hcju/proofwikiqa huggingface dataset:
From the definition of the Binomial distribution, $X$ has probability mass function: :$\map \Pr {X = k} = \dbinom n k p^k \paren {1 - p}^{n - k}$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^n \map \Pr {X = k} e^{t k}$ So:  {{begin-eqn}} {{eqn	| l = \map {M_X} t  	| r = \sum_{k \mathop = 0}^n \binom n k p^k \paren {1 - p}^{n - k} e^{t k} }} {{eqn	| r = \sum_{k \mathop = 0}^n \binom n k \paren {p e^t}^k \paren {1 - p}^{n - k} }} {{eqn	| r = \paren {1 - p + p e^t}^n 	| c = Binomial Theorem }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Binomial Distribution 09me4821a56apv712qyp0ij2zps2dww
From the definition of the Exponential distribution, $X$ has probability density function: :$\displaystyle \map {f_X} x = \frac 1 \beta e^{-\frac x \beta}$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$ Then:  {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \frac 1 \beta \int_0^\infty e^{x \paren {-\frac 1 \beta + t} } \rd x 	| c = Exponential of Sum }} {{eqn	| r = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \sqbrk {e^{x \paren {-\frac 1 \beta + t} } }_0^\infty 	| c = Primitive of Exponential Function }} {{end-eqn}} Note that if $t > \dfrac 1 \beta$, then $\displaystyle e^{x \paren {-\frac 1 \beta + t} } \to \infty$ as $x \to \infty$ by Exponential Tends to Zero and Infinity, so the integral diverges in this case.  If $t = \dfrac 1 \beta$ then the integrand is identically $1$, so the integral similarly diverges in this case.  If $t < \dfrac 1 \beta$, then $\displaystyle e^{x \paren {-\frac 1 \beta + t} } \to 0$ as $x \to \infty$ from Exponential Tends to Zero and Infinity, so the integral converges in this case.  Therefore, the function is only well defined for $t < \dfrac 1 \beta$. Proceeding: {{begin-eqn}} {{eqn	| l = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \sqbrk {e^{x \paren {-\frac 1 \beta + t} } }_0^\infty 	| r = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \paren {0 - 1} 	| c = Exponential Tends to Zero and Infinity, Exponential of Zero }} {{eqn	| r = \frac 1 {\beta \paren {\frac 1 \beta - t} } }} {{eqn	| r = \frac 1 {1 - \beta t} }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Exponential Distribution 9omebrbdg9jyyp53lp3kisnerl3psqr
From the definition of the geometric distribution, $X$ has probability mass function: :$\displaystyle \map \Pr {X = k} = \paren {1 - p}^k p$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}$ So: {{begin-eqn}} {{eqn	| l = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t} 	| r = \sum_{k \mathop = 0}^\infty \paren {1 - p}^k p e^{k t} }} {{eqn	| r = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k }} {{end-eqn}} By Sum of Infinite Geometric Sequence, for this sum to be convergent we must have:  :$\displaystyle \size {\paren {1 - p} e^t} < 1$ In the case $p = 1$, this demand is satisfied immediately regardless of $t$. Otherwise, as both $e^t$ and $1 - p$ are positive:  :$\displaystyle e^t < \frac 1 {1 - p}$ So, by Logarithm of Power:  :$\displaystyle t < -\map \ln {1 - p}$ is the range of $t$ for which the expectation is well-defined. Now applying Sum of Infinite Geometric Sequence, we have for this range of $t$:  :$\displaystyle \map {M_X} t = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k = \frac p {1 - \paren {1 - p} e^t}$ {{qed}} Category:Moment Generating Functions Category:Geometric Distribution j3aisrqpplynhjy8tsn3vgg9upy2uij
From the definition of the Discrete Uniform distribution, $X$ has probability mass function: :$\map \Pr {X = N} = \dfrac 1 n$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{N \mathop = 1}^n \map \Pr {X = N} e^{N t}$ So:  {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \frac 1 n \sum_{N \mathop = 1}^n \paren {e^t}^N }} {{eqn	| r = \frac {e^t} n \sum_{N \mathop = 0}^{n - 1} \paren {e^t}^N }} {{eqn	| r = \frac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} } 	| c = Sum of Geometric Sequence with $r = e^t$ }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Discrete Uniform Distribution dnfdu3348s4qk7brcq03pglci688tqw
