore from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$FH$ is rational and commensurable in length with $FG$ while from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$FK$ is rational and incommensurable in length with $FG$. Therefore by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}:  :$FH$ is incommensurable in length with $FK$. Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only. Therefore $KH$ is an apotome and $KF$ is the annex to $KH$. We have that: : $HF^2 = FK^2 + \lambda^2$ where either: :$\lambda$ is commensurable in length with $HF$ or: :$\lambda$ is incommensurable in length with $HF$. First suppose $\lambda$ is commensurable in length with $HF$. Then $HF$ is commensurable in length with the rational straight line $FG$. Therefore $KH$ is a first apotome. But from {{EuclidPropLink|book = X|prop = 91|title = Side of Area Contained by Rational Straight Line and First Apotome}}:  :the "side" of $LH$ is an apotome. Next suppose $\lambda$ is incommensurable in length with $HF$. Then $HF$ is incommensurable in length with the rational straight line $FG$. Therefore $KH$ is a fourth apotome. But from {{EuclidPropLink|book = X|prop = 94|title = Side of Area Contained by Rational Straight Line and Fourth Apotome}}:  :the "side" of $LH$ is minor. {{qed}} {{Euclid Note|108|X}}
:400px Let $BC$ be a medial area. Let the rational area $BD$ be subtracted from $BC$. It needs to be demonstrated that the "side" of the remainder $EC$ is either: :a first apotome of a medial straight line or: :a straight line which produces with a rational area a medial whole. Let $FG$ be a rational straight line. Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth. Let the area $GK$ equal to $BC$ be subtracted from $GH$. Then the remainder $EC$ is equal to $LH$. We have that $BC$ is medial and $BD$ is rational. We also have: :$BC = GH$ and: :$BD = GK$ Therefore $GH$ is medial and $GK$ is rational. But $GH$ and $GK$ are applied to a rational straight line $FG$. Therefore from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$FK$ is rational and commensurable in length with $FG$ while from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$FH$ is rational and incommensurable in length with $FG$. Therefore by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}:  :$FH$ is incommensurable in length with $FK$. Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only. Therefore $KH$ is an apotome and $KF$ is the annex to $KH$. We have that: :$HF^2 = FK^2 + \lambda^2$ where either: :$\lambda$ is commensurable in length with $HF$ or: :$\lambda$ is incommensurable in length with $HF$. First suppose $\lambda$ is commensurable in length with $HF$. Then $FK$ is commensurable in length with the rational straight line $FG$. Therefore $KH$ is a second apotome. But from {{EuclidPropLink|book = X|prop = 92|title = Side of Area Contained by Rational Straight Line and Second Apotome}}:  :the "side" of $LH$ is a first apotome of a medial straight line. Next suppose $\lambda$ is incommensurable in length with $HF$. Then $FK$ is incommensurable in length with the rational straight line $FG$. Therefore $KH$ is a fifth apotome. But from {{EuclidPropLink|book = X|prop = 95|title = Side of Area Contained by Rational Straight Line and Fifth Apotome}}:  :the "side" of $LH$ is a straight line which produces with a rational area a medial whole. {{qed}} {{Euclid Note|109|X}}
