hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | o =        | r = \begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + p_2}\right) \\ \left({y + q_2}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + p_2}\right) \\ \left({z + q_2}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + p_2}\right) \end{vmatrix}       | c =  }} {{eqn | r = \begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & \left({p_1 - q_2}\right) \left({x + q_3}\right) & \left({p_1 - q_3}\right) \left({x + p_2}\right) \\ \left({y + q_2}\right) \left({y + q_3}\right) & \left({p_1 - q_2}\right) \left({y + q_3}\right) & \left({p_1 - q_3}\right) \left({y + p_2}\right) \\ \left({z + q_2}\right) \left({z + q_3}\right) & \left({p_1 - q_2}\right) \left({z + q_3}\right) & \left({p_1 - q_3}\right) \left({z + p_2}\right) \end{vmatrix}       | c = Multiple of Row Added to Row of Determinant }} {{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & x + q_3 & x + p_2 \\ \left({y + q_2}\right) \left({y + q_3}\right) & y + q_3 & y + p_2 \\ \left({z + q_2}\right) \left({z + q_3}\right) & z + q_3 & z + p_2 \end{vmatrix}       | c = Determinant with Row Multiplied by Constant }} {{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \begin{vmatrix} x \left({x + q_3}\right) & x + q_3 & p_2 - q_3 \\ y \left({y + q_3}\right) & y + q_3 & p_2 - q_3 \\ z \left({z + q_3}\right) & z + q_3 & p_2 - q_3 \end{vmatrix}       | c = Multiple of Row Added to Row of Determinant }} {{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \begin{vmatrix} x \left({x + q_3}\right) & x + q_3 & 1\\ y \left({y + q_3}\right) & y + q_3 & 1\\ z \left({z + q_3}\right) & z + q_3 & 1 \end{vmatrix}       | c = Determinant with Row Multiplied by Constant }} {{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \begin{vmatrix} x^2 & x & 1\\ y^2 & y & 1\\ z^2 & z & 1 \end{vmatrix}       | c = Multiple of Row Added to Row of Determinant }} {{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \left({x - y}\right) \left({y - z}\right) \left({x - z}\right)       | c = Vandermonde Determinant }} {{end-eqn}} {{qed}} {{Namedfor|Christian Friedrich Krattenthaler|cat = Krattenthaler}}
The decimal expansion of $\sqrt 2$ is: :$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$ Thus: :$1 \le \sqrt 2 < 2$ Hence $1$ is the floor of $\sqrt 2$ by definition. {{qed}}
The decimal expansion of $\sqrt 2$ is: :$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$ Thus: :$1 < \sqrt 2 \le 2$ Hence $2$ is the ceiling of $\sqrt 2$ by definition. {{qed}}
We have that: :$0 \le \dfrac 1 2$ and: :$1 = \dfrac 2 2 > \dfrac 1 2$ Hence $0$ is the floor of $\dfrac 1 2$ by definition. {{qed}}
We have: :$0 \ge - \dfrac 1 2$ and: :$-1 = - \dfrac 2 2 < - \dfrac 1 2$ Hence $0$ is the ceiling of $- \dfrac 1 2$ by definition. {{qed}}
We have that: :$-1 = -\dfrac 2 2 \le -\dfrac 1 2$ and: :$0 > -\dfrac 1 2$ Hence $-1$ is the floor of $-\dfrac 1 2$ by definition. {{qed}}
