hcju/proofwikiqa huggingface dataset:
Let $x \in I_o$. Then by definition: :$0 < x < 1$ and so: :$0 \le x \le 1$ and so: :$x \in I_c$. Thus: :$I_o \subseteq I_c$ Consider: :$0 \in I_c$ by definition of closed interval. But it is not the case that $0 < 0$. So $0 \notin I_o$ and so $I_c \not \subseteq I_o$. Hence the result by definition of proper subset. {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b}       | r = \paren {\mathbf {a \cdot c} } \mathbf b - \paren {\mathbf {a \cdot b} } \mathbf c       | c =  }} {{eqn | o =        | ro= +       | r = \paren {\mathbf {b \cdot a} } \mathbf c - \paren {\mathbf {b \cdot c} } \mathbf a       | c =  }} {{eqn | o =       | ro= +       | r = \paren {\mathbf {c \cdot b} } \mathbf a - \paren {\mathbf {c \cdot a} } \mathbf b       | c = Lagrange's Formula }} {{eqn | r = \mathbf 0       | c = Dot Product Operator is Commutative }} {{end-eqn}} {{qed}} Category:Algebra Category:Vector Cross Product 6ec8ad0ti7cpssidwo3r3rb9zcdunlc
Let $X$ be finite subset of $S$ such that: :$X$ admits a supremum By definition of complete lattice: :$f \left[{X}\right]$ admits a supremum We will prove the result by induction on the cardinality of $X$. === Basis Case === :$\forall X \subseteq S: \left\vert{X}\right\vert = 0 \implies \sup \left({f \left[{X}\right]}\right) = f \left({\sup X}\right)$ where: :$\left\vert{X}\right\vert$ denotes the cardinality of $X$ Let $X \subseteq S$ such that: :$\left\vert{X}\right\vert = 0$ By Cardinality of Empty Set: :$X = \varnothing$ By definition of image of set: :$f \left[{X}\right] = \varnothing$ By definitions of bottom and smallest element: :$\bot \preceq x$ Thus: {{begin-eqn}} {{eqn | l = \sup \left({f \left[{X}\right]}\right)       | r = \bot       | c = Supremum of Empty Set is Smallest Element }} {{eqn | r = x \wedge \bot       | c = Preceding iff Meet equals Less Operand }} {{eqn | r = x \wedge \sup X       | c = Supremum of Empty Set is Smallest Element }} {{eqn | r = f \left({\sup X}\right)       | c = definition of $f$ }} {{end-eqn}} {{qed|lemma}} === Induction Hypothesis === :$\forall X \subseteq S: \left\vert{X}\right\vert = n \implies \sup \left({f \left[{X}\right]}\right) = f \left({\sup X}\right)$ === Induction Step === :$\forall X \subseteq S: \left\vert{X}\right\vert = n + 1 \implies \sup \left({f \left[{X}\right]}\right) = f \left({\sup X}\right)$ Let $X \subseteq S$ such that: :$\left\vert{X}\right\vert = n + 1$ Then $X = \left\{ {x_1, \dots, x_n, x_{n+1} }\right\}$. Thus: {{begin-eqn}} {{eqn | l = \sup \left({ f \left[{X}\right] }\right)       | r = \sup \left({ f \left[{\left\{ {x_1, \dots, x_n}\right\} \cup \left\{ {x_{n+1} }\right\} }\right] }\right)       | c = Union of Unordered Tuples }} {{eqn | r = \sup \left({ f \left[{ \left\{ {x_1, \dots, x_n}\right\} }\right] \cup f \left[{ \left\{ {x_{n+1} }\right\} }\right] }\right)       | c = Image of Union under Mapping }} {{eqn | r = \sup \left({f \left[{ \left\{ {x_1, \dots, x_n}\right\} }\right] }\right) \vee \sup \left({ f \left[{ \left\{ {x_{n+1} }\right\} }\right] }\right)       | c = Supremum of Suprema }} {{eqn | r = f \left({ \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee \sup \left({ f \left[{\left\{ {x_{n+1} }\right\} }\right] }\right)       | c = Induction Hypothesis }} {{eqn | r = f \left({ \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee \sup \left\{ { f \left({ x_{n+1} }\right) }\right\}       | c = Image of Singleton under Mapping }} {{eqn | r = f \left({ \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee f \left({ x_{n+1} }\right)       | c = Supremum of Singleton }} {{eqn | r = \left({ x \wedge \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee \left({ x \wedge x_{n+1} }\right)       | c = Definition of $f$ }} {{eqn | r = x \wedge \left({ \sup \left\{ {x_1, \dots, x_n}\right\} \vee x_{n+1} }\right)       | c = {{Defof|Distributive Lattice}} }} {{eqn | r = x \wedge \left({ \sup \left\{ {x_1, \dots, x_n}\right\} \vee \sup \left\{ {x_{n+1} }\right\} }\right)       | c = Supremum of Singleton }} {{eqn | r = x \wedge \sup X       | c = Supremum of Suprema }} {{eqn | r = f \left({\sup X}\right)       | c = definition of $f$ }} {{end-eqn}} {{qed}}
Proof by Counterexample: Let $S = \set {a, b}$. Let: :$A = \set {a, b}$ :$B = \set a$ :$C = \set b$ Then: {{begin-eqn}} {{eqn | l = A \cup B       | r = \set {a, b}       | c =  }} {{eqn | r = A \cup C       | c =  }} {{end-eqn}} but: {{begin-eqn}} {{eqn | l = B       | r = \set a       | c =  }} {{eqn | o = \ne       | r = \set b       | c =  }} {{eqn | r = C       | c =  }} {{end-eqn}} {{qed}}
Proof by Counterexample: Let $S = \left\{ {a, b, c}\right\}$. Let: :$A = \left\{ {a}\right\}$ :$B = \left\{ {a, b}\right\}$ :$C = \left\{ {a, c}\right\}$ Then: {{begin-eqn}} {{eqn | l = A \cap B       | r = \left\{ {a}\right\}       | c =  }} {{eqn | r = A \cap C       | c =  }} {{end-eqn}} but: {{begin-eqn}} {{eqn | l = B       | r = \left\{ {a, b}\right\}       | c =  }} {{eqn | o = \ne       | r = \left\{ {a, c}\right\}       | c =  }} {{eqn | r = C       | c =  }} {{end-eqn}} {{qed}}
Let $x, y \in I_1 \cap I_2$.  From the definition of a real interval, it suffices to show that: :for each $z \in \R$ with $x \le z \le y$ we have $z \in I_1 \cap I_2$. Let $z$ be a real number with: :$x \le z \le y$ Since $x, y \in I_1$, we have:  :$z \in I_1$ from the definition of a real interval. Similarly, since $x, y \in I_2$, we have: :$z \in I_2$ from the definition of a real interval. So: :$z \in I_1 \cap I_2$ as required. {{qed}}
Proof by Counterexample: Consider the real intervals: :$I_1 = \left({0 \,.\,.\, 2}\right)$ :$I_2 = \left({4 \,.\,.\, 6}\right)$ Then we have that: :$1 < 3 < 5$ where: :$1 \in I_1 \cup I_2$ :$5 \in I_1 \cup I_2$ but: :$3 \notin I_1 \cup I_2$ Thus $I_1 \cup I_2$ is not a real interval. {{qed}}
First it is shown that $f \cup g$ is well-defined. It is sufficient to check that $f \cup g$ maps any $x \in A \cap B$ to a single value in $Y$. Let $x \in A \cap B$. Because $x \in A$: :$\map {f \cup g} x = \map f x \in Y$ Because $x \in B$: :$\map {f \cup g} x = \map g x \in Y$ But by definition of agreement of mappings: :$\map f x = \map g x$ so $f \cup g$ maps $x$ to a unique value in $Y$, as was to be shown. {{qed|lemma}} Next it is shown that $f \cup g$ is continuous: We show that $\map {\paren {f \cup g}^{-1} } U$ is open in $A \cup B$ for every open $U$ in $Y$. {{tidy|The following needs to be progressive rather than regressive.}} Observe first that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$. For if $x \in (f \cup g)^{-1}(U) \subseteq A \cup B$, then $(f \cup g)(x) \in U$; but if $x \in A$, then $f(x) = (f \cup g)(x) \in U$ i.e. $x \in f^{-1}(U)$ or if $x \in B$, then $g(x) = (f \cup g)(x) \in U$ i.e. $x \in g^{-1}(U)$. In both cases, $x \in f^{-1}(U) \cup g^{-1}(U)$. Conversely if $x \in f^{-1}(U) \cup g^{-1}(U)$, then without loss of generality consider the case of $x \in f^{-1}(U) \subseteq A$; in this case $(f \cup g)(x) = f(x) \in U$ so that $x \in (f \cup g)^{-1}(U)$. Having established that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$, note that by continuity of $f$ and $g$, we have that $f^{-1}(U)$ is open in $A$ and $g^{-1}(U)$ is open in $B$. So there are an