hcju/proofwikiqa huggingface dataset:
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$ === Basis for the Induction === First from Power of Sum Modulo Prime we have that $\map P 1$ is true: :$\paren {a + b}^p \equiv a^p + b^p \pmod p$ 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: :$\paren {a + b}^{p^k} \equiv a^{p^k} + b^{p^k} \pmod p$ Then we need to show: :$\paren {a + b}^{p^{k + 1} } \equiv a^{p^{k + 1} } + b^{p^{k + 1} } \pmod p$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {a + b}^{p^k}       | o = \equiv       | r = a^{p^k} + b^{p^k}       | rr= \pmod p       | c = Induction Hypothesis }} {{eqn | ll= \leadsto       | l = \paren {\paren {a + b}^{p^k} }^p       | o = \equiv       | r = \paren {a^{p^k} + b^{p^k} }^p       | rr= \pmod p       | c = Congruence of Powers }} {{eqn | ll= \leadsto       | l = \paren {\paren {a + b}^{p^k} }^p       | o = \equiv       | r = \paren {a^{p^k} }^p + \paren {b^{p^k} }^p       | rr= \pmod p       | c = Basis for the Induction }} {{eqn | ll= \leadsto       | l = \paren {a + b}^{p^{k + 1} }       | o = \equiv       | r = a^{p^{k + 1} } + b^{p^{k + 1} }       | rr= \pmod p       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$ {{qed}}
From  Prime Power of Sum Modulo Prime we have: :$(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$ We can write this: :$\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$ By $(1)$ and Congruence of Powers, we therefore have: :$\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} + b^{p^n} }^k \pmod p$ The coefficient $\dbinom {p^n k} {p^n}$ is the binomial coefficient of $b^{p^n}$ in $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$. Expanding $\paren {a^{p^n} + b^{p^n} }^k$ using the Binomial Theorem, we find that the coefficient of $b^{p^n}$, the second term, is $\dbinom k 1 = k$. So: :$\dbinom {p^n k} {p^n} \equiv k \pmod p$ {{qed}}
Lucas' Theorem states that for $n, k, p \in \Z$ and $p$ be a prime number, such that: :$n = a_r p^r + \cdots + a_1 p + a_0$ :$k = b_r p^r + \cdots + b_1 p + b_0$ then: :$\displaystyle \binom n k \equiv \prod_{j \mathop = 0}^r \binom {a_j}{b_j} \pmod p$ Therefore: {{begin-eqn}} {{eqn | l = \binom {p^n k} {p^n}       | o = \equiv       | r = \binom k 1 \prod_{j \mathop = 0}^{n - 1} \binom 0 0 \pmod p       | c = Lucas' Theorem }} {{eqn | o = \equiv       | r = k \prod_{j \mathop = 0}^{n - 1} 1 \pmod p       | c = Binomial Coefficient with One, Binomial Coefficient with Zero }} {{eqn | o = \equiv       | r = k \pmod p       | c =  }} {{end-eqn}} {{qed}}
We see that: : $F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \left({-1}\right)^1$ so the proposition holds for $n = 1$. We also see that: : $F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \left({-1}\right)^2$ so the proposition holds for $n = 2$. Suppose the proposition is true for $n = k$, that is: : $F_{k + 1} F_{k - 1} - F_k^2 = \left({-1}\right)^k$ It remains to be shown that it follows from this that the proposition is true for $n = k + 1$, that is: : $F_{k + 2} F_k - F_{k + 1}^2 = \left({-1}\right)^{k + 1}$ So: {{begin-eqn}} {{eqn | l = F_{k + 2} F_k - F_{k + 1}^2       | r = \left({F_k + F_{k + 1} }\right) F_k - F_{k + 1}^2 }} {{eqn | r = F_k^2 + F_k F_{k + 1} - F_{k + 1}^2 }} {{eqn | r = F_k^2 + F_k F_{k + 1} - F_{k + 1} \left({F_k + F_{k - 1} }\right) }} {{eqn | r = F_k^2 + F_k F_{k + 1} - F_k F_{k + 1} - F_{k + 1} F_{k - 1} }} {{eqn | r = F_k^2 - F_{k + 1} F_{k - 1} }} {{eqn | r = \left({-1}\right) \left({F_{k + 1} F_{k - 1} - F_k^2}\right) }} {{eqn | r = \left({-1}\right) \left({-1}\right)^k }} {{eqn | r = \left({-1}\right)^{k + 1} }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}} Note that from the above we have that: : $F_{k + 2} F_k - F_{k + 1}^2 = \left({-1}\right)^{k + 1}$ from which: : $F_{n + 1}^2 - F_n F_{n + 2} = \left({-1}\right)^n$ follows immediately.