Let: : $-S = \left\{{-s: s \in S}\right\}$ : $-T = \left\{{-t: t \in T}\right\}$ Observe that: : $s \in S \iff -s \in -S$ : $t \in T \iff -t \in -T$ We know that $\inf S$ and $\inf T$ exist. The expression $\inf S \ge \inf T$ exists as $\inf S$ and $\inf T$ exist. In other words, for fixed sets $S$ and $T$, $\inf S \ge \inf T$ is either true or false. We find: {{begin-eqn}} {{eqn | l = \inf S       | o = \ge       | r = \inf T }} {{eqn | ll= \iff       | l = -\sup -S       | o = \ge       | r = -\sup -T       | c = by the lemma }} {{eqn | ll= \iff       | l = \sup -S       | o = \le       | r = \sup -T }} {{eqn | ll= \iff       | l = \forall \epsilon \in \R_{>0}: \forall x \in -S: \exists y \in -T: x       | o = <       | r = y + \epsilon       | c = Suprema of two Real Sets }} {{eqn | ll= \iff       | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \left({x: = -s}\right): \exists t \in T: \left({y: = -t}\right): x       | o = <       | r = y + \epsilon       | c = as $s \in S \iff -s \in −S$ and $t \in T \iff -t \in −T$ }} {{eqn | ll= \iff       | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: -s       | o = <       | r = -t + \epsilon }} {{eqn | ll= \iff       | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon       | o = >       | r = t }} {{end-eqn}} === Lemma === Let $X$ be a set of real numbers. Let $X$ admit an infimum. Let $-X = \left\{{-x: x \in X}\right\}$. Then: : $ \sup -X = -\inf X$ === Proof === Because $X$ admits an infimum, it follows that it is not empty. The result follows by Negative of Infimum is Supremum of Negatives. {{qed|lemma}} {{qed}}
By definition of floor function: :$\left\lfloor{x}\right\rfloor \le x < \left\lfloor{x}\right\rfloor + 1$ Thus by subtracting $1$: :$x - 1 < \left({\left\lfloor{x}\right\rfloor + 1}\right) - 1 = \left\lfloor{x}\right\rfloor$ So: :$\left\lfloor{x}\right\rfloor \le x$ and: :$x - 1 < \left\lfloor{x}\right\rfloor$ as required. {{qed}}
=== Definition 1 equals Definition 2 === Follows from Supremum of Set of Integers equals Greatest Element. {{qed|lemma}} === Definition 1 equals Definition 3 === Let $S$ be the set: :$S = \set {m \in \Z: m \le x}$ Let $n = \sup S$. By Supremum of Set of Integers is Integer, $n \in \Z$. By Supremum of Set of Integers equals Greatest Element, $n\in S$. Because $n \in S$, we have $n \le x$. Because $n + 1 > n$, we have by definition of supremum: :$n + 1 \notin S$ Thus $n + 1 > x$. Thus $n$ is an integer such that: :$n \le x < n + 1$ So $n$ is the floor function by definition 3. {{qed|lemma}} === Definition 3 equals Definition 2 === Let $n$ be an integer such that: :$n \le x < n + 1$ We show that $n$ is the greatest element of the set: :$S = \set {m \in \Z: m \le x}$ Let $m \in \Z$ such that $m \le x$. We show that $n \ge m$. {{AimForCont}} $m > n$. By Weak Inequality of Integers iff Strict Inequality with Integer plus One: :$m \ge n + 1$ and so from the definition of $g$ it follows that $m > x$. By Proof by Contradiction it follows that $m \le n$. Because $m \in S$ was arbitrary, $n$ is the greatest element of $S$. Thus $n$ is the floor function by definition 2. {{qed}}
Let $y = \ceiling x$. By Ceiling Function is Integer, $y$ is an integer. Then from Real Number is Integer iff equals Ceiling: :$\ceiling y = y$ So: :$\ceiling {\ceiling x} = \ceiling x$ {{qed}}
By Way Below implies Preceding: :$\forall x, y \in S: x \ll y \implies x \preceq y$ By Preceding and Way Below implies Way Below: :$\forall x, y, z, u \in S: x \preceq y \ll z \preceq u \implies x \ll u$ By Join is Way Below if Operands are Way Below: :$\forall x, y, z \in S: x \ll z \land y \ll z \implies x \vee y \ll z$ By Bottom is Way Below Any Element: :$\forall x: \bot \ll x$ where $\bot$ denotes the smallest element in $L$. Thus by definition: :$\ll$ is auxiliary relation. {{qed}}