:400px Let $BC$ be a medial area. Let the rational area $BD$ be subtracted from $BC$. It needs to be demonstrated that the "side" of the remainder $EC$ is either: : a second apotome of a medial straight line or: : a straight line which produces with a medial area a medial whole. Let $FG$ be a rational straight line. Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth. Let the area $GK$ equal to $BC$ be subtracted from $GH$. Then the remainder $EC$ is equal to $LH$. We have that: :$BC$ and $BD$ are medial. and: :$BC$ is incommensurable with $BD$. We also have: :$BC = GH$ and: :$BD = GK$ Therefore $GH$ and $GK$ are medial. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$FH$ and $FK$ are rational and incommensurable in length with $FG$. We have that $BC$ is incommensurable with $BD$. Therefore $GH$ is incommensurable with $GK$. From: :{{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}} and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}} we have that: :$HF$ is incommensurable in length with $FK$. Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only. Therefore $KH$ is an apotome and $KF$ is the annex to $KH$. We have that: : $HF^2 = FK^2 + \lambda^2$ where either: :$\lambda$ is commensurable in length with $HF$ or: :$\lambda$ is incommensurable in length with $HF$. First suppose $\lambda$ is commensurable in length with $HF$. We have that neither $FH$ and $FK$ is commensurable in length with the rational straight line $FG$. Therefore $KH$ is a third apotome. But from {{EuclidPropLink|book = X|prop = 93|title = Side of Area Contained by Rational Straight Line and Third Apotome}}:  :the "side" of $LH$ is a second apotome of a medial straight line. Next suppose $\lambda$ is incommensurable in length with $HF$. We have that neither $FH$ and $FK$ is commensurable in length with the rational straight line $FG$. Therefore $KH$ is a sixth apotome. But from {{EuclidPropLink|book = X|prop = 96|title = Side of Area Contained by Rational Straight Line and Sixth Apotome}}:  :the "side" of $LH$ is a straight line which produces with a medial area a medial whole. {{qed}} {{Euclid Note|110|X}}
:300px Let $AB$ be an apotome. Suppose $AB$ were the same with a binomial. Let $DC$ be a rational straight line. Let the rectangle $CE$ be applied to $CD$ equal to the square on $AB$ and producing $DE$ as breadth. We have that $AB$ is an apotome. Therefore from {{EuclidPropLink|book = X|prop = 97|title = Square on Apotome applied to Rational Straight Line}}: :$DE$ is a first apotome. Let $EF$ be the annex of $DE$. Therefore by {{EuclidDefLink|X (III)|1|First Apotome}}: :$DF$ and $EF$ are rational straight lines which are commensurable in square only :$DF^2 = FE^2 + \lambda^2$ where $\lambda$ is a straight line which is commensurable in length with $DF$ :$DF$ is commensurable in length with $DC$. We have that $AB$ is a binomial. Therefore by {{EuclidPropLink|book = X|prop = 60|title = Square on Binomial Straight Line applied to Rational Straight Line}}: :$DE$ is a first binomial. Let $DE$ be divided into its terms at $G$. Let $DG$ be the greater term. Then by {{EuclidDefLink|X (II)|1|First Binomial}}: :$DG$ and $GE$ are rational straight lines which are commensurable in square only :$DG^2 = GE^2 + \mu^2$ where $\mu$ is a straight line which is commensurable in length with $DF$ :$DG$ is commensurable in length with $DC$. Therefore by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$DG$ is commensurable in length with $DF$. Therefore by {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$GF$ is commensurable in length with $DG$. But $DF$ and $EF$ are rational straight lines which are commensurable in square only. So $DF$ is incommensurable in length with $EF$. Therefore by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$FG$ is incommensurable in length with $EF$. Therefore $GF$ and $EF$ are rational straight lines which are commensurable in square only. Therefore, by definition, $EG$ is an apotome. But $EG$ is also rational, which is impossible. Hence the result. {{qed}} {{Euclid Note|111|X}}