First we use this lemma: {{:Cassini's Identity/Lemma}} Then the determinant of both sides is taken. The {{LHS}} follows directly from the order 2 determinant: :$\begin{bmatrix} F_{n + 1}  & F_n     \\ F_n        & F_{n - 1} \end{bmatrix} = F_{n + 1} F_{n - 1} - F_n^2$ Now for the {{RHS}}: ==== Basis for the Induction ==== :$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \left({-1}\right)^1$ ==== Induction Hypothesis ==== For $k \in \Z_{>0}$, it is assumed that: :$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k = \left({-1}\right)^k$ It remains to be shown that: :$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k + 1} = \left({-1}\right)^{k + 1}$ ==== Induction Step ==== The induction step follows from Determinant of Matrix Product: :$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = \left({-1}\right)^k \left({-1}\right) = \left({-1}\right)^{k + 1}$ Hence by induction: :$\forall n \in \Z_{>0}: \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^n = \left({-1}\right)^n$ {{qed}}
The rational numbers are arranged thus: :$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3,  \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3, \dfrac {-4} 1, \dfrac {-4} 3 \ldots$ It is clear that every rational number will appear somewhere in this list. Thus it is possible to set up a bijection between each rational number and its position in the list, which is an element of $\N$. {{qed}}
Let us define the mapping $\phi: \Q \to \Z \times \N$ as follows: :$\forall \dfrac p q \in \Q: \phi \left({\dfrac p q}\right) = \left({p, q}\right)$ where $\dfrac p q$ is in canonical form. Then $\phi$ is clearly injective. From Cartesian Product of Countable Sets is Countable, we have that $\Z \times \N$ is countably infinite. The result follows directly from Domain of Injection to Countable Set is Countable. {{qed}}
For each $n \in \N$, define $S_n$ to be the set: :$S_n := \left\{{\dfrac m n: m \in \Z}\right\}$ By Integers are Countably Infinite, each $S_n$ is countably infinite. Because each rational number can be written down with a positive denominator, it follows that: :$\forall q \in \Q: \exists n \in \N: q \in S_n$ which is to say: :$\displaystyle \bigcup_{n \mathop \in \N} S_n = \Q$ By Countable Union of Countable Sets is Countable, it follows that $\Q$ is countable. Since $\Q$ is manifestly infinite, it is countably infinite. {{qed}}
Let $Q_\pm = \set {q \in \Q: \pm q > 0}$. For every $q \in Q_+$, there exists at least one pair $\tuple {m, n} \in \N \times \N$ such that $q = \dfrac m n$. Therefore, we can find an injection $i: Q_+ \to \N \times \N$. By Cartesian Product of Natural Numbers with Itself is Countable, $\N \times \N$ is countable. Hence $Q_+$ is countable, by Domain of Injection to Countable Set is Countable. The map $-: q \mapsto -q$ provides a bijection from $Q_-$ to $Q_+$, hence $Q_-$ is also countable. Hence $\Q$ is countable. {{qed}}
From Real Numbers form Field, we have that $\struct {\R, +, \times}$ forms a field. From Ordering Properties of Real Numbers we have that $\struct {\R, +, \times, \le}$ is a ordered field. {{qed}}
Recall that Rational Numbers form Ordered Field. Then from Rational Numbers form Subset of Real Numbers: :$\Q \subseteq \R$ Hence the result by definition of subfield. {{qed}}
=== Necessary Condition === Assume $x < y$. Then: {{begin-eqn}} {{eqn | l = x < y       | o = \implies       | r = x \times x < x \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | l = x < y       | o = \implies       | r = x \times y < y \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = \leadsto       | r = x^2 < y^2       | c = Real Number Ordering is Transitive }} {{end-eqn}} So: :$x < y \implies x^2 < y^2$ {{qed|lemma}} === Sufficient Condition === Assume $x^2 < y^2$. Then: {{begin-eqn}} {{eqn | l = x^2       | o = <       | r = y^2       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = y^2 - x^2       | c = Real Number Ordering is Compatible with Addition }} {{eqn | ll=\leadsto       | l = \paren {y - x} \paren {y + x}       | o = >       | r = 0       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = \paren {y - x} \paren {y + x} \paren {y + x}^{-1}       | o = >       | r = 0 \times \paren {y + x}^{-1}       | c = as $x + y > 0$ }} {{eqn | ll= \leadsto       | l = y - x       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = <       | r = y       | c =  }} {{end-eqn}} So: :$x^2 < y^2 \implies x < y$ {{qed}}