Fix $x \in \R$. Then: {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = \ceiling {\size x} }} {{eqn | ll= \leadsto       | l = n       | o = >       | r = -x       | c = Real Number is between Ceiling Functions and Negative of Absolute Value }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = \frac {-x} n       | c = dividing both sides by $n$ }} {{eqn | ll= \leadsto       | l = 1 + \frac x n       | o = >       | r = 0       | c = adding $\dfrac {-x} n$ to both sides }} {{eqn | ll= \leadsto       | l = \paren {1 + \frac x n}^n       | o = >       | r = 0       | c = Power of Strictly Positive Real Number is Strictly Positive: Positive Integer }} {{end-eqn}} {{qed}} Category:Exponential Function 2f98hvb2ixed1gcy50znmnpjs04sny9
By definition of ceiling of $x$: :$\forall x \in \R: \ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$ By definition of infimum: :$\ceiling x \ge x$ Also by definition of infimum: :$\ceiling x - 1 \not \ge x$ as $\ceiling x$ is the smallest integer with that property. That is: :$x > \ceiling x - 1$ Hence the result. {{qed}} Category:Ceiling Function l75iq9rcega0z865kapwqgmw7httq2r
From Number is between Ceiling and One Less: :$\ceiling x - 1 < x \le \ceiling x$ Thus by adding $1$: :$x + 1 > \paren {\ceiling x - 1} + 1 = \ceiling x$ So: :$x \le \ceiling x$ and: :$\ceiling x < x + 1$ as required. {{qed}}
{{begin-eqn}} {{eqn | l = \left \lceil {x + n} \right \rceil - 1       | o = <       | m = x + n        | mo= \le       | r = \left \lceil {x + n} \right \rceil       | c = Number is between Ceiling and One Less }} {{eqn | ll= \implies       | l = \left \lceil {x + n} \right \rceil - n - 1       | o = <       | m = x       | mo= \le       | r = \left \lceil {x + n} \right \rceil - n       | c =  }} {{eqn | ll= \implies       | l = \left \lceil {x} \right \rceil       | r = \left \lceil {x + n} \right \rceil - n       | c = Number is between Ceiling and One Less }} {{eqn | ll= \implies       | l = \left \lceil {x + n} \right \rceil       | r = \left \lceil {x} \right \rceil + n       | c = adding $n$ to both sides }} {{end-eqn}} {{qed}}
=== Non-empty === By definition of auxiliary relation: :$\left({\bot, x}\right) \in R$ where $\bot$ denotes the smallest element in $L$. By definition of relation segment: :$\bot \in x^R$ Thus by definition: :$x^R$ is non-empty. {{qed|lemma}} === Directed === Let $y, z \in x^R$. By definition of relation segment: :$\left({y, x}\right) \in R$ and $\left({z, x}\right) \in R$ By definition of auxiliary relation: :$\left({y \vee z, x}\right) \in R$ By definition of relation segment: :$y \vee z \in x^R$ Thus by Join Succeeds Operands: :$y \preceq y \vee z$ and $z \preceq y \vee z$ Thus by definition: :$x^R$ is directed. {{qed|lemma}} === Lower === Let $y \in x^R, z \in S$ such that :$z \preceq y$ By definition of relation segment: :$\left({y, x}\right) \in R$ By definition of reflexivity: :$x \preceq x$ By definition of auxiliary relation: :$\left({z, x}\right) \in R$ Thus by definition of relation segment: :$z \in x^R$ Thus by definition: :$x^R$ is lower set. {{qed|lemma}} Thus by definition: :$x^R$ is ideal in $L$. {{qed}}