open sets $P$ and $Q$ of $X$ such that $f^{-1}(U) = P \cap A$ and $g^{-1}(U) = Q \cap B$. Since $A$ and $B$ are open in $X$ and intersections of open sets are open, this implies that $f^{-1}(U)$ and $g^{-1}(U)$ are open in $X$. Furthermore, note that $P \cap A$ and $Q \cap B$ are both subsets of $A \cup B$. Hence, $f^{-1}(U) = f^{-1}(U) \cap (A \cup B)$ and $g^{-1}(U) = g^{-1}(U) \cap (A \cup B)$. Thus: :$(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U) = (f^{-1}(U) \cup g^{-1}(U)) \cap (A \cup B)$ demonstrating that $(f \cup g)^{-1}(U)$ is open in the subspace topology of $A \cup B$. {{qed}} Category:Continuous Mappings Category:Named Theorems 2oyqi031kxgrr5k967kl1gpj0jfzjph
The operation of real multiplication is defined on all real numbers. Thus: :$\forall x \in \R: \exists y \in \R: x^2 = y$ Hence the result by definition of domain. {{qed}}
From Square of Real Number is Non-Negative, the image of $f$ is $\R_{\ge 0}$. From Positive Real has Real Square Root: :$\forall x \in \R: \exists y \in \R: x^2 = y$ Hence the result by definition of image. {{qed}}
From Square of Real Number is Non-Negative: :$\forall x \in \R: x^2 \ge 0$ Hence the result by definition of domain. {{qed}}
{{begin-eqn}} {{eqn | l = a^{x + y}       | r = \map \exp {\paren {x + y} \ln a}       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp {x \ln a + y \ln a}       | c =  }} {{eqn | r = \map \exp {x \ln a} \, \map \exp {y \ln a}       | c = Exponential of Sum }} {{eqn | r = a^x a^y       | c = Definition of Power to Real Number }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | n = 1       | o =        | r = 0 < a < b       | c =  }} {{eqn | n = 2       | o = \leadsto       | r = 0 < b       | c = Ordering is Transitive }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 3       | o =        | r = 0 < c < d       | c =  }} {{eqn | n = 4       | o = \leadsto       | r = 0 < c       | c = Ordering is Transitive }} {{end-eqn}} {{begin-eqn}} {{eqn | o = \leadsto       | r = 0 < a c < b c       | c = $(1)$ and $(4)$: Real Number Axioms: $\R \text O 2$: Usual ordering is compatible with multiplication }} {{eqn | o =        | r = 0 < b c < b d       | c = $(2)$ and $(3)$: Real Number Axioms: $\R \text O 2$: Usual ordering is compatible with multiplication }} {{eqn | o = \leadsto       | r = 0 < a c < b d       | c = Ordering is Transitive }} {{end-eqn}} {{qed}} Category:Inequalities Category:Real Multiplication 8tqah01rau1ebolhjvbwd6gc0gn332y
{{begin-eqn}} {{eqn | l = \left\vert{x - z}\right\vert       | o = <       | m = y       | c =  }} {{eqn | ll= \iff       | l = -y       | o = <       | m = x - z       | mo= <       | r = y       | c = Negative of Absolute Value: Corollary 1 }} {{eqn | ll= \iff       | l = z - y       | o = <       | m = x       | mo= <       | r = z + y       | c = Real Number Ordering is Compatible with Addition }} {{end-eqn}} {{qed}} Category:Negative of Absolute Value bectnqs04v7m7wndbdbq7o8oomwoumw
First: {{begin-eqn}} {{eqn | l = \epsilon       | o = <       | r = \min \set {y_1, y_2, 1} }} {{eqn | ll= \leadsto       | l = \epsilon       | o = <       | r = y_1       | c = {{Defof|Min Operation}} }} {{eqn | ll= \leadsto       | l = \epsilon - \epsilon       | o = <       | r = y_1 - \epsilon       | c = subtracting $\epsilon$ from both sides }} {{eqn | n = 1       | ll= \leadsto       | l = 0       | o = <       | r = y_1 - \epsilon }} {{end-eqn}} The same logic, mutatis mutandis, shows that $0 < y_2 - \epsilon$. From Negative of Absolute Value: Corollary 3: {{begin-eqn}} {{eqn | l = \size {x_1 - y_1} < \epsilon       | o = \implies       | r = y_1 - \epsilon < x_1 < y_1 - \epsilon }} {{eqn | l = \size {x_2 - y_2} < \epsilon       | o = \implies       | r = y_2 - \epsilon < x_2 < y_2 - \epsilon }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \paren {y_1 - \epsilon} \paren {y_2 - \epsilon}       | o = <       | m = x_1 x_2       | mo= <       | r = \paren {y_1 + \epsilon} \paren {y_1 + \epsilon}       | c = Inequality of Product of Unequal Numbers: from $(1)$ }} {{eqn | ll= \leadsto       | l = y_1 y_2 - \epsilon \paren {y_1 + y_2} + \epsilon^2       | o = <       | m = x_1 x_2       | mo= <       | r = y_1 y_2 + \epsilon \paren {y_1 + y_2} + \epsilon^2       | c = Multiplication of Numbers Distributes over Addition }} {{eqn | ll= \leadsto       | l = y_1 y_2 - \epsilon \paren {y_1 + y_2} - \epsilon^2       | o = <       | m = x_1 x_2       | mo= <       | r = y_1 y_2 + \epsilon \paren {y_1 + y_2} + \epsilon^2       | c = Square of Non-Zero Real Number is Strictly Positive }} {{end-eqn}} Subtracting $y_1 y_2$ from all sections of the inequality: :$-\epsilon \paren {y_1 + y_2} - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \paren {y_1 + y_2} + \epsilon^2$ If follows that: {{begin-eqn}} {{eqn | l = \size {x_1 x_2 - y_1 y_2}       | o = <       | r = \epsilon \paren {y_1 + y_2} + \epsilon^2       | c = Negative of Absolute Value: Corollary 1 }} {{eqn | o = <       | r = \epsilon \paren {y_1 + y_2} + \epsilon       | c = Since $\epsilon < \min \size {y_1, y_2, 1} < 1$, we may apply Real Number between Zero and One is Greater than Square }} {{eqn | o = <       | r = \epsilon \paren {y_1 + y_2 + 1}       | c = Multiplication of Numbers Distributes over Addition }} {{end-eqn}} Hence the result. {{qed}} Category:Exponent Combination Laws j6l3ndjrgtgkvp631xr58058fdbl8x6
Let $a$ be arbitrary. :$a \in \set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m}$ {{iff}} :$a \in \set {x_1, \dots, x_n}$ or $a \in \set {x_{n + 1}, \dots, x_m}$ by definition of union {{iff}} :$a = x_1 \lor \dots \lor a = x_n$ or $a = x_{n + 1} \lor \dots \lor a = x_m$ by definition of unordered tuple {{iff}} :$a = x_1 \lor \dots \lor a = x_n \lor a = x_{n + 1} \lor \dots \lor a = x_m$ {{iff}} :$a \in \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$ by definition of unordered tuple Thus result follows by definition of set equality. {{qed}} Category:Set Theory e5gxlqozltwspxusogyf4rkscz2r779
By Singleton is Directed and Filtered Subset: :$\left\{ {y}\right\}$ is directed. By Supremum of Singleton: :$\left\{ {y}\right\}$ admits a supremum and $\sup \left\{ {y}\right\} = y$ By definition of reflexivity: :$y \preceq \sup \left\{ {y}\right\}$ By definition of way below: :$\exists d \in \left\{ {y}\right\}: x \preceq d$ Thus by definition of singleton: :$x \preceq y$ {{qed}}
Let $D$ be a directed subset of $S$ such that :$D$ admits a supremum and :$z \preceq \sup D$ By definition of transitivity: :$y \preceq \sup D$ By definition of way below relation: :$\exists d \in D: x \preceq d$ Thus by definition of transitivity: :$\exists d \in D: u \preceq d$ Thus by definition of way below relation: :$u \ll z$ {{qed}}
Let $D$ be a directed subset of $S$ such that :$D$ admits a supremum and :$z \preceq \sup D$ By definition of way below relation: :$\exists d_1 \in D: x \preceq d_1$ and :$\exists d_2 \in D: y \preceq d_2$ By definition of directed subset: :$\exists d \in D: d_1 \preceq d$ and $d_2 \preceq d$ By definition of transitivity: :$x \preceq d$ and $y \preceq d$ Thus by definition of supremum: :$x \vee y \preceq d$ Thus by definition way below relation: :$x \vee y \ll z$ {{qed}}