From Real Numbers form Ordered Field, the real numbers form an ordered field. The result follows from Order of Squares in Ordered Field. {{qed}}
From Real Numbers form Ordered Field, the real numbers form an ordered field. By definition, an ordered field is a totally ordered ring without proper zero divisors. The result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors. {{qed}}
=== Necessary Condition === Let $x < y$. Then: {{begin-eqn}} {{eqn | l = x < y       | o = \implies       | r = x \times x < x \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | l = x < y       | o = \implies       | r = x \times y < y \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = \leadsto       | r = x^2 < y^2       | c = Real Number Ordering is Transitive }} {{end-eqn}} So: :$x < y \implies x^2 < y^2$ {{qed|lemma}} === Sufficient Condition === Let $x^2 < y^2$. {{AimForCont}} $x \ge y$. Then: {{begin-eqn}} {{eqn | l = x \ge y       | o = \implies       | r = x \times x \ge x \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | l = x \ge y       | o = \implies       | r = x \times y \ge y \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = \leadsto       | r = x^2 \ge y^2       | c = Real Number Ordering is Transitive }} {{end-eqn}} But this contradicts our assertion that $x^2 < y^2$. Hence by Proof by Contradiction it follows that: :$x < y$ {{qed}}
{{AimForCont}} $a > b$. Then: :$a - b > 0$ By hypothesis, we have: :$\forall \epsilon > 0: a < b + \epsilon$ Let $\epsilon = a - b$. Then: :$a < b + \paren {a - b} \implies a < a$ The result follows by Proof by Contradiction. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {x + y} \circ \paren {x + \paren {-y} }       | r = x \circ x + y \circ x + x \circ \paren {-y} + y \circ \paren {-y}       | c = Distributivity of $\circ$ over $+$ in a ring }} {{eqn | r = x \circ x + x \circ y + x \circ \paren {-y} + y \circ \paren {-y}       | c = $R$ is a commutative ring }} {{eqn | r = x \circ x + x \circ \paren {y + \paren {-y} } + \paren {-\paren {y \circ y} }       | c = various ring properties }} {{eqn | r = x \circ x + x \circ 0_R + \paren {-\paren {y \circ y} }       | c =  }} {{eqn | r = x \circ x + \paren {-\paren {y \circ y} }       | c =  }} {{end-eqn}} {{qed}}
This is a special case of Difference of Two Powers: :$\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \cdots + a b^{n - 2} + b^{n - 1} }\right) = \left({a - b}\right) \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ The result follows by setting $n = 2$. {{qed}}
{{:Euclid:Proposition/II/5}} :400px Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$. Then the rectangle contained by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$. (That is, let $x = AC, y = CD$. Then $\paren {x + y} \paren {x - y} + y^2 = x^2$.) This is proved as follows. Construct the square $CBFE$ on $CB$, and join $BE$. Construct $DG$ parallel to $CE$ through $G$, and let $DG$ cross $BE$ at $H$. Construct $KM$ parallel to $AB$ through $H$. Construct $AK$ parallel to $BF$ through $A$. From Complements of Parallelograms are Equal: :$\Box CDHL = \Box FGHM$. Add the square $DBMH$ to each. Then $\Box CBML = \Box DBFG$. But as $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that: :$\Box ACLK = \Box CBML$ Add $\Box CDHL$ to each. Then $\Box ADHK$ is equal in area to the gnomon $CBFGHL$. But $\Box ADHK$ is the rectangle contained by $AD$ and $DB$, because $DB = DH$. So the gnomon $CBFGHL$ is equal in area to the rectangle contained by $AD$ and $DB$. Now $\Box LHGE$ is equal to the square on $CD$. Add $\Box LHGE$ to each of the gnomon $CBFGHL$ and $\Box ADHK$. Then the gnomon $CBFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $DB$ and the square on $CD$. But the gnomon $CBFGHL$ together with $\Box LHGE$ is the square $CBFE$. Hence the result. {{qed}}