From Real Number minus Floor: :$\dfrac x y - \left \lfloor {\dfrac x y}\right \rfloor \in \left[{0 \,.\,.\, 1}\right)$ Thus by definition of half-open real interval: :$0 \le \dfrac x y - \left \lfloor {\dfrac x y}\right \rfloor < 1$ Then: {{begin-eqn}} {{eqn | l = x \bmod y       | r = x - y \left \lfloor {\frac x y}\right \rfloor       | c = Definition of Modulo Operation }} {{eqn | ll= \leadsto       | l = \frac {x \bmod y} y       | r = \frac {x - y \left \lfloor {\dfrac x y}\right \rfloor} y       | c =  }} {{eqn | r = \frac x y - \frac y y \left \lfloor {\dfrac x y}\right \rfloor       | c =  }} {{eqn | r = \frac x y - \left \lfloor {\dfrac x y}\right \rfloor       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = 0       | o = \le       | m = \frac {x \bmod y} y       | mo= <       | r = 1       | c = Quotient of Modulo Operation with Modulus }} {{eqn | ll= \leadsto       | l = 0       | o = \le       | m = \frac {x \bmod y} y \times y       | mo= <       | r = 1 \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = 0       | o = \le       | m = x \bmod y       | mo= <       | r = y       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = 0       | o = \le       | m = \frac {x \bmod y} y       | mo= <       | r = 1       | c = Quotient of Modulo Operation with Modulus }} {{eqn | ll= \leadsto       | l = 0       | o = \ge       | m = \frac {x \bmod y} y \times y       | mo= >       | r = 1 \times y       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = 0       | o = \ge       | m = x \bmod y       | mo= >       | r = y       | c =  }} {{end-eqn}} Hence the result. {{qed}}
We have that: :$0 \times 0 = 0$ The result follows by definition of integer multiple. {{qed}} Category:Number Theory cr4qeo16keg2mqnhx82y0t2xypme5gz
When $y = 0$ we have: :$x \bmod y := x$ Thus: :$x - \left({x \bmod y}\right) = 0$ From Zero is Integer Multiple of Zero it follows that: :$x - \left({x \bmod y}\right)$ is an integer multiple of $y$. Let $y \ne 0$. Then: {{begin-eqn}} {{eqn | l = x \bmod y       | r = x - y \left \lfloor {\dfrac x y}\right \rfloor       | c = Definition of Modulo Operation }} {{eqn | ll= \leadsto       | l = x - \left({x \bmod y}\right)       | r = y \left \lfloor {\dfrac x y}\right \rfloor       | c = Definition of Modulo Operation }} {{end-eqn}} From Floor Function is Integer, $\left \lfloor {\dfrac x y}\right \rfloor$ is an integer. Thus: :$\exists n \in \Z: x - \left({x \bmod y}\right) = n y$ where in the case $n = \left \lfloor {\dfrac x y}\right \rfloor$ Hence the result by definition of integer multiple. {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac 5 3 = 1 + \dfrac 2 3$ and so: :$\left\lfloor{\dfrac 5 3}\right\rfloor = 1$ Thus: :$5 \bmod 3 = 5 - 3 \times \left\lfloor{\dfrac 5 3}\right\rfloor = 5 - 3 \times 1 = 2$ {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {18} 3 = 6 + \dfrac 0 3$ and so: :$\left\lfloor{\dfrac {18} 3}\right\rfloor = 6$ Thus: :$18 \bmod 3 = 18 - 3 \times \left\lfloor{\dfrac {18} 3}\right\rfloor = 18 - 3 \times 6 = 0$ {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {-2} 3 = -1 + \dfrac 1 3$ and so: :$\left\lfloor{\dfrac {-2} 3}\right\rfloor = -1$ Thus: :$-2 \bmod 3 = -2 - 3 \times \left\lfloor{\dfrac {-2} 3}\right\rfloor = -2 - 3 \times \left({-1}\right) = 1$ {{qed}}