By Way Below implies Preceding: :$x \preceq y$ By definition of reflexivity: :$z \preceq z$ Thus by Preceding and Way Below implies Way Below: :$x \ll z$ {{qed}}
By Way Below implies Preceding: :$x \preceq y$ and $y \preceq x$ Thus by definition of antisymmetry: :$x = y$ {{qed}}
Let: :$x = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$ Then: {{begin-eqn}} {{eqn | l = x       | r = 1 + \frac 1 x       | c = substituting for $x$ }} {{eqn | ll= \leadsto       | l = x^2       | r = x + 1       | c =  }} {{eqn | ll= \leadsto       | l = x^2 - x - 1       | r = 0       | c =  }} {{end-eqn}} The result follows from Golden Mean as Root of Quadratic. {{qed}}
=== Definition 1 is equivalent to Definition 3 === Let $AB : AC = AC : BC$. Let $\dfrac {AB} {AC} = \dfrac {AC} {BC} = \phi$. Then: {{begin-eqn}} {{eqn | l = \phi       | r = \frac {AC + BC} {AC}       | c = as $AB = AC + BC$ }} {{eqn | r = 1 + \frac {BC} {AC}       | c =  }} {{eqn | r = 1 + \frac 1 \phi       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \phi - 1       | r = \frac 1 \phi       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 {\phi - 1}       | r = \phi       | c =  }} {{end-eqn}} {{qed|lemma}} === Definition 2 equivalent to Definition 3 === {{begin-eqn}} {{eqn | l = \phi       | r = \frac 1 {\phi - 1}       | c = {{Defof|Golden Mean|index = 3}} }} {{eqn | ll= \leadstoandfrom       | l = \phi \paren {\phi - 1}       | r = 1       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \phi^2 - \phi - 1       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \phi       | r = \frac {1 \pm \sqrt {1^2 - 4 \times 1 \times \paren {-1} } } 2       | c = Quadratic Formula }} {{eqn | r = \frac {1 \pm \sqrt 5} 2       | c =  }} {{end-eqn}} Of these two roots, only $\dfrac {1 + \sqrt 5} 2$ is positive. {{qed}} Category:Golden Mean gc27ire02t5vge6vmcrgk8dkf1l2l41
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$F_n \le \phi^{n - 1}$ === Basis for the Induction === $P \left({1}\right)$ is true, as this just says: :$F_1 = 1 = \phi^0 = \phi^{1 - 1}$ It is also necessary to demonstrate $P \left({2}\right)$ is true: :$F_2 = 1 \le \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{2 - 1}$ This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, for all $1 \le k \le n$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$F_k \le \phi^{k - 1}$ from which it is to be shown that: :$F_{k + 1} \le \phi^k$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = F_{k + 1}       | r = F_{k - 1} + F_k       | c = Definition of Fibonacci Numbers }} {{eqn | o = \le       | r = \phi^{k - 2} + \phi^{k - 1}       | c = Induction Hypothesis }} {{eqn | r = \phi^{k - 2} \left({1 + \phi}\right)       | c =  }} {{eqn | r = \phi^{k - 2} \left({\phi^2}\right)       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \phi^k       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{> 0}: F_n \le \phi^{n - 1}$ {{qed}}
{{begin-eqn}} {{eqn | l = \phi       | r = \frac 1 {\phi - 1}       | c = Definition of Golden Mean }} {{eqn | ll= \leadstoandfrom       | l = \phi \paren {\phi - 1}       | r = 1       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \phi^2       | r = \phi + 1       | c =  }} {{end-eqn}} {{qed}}