Let $r, s \in \R$ be such that: :$r < s$ and: :$r = \dfrac a b, s = \dfrac c d$ where $a, b, c, d$ are real numbers such that $b > 0, d > 0$. Because $b, d > 0$, it follows from Real Number Ordering is Compatible with Multiplication that: :$b d > 0$ Thus: {{begin-eqn}} {{eqn | l = \frac a b       | o = <       | r = \frac c d       | c =  }} {{eqn | ll= \leadsto       | l = \frac a b \paren {b d}       | o = <       | r = \frac c d \paren {b d}       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = a d       | o = <       | r = b c       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = a \paren {b + d}       | r = a b + a d       | c =  }} {{eqn | o = <       | r = a b + b c       | c = Real Number Ordering is Compatible with Addition, as $a d < b c$ from above }} {{eqn | r = \paren {a + c} b       | c =  }} {{end-eqn}} From Reciprocal of Strictly Positive Real Number is Strictly Positive: :$\paren {a + c}^{-1} > 0$ and: :$b^{-1} > 0$ It follows from Real Number Ordering is Compatible with Multiplication that: :$\dfrac a b < \dfrac {a + c} {b + d}$ The other half is proved similarly. {{Qed}}
From Real Plus Epsilon: :$b < a + \epsilon \implies b \le a$ From Real Number Ordering is Compatible with Addition: :$a - \epsilon < b \implies a < b + \epsilon$ Then from Real Plus Epsilon: :$a < b + \epsilon \implies a \le b$ The result follows. {{qed}}
Clearly we may assume without loss of generality that $x < 0$. From the Archimedean Principle: :$\exists m \in \N: m > -x$ By Real Numbers form Ordered Field, we have that $\R$ is an ordered field. Therefore by property $(3)$ of Properties of Ordered Field, $\Z \owns -m < x$. {{qed}} Category:Analysis 279b5qb10xgfmugmysigl3ksxrndsf3
Let $x \in \R$. Let $S$ be the set of all natural numbers less than or equal to $x$: :$S = \set {a \in \N: a \le x}$ It is possible that $S = \O$. Suppose $0 \le x$. Then by definition, $0 \in S$. But $S = \O$, so this is a contradiction. From the Trichotomy Law for Real Numbers it follows that $0 > x$. Thus we have the element $0 \in \N$ such that $0 > x$. Now suppose $S \ne \O$. Then $S$ is bounded above (by $x$, for example). Thus by the Continuum Property of $\R$, $S$ has a supremum in $\R$. Let $s = \map \sup S$. Now consider the number $s - 1$. Since $s$ is the supremum of $S$, $s - 1$ cannot be an upper bound of $S$ by definition. So $\exists m \in S: m > s - 1 \implies m + 1 > s$. But as $m \in \N$, it follows that $m + 1 \in \N$. Because $m + 1 > s$, it follows that $m + 1 \notin S$ and so $m + 1 > x$.
$\floor x$ is defined as: :$\floor x = \sup \set {m \in \Z: m \le x}$ So $\floor x \le x$ by definition. From Floor plus One: :$\floor {x + 1} > \floor x$ Hence by the definition of the supremum: :$\floor {x + 1} > x$ The result follows. {{qed}} Category:Floor Function akn94fzyqk42jhnqcxjm9xqmvzfon8w
$\left \lceil {x} \right \rceil$ is defined as: :$\left \lceil {x} \right \rceil = \inf \left({\left\{{m \in \Z: m \ge x}\right\}}\right)$ So $\left \lceil {x} \right \rceil \ge x$ by definition. Now $\left \lceil {x - 1} \right \rceil < \left \lceil {x} \right \rceil$, so by the definition of the infimum: :$\left \lceil {x - 1} \right \rceil > x$ The result follows. {{qed}} Category:Ceiling Function d8pg5fzw8p22asnvcvmbkt769zo7dcs