=== Reflexivity === Let $f \in F$. By definition of reflexivity: :$\forall x \in S: f\left({x}\right) \precsim f\left({x}\right)$ Thus by definition of ordering on mappings: :$f \preccurlyeq f$ {{qed|lemma}} === Transitivity === Let $f, g, h \in F$ such that :$f \preccurlyeq g \preccurlyeq h$ By definition of ordering on mappings: :$\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \precsim h\left({x}\right)$ By definition of transitivity: :$\forall x \in S: f\left({x}\right) \precsim h\left({x}\right)$ Thus by definition of ordering on mappings: :$f \preccurlyeq h$ {{qed|lemma}} === Antisymmetry === Let $f, g \in F$ such that :$f \preccurlyeq g$ and $g \preccurlyeq f$ By definition of ordering on mappings: :$\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \land g\left({x}\right) \precsim f\left({x}\right)$ By definition of antisymmetry: :$\forall x \in S: f\left({x}\right) = g\left({x}\right)$ Thus by equality of mappings: :$f = g$ {{qed|lemma}} Thus by definition: :$M$ is an ordered set. {{qed}}
=== Necessary Condition === Let $x < n$. By definition of the floor of $x$: :$\left \lfloor {x} \right \rfloor \le x$ Hence: :$\left \lfloor {x} \right \rfloor < n$ {{qed|lemma}} === Sufficient Condition === Let $\left \lfloor{x}\right \rfloor < n$. We have that: :$\forall m, n \in \Z: m < n \iff m + 1 \le n$ and so: :$(1): \quad \left \lfloor{x}\right \rfloor + 1 \le n$ Then: {{begin-eqn}} {{eqn | l = x       | o = <       | r = \left \lfloor{x}\right \rfloor + 1       | c = Definition of Floor Function }} {{eqn | o = \le       | r = n       | c = from $(1)$ }} {{end-eqn}} {{qed|lemma}} Hence the result: :$\left \lfloor{x}\right \rfloor < n \iff x < n$ {{qed}}
=== Necessary Condition === Let $\left \lfloor{x}\right \rfloor \ge n$. By definition of the floor of $x$: :$x \ge \left \lfloor{x}\right \rfloor$ Hence: :$x \ge n$ {{qed|lemma}} === Sufficient Condition === Let $x \ge n$. {{AimForCont}} $\left \lfloor{x}\right \rfloor < n$. We have that: :$\forall m, n \in \Z: m < n \iff m + 1 \le n$ Hence: :$\left \lfloor{x}\right \rfloor + 1 \le n$ and so by hypothesis: :$\left \lfloor{x}\right \rfloor + 1 \le x$ This contradicts the definition of the floor of $x$: :$\left \lfloor{x}\right \rfloor + 1 > x$ Thus by Proof by Contradiction: :$\left \lfloor{x}\right \rfloor \ge n$ {{qed|lemma}} Hence the result: :$\left \lfloor{x}\right \rfloor \ge n \iff x \ge n$ {{qed}}
=== Necessary Condition === Let $\left \lceil{x}\right \rceil \le n$. By Number is between Ceiling and One Less: :$x \le \left \lceil{x}\right \rceil$ Hence: :$x \le n$ {{qed|lemma}} === Sufficient Condition === Let $x \le n$. {{AimForCont}} $\left \lceil{x}\right \rceil > n$. We have that: :$\forall m, n \in \Z: m < n \iff m \le n - 1$ Hence: :$\left \lceil{x}\right \rceil - 1 \ge n$ and so by hypothesis: :$\left \lceil{x}\right \rceil - 1 \ge x$ This contradicts the result Number is between Ceiling and One Less: :$\left \lceil{x}\right \rceil - 1 < x$ Thus by Proof by Contradiction: :$\left \lceil{x}\right \rceil \le n$ {{qed|lemma}} Hence the result: :$\left \lceil{x}\right \rceil \le n \iff x \le n$ {{qed}}
=== Necessary Condition === Let $x > n$. By Number is between Ceiling and One Less: :$\left \lceil{x}\right \rceil \ge x$ Hence: :$\left \lceil{x}\right \rceil > n$ {{qed|lemma}} === Sufficient Condition === Let $\left \lceil{x}\right \rceil > n$. We have that: :$\forall m, n \in \Z: m < n \iff m \le n - 1$ and so: :$(1): \quad \left \lceil{x}\right \rceil - 1 \ge n$ Then: {{begin-eqn}} {{eqn | l = x       | o = >       | r = \left \lceil{x}\right \rceil - 1       | c = Number is between Ceiling and One Less }} {{eqn | o = \ge       | r = n       | c = from $(1)$ }} {{end-eqn}} {{qed|lemma}} Hence the result: :$\left \lceil{x}\right \rceil > n \iff x > n$ {{qed}}