Let $J$ be a maximal ideal. Because $J \subset R$, it follows from Quotient Ring of Commutative Ring is Commutative and Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a commutative ring with unity. We now need to prove that every non-zero element of $\struct {R / J, +,  \circ}$ has a product inverse in $R / J$. Let $x \in R$ such that $x + J \ne J$, that is: $x \notin J$. Thus $x + J \in R / J$ is not the zero element of $R / J$. Take $K \subseteq R$ such that: :$K = \set {j + r \circ x: j \in J, r \in R}$ that is, the subset of $R$ which can be expressed as a sum of an element of $J$ and a product in $R$ of $x$. Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$. So: :$(1): \quad K \ne \O$ Now let $g, h \in K$. That is: :$g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$ Then: :$-h = -j_2 + \paren {-r_2} \circ x$ But $j_1 - j_2 \in J$ from Test for Ideal. Similarly $-r_2 \in R$. So $-h \in K$ and we have: :$(2) \quad g + \paren {-h} = \paren {j_1 - j_2} + \paren {r_1 - r_2} \circ x$ Now consider $g \in J, y \in R$. Then: :$g \circ y = \paren {j_1 + r_1 \circ x} \circ y = \paren {j_1 \circ y} + \paren {r_1 \circ y} \circ x$ which is valid by the fact that $R$ is a commutative ring. But as $J$ is an ideal: :$\paren {j_1 \circ y} \in J$ while $r_1 \circ y \in R$. Thus: :$(3) \quad g \circ y \in K$ and similarly: :$(3) \quad y \circ g \in K$ So Test for Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is an ideal of $R$. Now: {{begin-eqn}} {{eqn | l = j       | o = \in       | r = J       | c =  }} {{eqn | ll= \leadsto       | l = j + 0_R \circ x       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = j       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = J       | o = \subseteq       | r = K       | c =  }} {{end-eqn}} and because $x = 0_R + 1_R \circ x$ (remember $0_R \in J$), then $x \in K$ too. So, because $x \notin J$, $K$ is an ideal such that $J \subset K \subseteq R$. Because $J$ is a maximal ideal, then $K = R$. Thus $1_R \in K$ and thus: :$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$ So: :$1_R + \paren {-s \circ x} = j_0 \in J$ Hence: :$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$ So in the commutative ring $\struct {R / J, +,  \circ}$, the product inverse of $x + J$ is $s + J$. The result follows.
Let $R / J$ be a field. Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subseteq R$. We have that $J$ is the zero of $R / J$. Let $x \in K \setminus J$. Because $x \notin J$ then $x + J \ne J$. Because $R / J$ is a field then $x + J \in R / J$ has a product inverse, say $s + J$. Hence: :$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$ By Left Cosets are Equal iff Product with Inverse in Subgroup: :$1_R - s \circ x \in J \subsetneq K$ By the definition of an ideal: :$x \in K, s \in R \implies s \circ x \in K$ :$1_R - s \circ x \in K, s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x} = 1_R \in K$ :$1_R \in K \implies \forall y \in R:  y \circ 1_R = y \in K$ Hence $K = R$. The result follows.
Proof by Counterexample: Consider the integral domain $\struct {\Z_2, +, \times}$. From Ring of Integers Modulo Prime is Integral Domain, it is seen that $\struct {\Z_2, +, \times}$ is indeed an integral domain. Consider the ring of polynomial forms $\Z_2 \sqbrk X$. This is an infinite ring, as it can be seen that $S \subseteq \Z_2 \sqbrk X$ where: :$S = \set {1, X, X^2, X^3, \dotsc}$ But the ring of polynomial functions $\map P D$ is finite, as: :$\map P D \subseteq \Z_2^{Z_2}$ where $\Z_2^{Z_2}$ is the set of all mappings from $\Z_2$ to $\Z_2$, and has $4$ elements. {{qed}}
From the definition of the Bernoulli distribution, $X$ has probability mass function: :$\map \Pr {X = n} = \begin{cases} q & : n = 0 \\ p & : n = 1 \\ 0 & : n \notin \set {0, 1} \\ \end{cases}$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^1 \map \Pr {X = n} e^{t n}$ So: {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \map \Pr {X = 0} e^0 + \map \Pr {X = 1} e^t }} {{eqn	| r = q + p e^t 	| c = Exponential of Zero }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Bernoulli Distribution 5ivqydxhdfwngfzylmj8i7nvjxh7eef