{{begin-eqn}} {{eqn | l = \left \lceil {x} \right \rceil - 1       | o = <       | r = x \le \left \lceil {x} \right \rceil       | c = Real Number is between Ceiling Functions }} {{eqn | ll= \implies       | l = \left \lceil {x} \right \rceil - 1 - \left \lceil {x} \right \rceil       | o = <       | r = x - \left \lceil {x} \right \rceil \le \left \lceil {x} \right \rceil - \left \lceil {x} \right \rceil       | c =  }} {{eqn | ll= \implies       | l = -1       | o = <       | r = x - \left \lceil {x} \right \rceil \le 0       | c =  }} {{eqn | ll= \implies       | l = 1       | o = >       | r = \left \lceil {x} \right \rceil - x \ge 0       | c =  }} {{eqn | ll= \implies       | l = \left \lceil {x} \right \rceil - x       | o = \in       | r = \left[{0 \,.\,.\, 1}\right)       | c =  }} {{end-eqn}} {{qed}} Category:Ceiling Function f6xkmvsorteylqbus2ipuydihvfmd4a
=== Sufficient Condition === Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$. We have that $1 - t > 0$. Thus: :$0 \le x - n < 1$ Thus: : $n \le x < n + 1$ That is, $n$ is the floor of $x$. {{qed|lemma}} === Necessary Condition === Let $n = \floor x$. Let $t = x - \floor x$. Then $x = n + t$. From Real Number minus Floor: :$t = x - \floor x \in \hointr 0 1$ and so: :$x = n + t: t \in \hointr 0 1$ {{qed}}
{{begin-eqn}} {{eqn | l = \left \lfloor {x + 1} \right \rfloor       | r = n       | c =  }} {{eqn | ll= \implies       | l = n       | o = \le       | r = x + 1 < n + 1       | c = Definition of Floor Function }} {{eqn | ll= \implies       | l = n - 1       | o = \le       | r = x < n       | c =  }} {{eqn | ll= \implies       | l = \left \lfloor {x} \right \rfloor       | r = n - 1       | c = Definition of Floor Function }} {{eqn | ll= \implies       | l = \left \lfloor {x + 1} \right \rfloor       | r = \left \lfloor {x} \right \rfloor + 1       | c =  }} {{end-eqn}} {{qed}}
Let $x \in \Z$. Then from Real Number is Integer iff equals Floor: :$x = \floor x$ Now $x \in \Z \implies -x \in \Z$, so: :$\floor {-x} = -x$ Thus: :$\floor x + \floor {-x} = x + \paren {-x} = x - x = 0$ Now let $x \notin \Z$. From Real Number is Floor plus Difference: :$x = n + t$ where $n = \floor x$ and $t \in \hointr 0 1$. Thus: :$-x = -\paren {n + t} = -n - t = -n - 1 + \paren {1 - t}$ As $t \in \hointr 0 1$, we have: :$1 - t \in \hointr 0 1$ Thus: :$\floor {-x} = -n - 1$ So: :$\floor x + \floor {-x} = n + \paren {-n - 1} = n - n - 1 = -1$ {{qed}}
Checking in turn each of the critera for equivalence: === Reflexivity === :$\forall x \in \R: \floor x = \floor x$ Thus the floor function is reflexive. {{qed|lemma}} === Symmetry === :$\forall x, y \in \R: \floor x = \floor y \implies \floor y = \floor x$ Thus the floor function is symmetric. {{qed|lemma}} === Transitivity === Let $\floor x = \floor y$ and $\floor y = \floor z$. Let $n = \floor x = \floor y = \floor z$, which follows from transitivity of $=$. Thus from Real Number is Floor plus Difference: :$x = n + t_x, y = n + t_y, z = n + t_z: t_x, t_y, t_z \in \hointr 0 1$ Thus: :$x = n + t_x, z = n + t_z$ and: :$\floor x = \floor z$ Thus the floor function is transitive. {{qed|lemma}} Thus we have shown that $\RR$ is an equivalence relation. {{qed|lemma}} Now we show that the $\RR$-class of $n$ is the interval $\hointr n {n + 1}$. Defining $\RR$ as above, with $n \in \Z$: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \eqclass n \RR       | c =  }} {{eqn | ll= \leadsto       | l = \floor x       | r = \floor n       | c =  }} {{eqn | r = n       | c =  }} {{eqn | ll= \leadsto       | l = \exists t \in \hointr 0 1: x       | r = n + t       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \hointr n {n + 1}       | c =  }} {{end-eqn}} {{qed}}