=== Necessary Condition === Let $x - 1 < n \le x$. From $n \le x$, we have by Number not less than Integer iff Floor not less than Integer: :$n \le \floor x$ From $x - 1 < n$: :$x < n + 1$ Hence by Number less than Integer iff Floor less than Integer: :$\floor x < n + 1$ We have that: :$\forall m, n \in \Z: m \le n \iff m < n + 1$ and so: :$\floor x \le n$ Thus as: :$n \le \floor x$ and: :$\floor x \le n$ it follows that: :$\floor x = n$ {{qed|lemma}} === Sufficient Condition === Let $\floor x = n$. Then: :$\floor x \ge n$ By Number not less than Integer iff Floor not less than Integer: :$n \le x$ By definition of the floor of $x$: :$x < \floor x + 1$ and so subtracting $1$ from both sides: :$x - 1 < \floor x$ and so by hypothesis: :$x - 1 < n$ So: :$\floor x = n \implies x - 1 < n \le x$ {{qed|lemma}} Hence the result: :$\floor x = n \iff x - 1 < n \le x$ {{qed}}
=== Necessary Condition === Let $n \le x < n + 1$. From Number not less than Integer iff Floor not less than Integer: :$n \le x \implies n \le \floor x$ By definition of the floor of $x$: :$\floor x \le x$ and so by hypothesis: :$\floor x < n + 1$ We have that: :$\forall m, n \in \Z: m \le n \iff m < n + 1$ and so: :$\floor x \le n$ So we have: :$n \le \floor x$ and: :$\floor x \le n$ Thus: :$n \le x < n + 1 \implies \floor x = n$ {{qed|lemma}} === Sufficient Condition === Let $\floor x = n$. By definition of the floor of $x$: :$\floor x \le x$ and so by hypothesis: :$n \le x$ Also by definition of the floor of $x$: :$x < \floor x + 1$ and so by hypothesis: :$x < n + 1$ Thus: :$\floor x = n \implies n \le x < n + 1$. {{qed|lemma}} Hence the result: :$\floor x = n \iff n \le x < n + 1$ {{qed}}
=== Necessary Condition === Let $x \le n < x + 1$. From $x \le n$, we have by Number not greater than Integer iff Ceiling not greater than Integer: :$\left \lceil{x}\right \rceil \le n$ From $n < x + 1$: :$n - 1 < x$ Hence by Number greater than Integer iff Ceiling greater than Integer: :$n - 1 < \left \lceil{x}\right \rceil$ We have that: :$\forall m, n \in \Z: m < n \iff m \le n - 1$ and so: :$n \le \left \lceil{x}\right \rceil$ Thus as: :$\left \lceil{x}\right \rceil \le n$ and: :$\left \lceil{x}\right \rceil \ge n$ it follows that: :$\left \lceil{x}\right \rceil = n$ {{qed|lemma}} === Sufficient Condition === Let $\left \lceil{x}\right \rceil = n$. Then: :$\left \lceil{x}\right \rceil \le n$ By Number not greater than Integer iff Ceiling not greater than Integer: :$x \le n$ From Number is between Ceiling and One Less: :$\left \lceil{x}\right \rceil - 1 < x$ and so adding $1$ to both sides: :$\left \lceil{x}\right \rceil < x + 1$ and so by hypothesis: :$n < x + 1$ So: :$\left \lceil{x}\right \rceil = n \implies x \le n < x + 1$ {{qed|lemma}} Hence the result: :$\left \lceil{x}\right \rceil = n \iff x \le n < x + 1$ {{qed}}