It is taken as read that $X^2 + 1$ is irreducible in $\R \sqbrk X$. Hence by Polynomial Forms over Field form Principal Ideal Domain: Corollary 1, $\R \sqbrk X / \ideal p$ is indeed a field. Let $\nu$ be the quotient epimorphism from $\R \sqbrk X$ onto $\R \sqbrk X / \ideal p$. From Quotient Ring Epimorphism is Epimorphism: :$\map \ker {\map \nu \R} = \R \cap \ideal p = \set 0$ So $\map \nu \R$ is a monomorphism from $\R \sqbrk X$ to $\R \sqbrk X / \ideal p$. Thus $\map \nu \R$ is an isomorphic copy of $\R$ inside $\R \sqbrk X / \ideal p$. We identify this isomorphic copy of $\R$ with $\R$ itself, ignoring the difference between $\map \nu x$ and $x$ when $x = \R$. Hence: :$(1): \quad \R \subseteq \R \sqbrk X / \ideal p$ Let $f \in \R \sqbrk X$ be arbitrary. By Division Theorem for Polynomial Forms over Field: :$\exists q, r \in \R \sqbrk X: f = q p + r$ where $r =a + b X$ for some $a, b \in \R$. Hence: {{begin-eqn}} {{eqn | l = \map \nu f       | r = \map \nu q \, \map \nu p + \map \nu r       | c = as $\nu$ is a ring honomorphism }} {{eqn | r = \map \nu r       | c = as $p \in \map \ker \nu$ }} {{eqn | r = \map \nu a + \map \nu b \, \map \nu X       | c =  }} {{eqn | r = a + b \, \map \nu X       | c = as we have identified $\map \nu \R$ with $\R$ }} {{eqn | r = a + b i       | c = where $i := \map \nu X$ }} {{end-eqn}} As $\nu$ is an epimorphism, it is a fortiori surjection. Hence: :$(2): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed in the form: ::$w = a + b i$ :for some $a, b \in \R$. Because $X^2 + 1 \in \ker \nu$, we have: {{begin-eqn}} {{eqn | l = 0       | r = \map \nu {X^2 + 1}       | c =  }} {{eqn | r = \paren {\map \nu X}^2 + \map \nu 1       | c =  }} {{eqn | r = i^2 + 1       | c =  }} {{end-eqn}} Hence we have that: :$(3): \quad$ In $\R \sqbrk X / \ideal p$, $i^2 = -1$ Thus $(2)$ can be improved to: :$(4): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed '''uniquely''' in the form: ::$w = a + b i$ :for some $a, b \in \R$. From $(1)$, $(3)$ and $(4)$, the field $\R \sqbrk X / \ideal p$ is recognised as the field of complex numbers. {{Qed}}
Let $X = \C$. By Non-Zero Complex Numbers Closed under Multiplication we have that $2 L \in X_{\ne 0}$. Then: {{begin-eqn}} {{eqn | l = \map f {x + 2L}       | r = \map f {x + \paren {L + L} } }} {{eqn | r = \map f {\paren {x + L} + L}       | c = Complex Addition is Associative }} {{eqn | r = -\map f {x + L} }} {{eqn | r = \map f x       | c = Negative of Negative Complex Number }} {{end-eqn}} The proof for when $X = \R$ is nearly identical. {{qed}} Category:Antiperiodic Functions s2rr55kih3dzer6rke4k22dwmvhp3sz
Let $0_R$ denote the zero of $R$. Let $x \in R$. Then: {{begin-eqn}} {{eqn | l = x + x       | r = 0_R       | c = Idempotent Ring has Characteristic Two }} {{eqn | ll= \leadsto       | l = -x + x + x       | r = -x + 0_R       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = -x       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From the definition of skewness:  :$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$ From the definition of the Gaussian distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} }$ So, from Expectation of Function of Continuous Random Variable: :$\displaystyle \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac {x - \mu} \sigma}^3 \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} } \rd x$ Making a substitution of $u = x - \mu$: :$\displaystyle \gamma_1 = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u$ We have that:  :$\paren {-u}^3 \map \exp {-\dfrac {\paren {-u}^2} {2 \sigma^2} } = -u^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} }$ So we can see that the integrand is odd. So, by Definite Integral of Odd Function: :$\displaystyle \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u = 0$ giving:  :$\gamma_1 = 0$  {{qed}} Category:Skewness Category:Gaussian Distribution i7arvorvfenqsg3oqny9d8bnl36n5ql
Let $u \in R$ be a unit of $R$. Let $u \in \map {C_R} S$. Then from Commutation with Inverse in Monoid: :$u^{-1} \in \map {C_R} S$ {{qed}}
Follows directly from the definition of center and Inverse of Unit in Centralizer of Ring is in Centralizer. {{qed}}