The proof proceeds by induction. For all $n \in \N_{\ge 2}$, let $P \left({n}\right)$ be the proposition: :$F_n \ge \phi^{n - 2}$ === Basis for the Induction === $P \left({2}\right)$ is true, as this just says: :$F_2 = 1 = \phi^0 = \phi^{2 - 2}$ It is also necessary to demonstrate $P \left({3}\right)$ is true: :$F_3 = 2 \ge \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{3 - 1}$ This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, for all $1 \le k \le n$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$F_k \ge \phi^{k - 2}$ from which it is to be shown that: :$F_{k + 1} \ge \phi^{k - 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = F_{k + 1}       | r = F_{k - 1} + F_k       | c = Definition of Fibonacci Numbers }} {{eqn | o = \ge       | r = \phi^{k - 3} + \phi^{k - 2}       | c = Induction Hypothesis }} {{eqn | r = \phi^{k - 3} \left({1 + \phi}\right)       | c =  }} {{eqn | r = \phi^{k - 3} \left({\phi^2}\right)       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \phi^{k - 1}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{\ge 2}: F_n \ge \phi^{n - 2}$ {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \frac {n \paren {n + 1} } 2 = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \frac {n \paren {n + 1} } 2       | r = 1       | c =  }} {{eqn | r = 1^2       | c =  }} {{end-eqn}} which 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: :$\displaystyle \frac {k \paren {k + 1} } 2 = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2$ from which it is to be shown that: :$\displaystyle \frac {\paren {k + 1} \paren {k + 2} } 2 = \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2       | r = \paren {k + 1}^2 - \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2}       | c =  }} {{eqn | r = \paren {k + 1}^2 - \frac {k \paren {k + 1} } 2       | c = Induction Hypothesis }} {{eqn | r = \frac {2 k^2 + 4 k + 2 - k^2 - k} 2       | c = multiplying everything out }} {{eqn | r = \frac {k^2 + 3 k + 2} 2       | c = simplifying }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} } 2       | c = factoring }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
A visual illustration of the proof for $n = 5$: :500px The number of squares of side $n$ is seen to be $4 n$. To go from $n$ to $n + 1$, a ring of $4 \paren {n + 1}$ squares of side $n + 1$ is to be added: :$4$ for the one at each corner :$4 n$ for the ones that abut the sides of the $n + 1$ squares of side $n$. The length of one side is given by: :$S = 2 \paren {1 + 2 + \cdots + n}$ The length of one side is also given by: :$S = n \paren {n + 1}$ The area is therefore given in two ways as: :$A = 4 \paren {1 + 2 + \cdots + n}^2 = \paren {n \paren {n + 1} }^2$ and also as: {{begin-eqn}} {{eqn | l = A       | r = 4 \times 1^2 + 4 \times 2 \times 2^2 + 4 \times 3 \times 3^2 + \cdots + 4 \times n \times n^2       | c =  }} {{eqn | r = 4 \paren {1^3 + 2^2 + 3^3 + \cdots + n^3}       | c =  }} {{end-eqn}} The result follows by equating the expressions for area. {{qed}}
Let $0 < x < 1$. Let $y = -x$. Then $y > -1$ and by Bernoulli's Inequality: :$\paren {1 + y}^n \ge 1 + n y$ Thus: :$\paren {1 + \paren {-x} }^n \ge 1 + n \paren {-x}$ Hence the result. {{qed}}
Proof by induction: Let $0 < x < 1$. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\left({1 - x}\right)^n \ge 1 - n x$ === Basis for the Induction === $P \left({0}\right)$ is the case: :$\left({1 - x}\right)^0  = 1 \ge 1 - 0 x = 1$ so $P \left({0}\right)$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\left({1 - x}\right)^k \ge 1 - k x$ We need to show that: :$\left({1 - x}\right)^{k+1} \ge 1 - \left({k + 1}\right) x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({1 - x}\right)^{k + 1}       | r = \left({1 - x}\right)^k \left({1 - x}\right) }} {{eqn | o = \ge       | r = \left({1 - k x}\right) \left({1 - x}\right)       | c = Induction Hypothesis }} {{eqn | r = 1 - \left({k + 1}\right) x + k x^2 }} {{eqn | o = \ge       | r = 1 - \left({k + 1}\right) x }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. {{qed}}