=== Necessary Condition === Let $n - 1 < x \le n$. From Number not greater than Integer iff Ceiling not greater than Integer: :$x \le n \implies \ceiling x \le n$ From Number is between Ceiling and One Less: :$x \le \ceiling x$ and so: :$n - 1 < \ceiling x$ We have that: :$\forall m, n \in \Z: m - 1 < n \iff m \le n$ and so: :$n \le \ceiling x$ So we have: :$\ceiling x \le n$ and: :$n \le \ceiling x$ Thus: :$n - 1 < x \le n \implies \ceiling x = n$ {{qed|lemma}} === Sufficient Condition === Let $\ceiling x = n$. From Number is between Ceiling and One Less: :$x \le \ceiling x$ and so: :$x \le n$ Also from Number is between Ceiling and One Less: :$\ceiling x - 1 < x$ and so: :$n - 1 < x$ Thus: :$\ceiling x = n \implies n - 1 < x \le n$. {{qed|lemma}} Hence the result: :$\ceiling x = n \iff n - 1 < x \le n$ {{qed}}
{{begin-eqn}} {{eqn | l = n       | m = \floor {\sqrt x}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = n       | o = \le       | m = \sqrt x       | mo= <       | r = n + 1       | c = Integer equals Floor iff Number between Integer and One More }} {{eqn | ll= \leadstoandfrom       | l = n^2       | o = \le       | m = x       | mo= <       | r = \paren {n + 1}^2       | c = Order is Preserved on Positive Reals by Squaring }} {{eqn | ll= \leadstoandfrom       | l = n^2       | o = \le       | m = \floor x       | mo= <       | r = \paren {n + 1}^2       | c = Number not less than Integer iff Floor not less than Integer }} {{eqn | ll= \leadstoandfrom       | l = n       | o = \le       | m = \sqrt {\floor x}       | mo= <       | r = n + 1       | c = Order is Preserved on Positive Reals by Squaring }} {{eqn | ll= \leadstoandfrom       | l = n       | m = \floor {\sqrt {\floor x} }       | c = Integer equals Floor iff Number between Integer and One More }} {{end-eqn}} {{qed}} {{expand|Add a proof based on McEliece's Theorem (Integer Functions)}}
{{begin-eqn}} {{eqn | l = n       | m = \ceiling {\sqrt x}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = n - 1       | o = <       | m = \sqrt x       | mo= \le       | r = n       | c = Integer equals Ceiling iff Number between Integer and One Less }} {{eqn | ll= \leadstoandfrom       | l = \paren {n - 1}^2       | o = <       | m = x       | mo= \le       | r = n^2       | c = Order is Preserved on Positive Reals by Squaring }} {{eqn | ll= \leadstoandfrom       | l = \paren {n - 1}^2       | o = <       | m = \ceiling x       | mo= \le       | r = n^2       | c = Number not greater than Integer iff Ceiling not greater than Integer }} {{eqn | ll= \leadstoandfrom       | l = n - 1       | o = <       | m = \sqrt {\ceiling x}       | mo= \le       | r = n       | c = Order is Preserved on Positive Reals by Squaring }} {{eqn | ll= \leadstoandfrom       | l = n       | m = \ceiling {\sqrt {\ceiling x} }       | c = Integer equals Ceiling iff Number between Integer and One Less }} {{end-eqn}} {{qed}} {{expand|Add a proof based on McEliece's Theorem (Integer Functions)}}
{{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod {r s}       | c =  }} {{eqn | ll= \implies       | l = a - b       | r = q r s       | c = {{Defof|Congruence Modulo Integer}} }} {{eqn | ll= \implies       | l = a - b       | r = \paren {q r} s       | c =  }} {{eqn | lo= \land       | l = a - b       | r = \paren {q s} r       | c =  }} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod r       | c = {{Defof|Congruence Modulo Integer}}: $q s$ is an integer }} {{eqn | lo= \land       | l = a       | o = \equiv       | r = b       | rr= \pmod s       | c = {{Defof|Congruence Modulo Integer}}: $q r$ is an integer }} {{end-eqn}} {{qed}}