:350px Let the rectangle $AB \cdot CD$ be contained by the apotome $AB$ and the binomial straight line $CD$. Let $CE$ be the greater term of $CD$. Let: :$CE$ be commensurable in length with $AF$ :$ED$ be commensurable in length with $FB$ :$CE : ED = AF : FB$ Let the "side" of the rectangle $AB \cdot CD$ be $G$. It is to be demonstrated that $G$ is rational. Let $H$ be a rational straight line. Let a rectangle equal to $H^2$ be applied to $CD$ which produces $KL$ as breadth. By definition, $KL$ is an apotome. Let the terms of $KL$ be $KM$ and $ML$. Let $KM$ and $ML$ be commensurable with the terms $CE$ and $ED$ of the binomial straight line $CD$. But from  {{EuclidPropLink|book = X|prop = 112|title = Square on Rational Straight Line applied to Binomial Straight Line}}: :$CE$ and $ED$ are commensurable with the terms $AF$ and $FB$ of the apotome $CD$. Therefore: :$AF : FB = KM : ML$ and so: :$AF : KM = BF : LM$ and so by {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$AB : KL = AF : KM$ But from {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$AF$ is commensurable in length with $KM$. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AB$ is commensurable in length with $KL$. From {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$AB : KL = CD \cdot AB : CD \cdot KL$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$CD \cdot AB$ is commensurable with $CD \cdot KL$. But $CD \cdot KL = H^2$. Therefore $CD \cdot AB$ is commensurable with $H^2$. But $G^2 = CD \cdot AB$. Therefore $G^2$ is commensurable with $H^2$. But $H^2$ is rational. Therefore $G^2$ is also rational. Therefore $G$ is rational straight line. But $G^2$ is the "side" of the rectangle $AB \cdot CD$. Hence the result. {{qed}} {{Euclid Note|114|X}}
:350px Let $A$ be a medial straight line. It is to be shown that there is an infinite number of irrational straight lines, and none of them are: :Binomial :First bimedial :Second bimedial :Major :The side of a rational plus a medial area :The side of the sum of two medial areas :Apotome :First apotome of a medial :Second apotome of a medial :Minor :That which produces a medial whole with a rational area :That which produces a medial whole with a medial area Let the above set be called $\mathcal K$. Let $B$ be a rational straight line. Let $C^2$ be equal to $A \cdot B$. Therefore by: :{{EuclidDefLink|X|4|Rational Area}} and from: :{{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}} it follows that: :$C$ is irrational. Consider a square on any of $\mathcal K$. None of them, when applied to a rational straight line, produces as breadth a medial straight line. Let $D^2$ be equal to $B \cdot C$. Therefore by: :{{EuclidDefLink|X|4|Rational Area}} and from: :{{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}} it follows that: :$D$ is irrational. Consider a square on any of $\mathcal K$. None of them, when applied to a rational straight line, produces $C$ as breadth. This process can be continued ad infinitum. {{qed}} {{Euclid Note|115|X}}
Let $x > 0$. {{begin-eqn}} {{eqn | l = x - 1        | o = \ge       | r = \ln x       | c = Upper Bound of Natural Logarithm }} {{eqn | ll= \leadsto       | l = \frac 1 x -1       | o = \ge       | r = \ln \frac 1 x       | c = putting $\frac 1 x$ into the above inequality }} {{eqn | r = -\ln x       | c = Logarithm of Reciprocal }} {{eqn | ll= \leadsto       | l = 1 - \frac 1 x       | o = \le       | r = \ln x       | c = multiplying throughout by $-1$ }} {{end-eqn}} {{qed}}