Checking in turn each of the critera for equivalence: === Reflexivity === :$\forall x \in \R: \left \lceil {x}\right \rceil = \left \lceil {x}\right \rceil$ Thus the ceiling function is reflexive. {{qed|lemma}} === Symmetry === :$\forall x, y \in \R: \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil \implies \left \lceil {y}\right \rceil = \left \lceil {x}\right \rceil$ Thus the ceiling function is symmetric. {{qed|lemma}} === Transitivity === Let: :$\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$ :$\left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$ Let: :$n = \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$ which follows from transitivity of $=$. Thus: :$x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \left[{0 \,.\,.\, 1}\right)$ from Real Number is Ceiling minus Difference. So: :$x = n - t_x$ and: :$z = n - t_z$ and so: :$\left \lceil {x}\right \rceil = \left \lceil {z}\right \rceil$ Thus the ceiling function is transitive. {{qed|lemma}} Thus we have shown that $\mathcal R$ is an equivalence. Now we show that the $\mathcal R$-class of $n$ is the interval $\left({n - 1 \,.\,.\, n}\right]$. Defining $\mathcal R$ as above, with $n \in \Z$: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \left[\!\left[{n}\right]\!\right]_\mathcal R       | c =  }} {{eqn | ll= \implies       | l = \left \lceil {x}\right \rceil       | r = \left \lceil {n}\right \rceil = n       | c =  }} {{eqn | ll= \implies       | l = \exists t \in \left({0 \,.\,.\, 1}\right]: x       | r = n - t       | c =  }} {{eqn | ll= \implies       | l = x       | o = \in       | r = \left({n - 1 \,.\,.\, n}\right]       | c =  }} {{end-eqn}} {{qed}} Category:Ceiling Function Category:Equivalence Relations kktdzbepmt2otz9fgu475n43d93twsn
=== 1 implies 2 === Let $x = n - t$, where $t \in \hointr 0 1$. Because $0 \le t < 1$, we have: :$0 \leq n - x < 1$ Thus: :$n - 1 < x \le  n$ That is, $n$ is the ceiling of $x$. {{qed|lemma}} === 2 implies 1 === Now let $n = \ceiling x$. Let $t = \ceiling x - x$. Then $x = n - t$. From Ceiling minus Real Number, $t = \ceiling x - x \in \hointr 0 1$ {{qed}}
For any $\lambda \in \R$, we define $f: \R \to \R$ as the function: :$\displaystyle \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$ Now: :$\map f \lambda \ge 0$ because it is the sum of squares of real numbers. Hence: {{begin-eqn}} {{eqn | l = \forall \lambda \in \R: \map f \lambda       | o = \equiv       | r = \sum {\paren {r_i^2 + 2 \lambda r_i s_i + \lambda^2 s_i^2} } \ge 0       | c =  }} {{eqn | o = \equiv       | r = \sum {r_i^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum {s_i^2} \ge 0       | c =  }} {{end-eqn}} This is a quadratic equation in $\lambda$. From Solution to Quadratic Equation: :$\displaystyle a \lambda^2 + b \lambda + c = 0: a = \sum {s_i^2}, b = 2 \sum {r_i s_i}, c = \sum {r_i^2}$ The discriminant of this equation (that is $b^2 - 4 a c$) is: :$D := \displaystyle 4 \paren {\sum {r_i s_i} }^2 - 4 \sum {r_i^2} \sum {s_i^2}$ {{AimForCont}} $D$ is (strictly) positive. Then $\map f \lambda = 0$ has two distinct real roots, $\lambda_1 < \lambda_2$, say. From Sign of Quadratic Function Between Roots, it follows that $f$ is (strictly) negative somewhere between $\lambda_1$ and $\lambda_2$. But we have: :$\forall \lambda \in \R: \map f \lambda \ge 0$ From this contradiction it follows that: :$D \le 0$ which is the same thing as saying: :$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \paren {\sum {r_i s_i} }^2$ {{qed}}
From the Complex Number form of the Cauchy-Schwarz Inequality, we have: :$\displaystyle \sum \left|{w_i}\right|^2 \left|{z_i}\right|^2 \ge \left|{\sum w_i z_i}\right|^2$ where all of $w_i, z_i \in \C$. As elements of $\R$ are also elements of $\C$, it follows that: :$\displaystyle \sum \left|{r_i}\right|^2 \left|{s_i}\right|^2 \ge \left|{\sum r_i s_i}\right|^2$ where all of $r_i, s_i \in \R$. But from the definition of modulus, it follows that: :$\displaystyle \forall r_i \in \R: \left|{r_i}\right|^2 = r_i^2$ Thus: :$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$ where all of $r_i, s_i \in \R$. {{qed}}