Let $x > 0$. Note that: :$1 - \dfrac 1 x \le \ln x$ is logically equivalent to: :$1 - \dfrac 1 x - \ln x \le 0$ Let $f \left({x}\right) = 1 - \dfrac 1 x - \ln x$. Then: {{begin-eqn}} {{eqn | l = f \left({x}\right)       | r = 1 - \dfrac 1 x - \ln x }} {{eqn | ll= \implies       | l = f' \left({x}\right)       | r = \frac 1 {x^2} - \frac 1 x       | c = Derivative of Constant, Power Rule for Derivatives, Derivative of Natural Logarithm Function }} {{eqn | r = \frac {1 - x} {x^2} }} {{eqn | ll= \implies       | l = f'' \left({x}\right)       | r = - \frac 2 {x^3} + \frac 1 {x^2}       | c = Power Rule for Derivatives }} {{end-eqn}}  Note that $f' \left({1}\right) = 0$. Also, $f'' \left({1}\right) < 0$. So by the Second Derivative Test, $x = 1$ is a local maximum. On $\left ({0 \,.\,.\, 1} \right)$: : $f' \left({x}\right) > 0$ By Derivative of Monotone Function, $f$ is strictly increasing on that interval. On $\left ({1 \,.\,.\, +\infty} \right)$: :$f'\left({x}\right) < 0$ By Derivative of Monotone Function, $f$ is strictly decreasing on that interval. So $x = 1$ yields a global maximum, at which by Logarithm of 1 is 0: :$f \left({1}\right) = 1 - 1 - 0 = 0$ That is: : $\forall x > 0: f \left({x}\right) \le 0$ and so by definition of $f \left({x}\right)$: :$1 - \dfrac 1 x - \ln x \le 0$ {{qed}}
Let $\sequence {f_n}$ be the sequence of mappings $f_n: \R_{>0} \to \R$ defined as: :$\map {f_n} x = n \paren {\sqrt [n] x - 1 }$ Fix $x \in \R_{>0}$. We first show that $\forall n \in \N : 1 - \dfrac 1 x \le < n \paren {\sqrt [n] x - 1}$ Let $n \in \N$. From Sum of Geometric Sequence: :$\sqrt [n] x - 1 = \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }$ === Case 1: $0 < x < 1$ === {{begin-eqn}} {{eqn | l = 0 < x < 1       | o = \leadsto       | r = \forall k < n: \sqrt [n] x^{n - k} > x > 0       | c = Power Function on Base between Zero and One is Strictly Decreasing/Rational Number }} {{eqn | o = \leadsto       | r = 0 < n x < \sum_{k = 0}^{n - 1} \sqrt [n] x^{n - k}       | c = Real Number Ordering is Compatible with Addition }} {{eqn | o = \leadsto       | r = \dfrac 1 {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} } > \dfrac 1 {n x}       | c = Ordering of Reciprocals }} {{eqn | o = \leadsto       | r = \dfrac {x - 1} {n x} < \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{eqn | o = \leadsto       | r = \dfrac {x - 1} {n x} < \sqrt [n] x - 1       | c = Sum of Geometric Sequence }} {{eqn | o = \leadsto       | r = 1 - \dfrac 1  x < n \paren {\sqrt [n] x - 1}       | c = Real Number Ordering is Compatible with Multiplication }} {{end-eqn}} {{qed|lemma}} === Case 2: $x = 1$ === {{begin-eqn}} {{eqn | l = \dfrac {x - 1} x       | r = 0 }} {{eqn | r = \sqrt [n] 1 - 1 }} {{end-eqn}} {{qed|lemma}} === Case 3: $x > 1$ === {{begin-eqn}} {{eqn | l = x > 1       | o = \leadsto       | r = \forall k < n: 1 < \sqrt [n] x^{n - k} < x       | c = Power Function on Base Greater than One is Strictly Increasing/Rational Number }} {{eqn | o = \leadsto       | r = 0 < \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k} < n x       | c = Real Number Ordering is Compatible with Addition }} {{eqn | o = \leadsto       | r = 0 < \dfrac 1 {n x} < \dfrac 1 {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }       | c = Ordering of Reciprocals }} {{eqn | o = \leadsto       | r = \dfrac {x - 1} {n x} < \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = \leadsto       | r = \dfrac {x - 1} {n x} < \sqrt [n] x - 1       | c = Sum of Geometric Sequence }} {{eqn | o = \leadsto       | r = 1 - \dfrac 1 x < n \paren {\sqrt [n] x - 1 }       | c = Real Number Ordering is Compatible with Multiplication }} {{end-eqn}} {{qed|lemma}} Thus: :$\forall n \in \N : 1 - \dfrac 1 x \le n \paren {\sqrt [n] x - 1 }$ by Proof by Cases. Thus: :$\displaystyle 1 - \dfrac 1 x \le \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1 }$ from Limit of Bounded Convergent Sequence is Bounded. Hence the result, from the definition of $\ln$. {{qed}}
* Upper Bound of Natural Logarithm: $\ln y \le y - 1$ * Lower Bound of Natural Logarithm: $1 - \dfrac 1 y \le \ln y$ {{qed}}