{{begin-eqn}} {{eqn	| l = \gamma_1 	| r = \expect {\paren {\dfrac {X - \mu} \sigma}^3} 	| c = {{Defof|Skewness}} }} {{eqn	| r = \frac {\expect {X^3 - 3 \mu X^2 + 3 \mu^2 X - \mu^3} } {\sigma^3}  	| c = Linearity of Expectation Function, Cube of Difference }} {{eqn	| r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect X - \mu^3} {\sigma^3} 	| c = Linearity of Expectation Function }} {{eqn	| r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \mu \expect X} - \mu^3} {\sigma^3} }} {{eqn	| r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \paren {\expect X}^2} - \mu^3} {\sigma^3} 	| c = $\mu = \expect X$ }} {{eqn	| r = \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3} 	| c = Variance as Expectation of Square minus Square of Expectation }} {{end-eqn}} {{qed}} Category:Skewness pm850rumjl7z7uv4xmbn26t3vt0viye
Let $L = \bigcap \mathbb L$. From Intersection of Subrings is Subring, we have that $L$ is a subring of $R$. Let $x \in L$ and $y \in R$. Then: :$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$ as every element of $\bigcap \mathbb L$, including $T$, is an ideal of $R$. If $y \in R$, then $x \circ y$ and $y \circ x$ are in every element of $\mathbb L$, and hence in $L$. So $L$ is an ideal of $R$. {{qed}}
By definition of ring zero: :$\forall x \in R: x \circ 0_R = 0_R$ Hence $0_R \in A$ and so $A \ne \O$. Let $a, b \in A$. {{begin-eqn}} {{eqn | ll= \forall x \in R:       | l = x \circ b       | r = 0_R       | c =  }} {{eqn | lll=\leadsto       | l = -\paren {x \circ b}       | r = 0_R       | c =  }} {{eqn | lll=\leadsto       | l = x \circ \paren {-b}       | r = 0_R       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | ll= \forall x \in R:       | l = x \circ a       | r = 0_R       | c =  }} {{eqn | lo= \land       | l = x \circ \paren {-b}       | r = 0_R       | c =  }} {{eqn | lll=\leadsto       | l = \paren {x \circ a} + \paren {x \circ \paren {-b} }       | r = 0_R       | c = as $0_R$ is identity of $\struct {R, +}$ }} {{eqn | lll=\leadsto       | l = x \circ \paren {a + \paren {-b} }       | r = 0_R       | c = {{Ring-axiom|D}} }} {{eqn | lll=\leadsto       | l = a + \paren {-b}       | o = \in       | r = A       | c = Definition of $A$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | ll= \forall x \in R:       | l = x \circ a       | r = 0_R       | c =  }} {{eqn | lo= \land       | l = x \circ b       | r = 0_R       | c =  }} {{eqn | lll=\leadsto       | l = \paren {x \circ a} \circ b       | r = 0_R       | c = {{Defof|Ring Zero}} }} {{eqn | lll=\leadsto       | l = x \circ \paren {a \circ b}       | r = 0_R       | c = {{Ring-axiom|M1}} }} {{eqn | lll=\leadsto       | l = a \circ b       | o = \in       | r = A       | c = Definition of $A$ }} {{end-eqn}} Hence the result, from Test for Ideal: {{qed}}
From Skewness in terms of Non-Central Moments:  :$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$  where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation. We have, by Expectation of Bernoulli Distribution: :$\mu = p$ By Variance of Bernoulli Distribution, we also have:  :$\var X = \sigma^2 = p \paren {1 - p}$ so:  :$\sigma = \sqrt {p \paren {1 - p} }$ By Raw Moment of Bernoulli Distribution, we have:  :$\expect {X^3} = p$ So: {{begin-eqn}} {{eqn	| l = \gamma_1 	| r = \frac {p - 3 p^2 \paren {1 - p} - p^3} {p^{3/2} \paren {1 - p}^{3/2} } }} {{eqn	| r = \frac {p \paren {1 - p^2} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} } }} {{eqn	| r = \frac {p \paren {1 - p} \paren {1 + p} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} } 	| c = Difference of Two Squares }} {{eqn	| r = \frac {p \paren {1 - p} \paren {1 + p - 3p} } {p^{3/2} \paren {1 - p}^{3/2} } }} {{eqn	| r = \frac {1 - 2 p} {\sqrt {p \paren {1 - p} } } }} {{eqn	| r = \frac {1 - 2 p} {\sqrt {p q} } 	| c = $q = 1 - p$ }} {{end-eqn}} {{qed}} Category:Bernoulli Distribution Category:Skewness sdipu0lqoiyvtqullidfadba1bjeyul
Let $\displaystyle L = \bigcap \mathbb L$. By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is a subgroup of $\struct {R, +}$. By the One-Step Subgroup Test: :$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$ By Intersection of Subsemigroups, $\struct {L, \circ}$ a subsemigroup of $\struct {R, \circ}$. So by definition of subsemigroup: :$\forall x, y \in \struct {L, \circ}: x \circ y \in L$ By the Subring Test it follows that $\struct {L, +, \circ}$ is a subring of $\struct {R, +, \circ}$. {{qed}}
Let $M$ be the median of $X$. From the definition of the exponential distribution, $X$ has probability density function:  :$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$ Note that $f_X$ is non-zero, so the median is unique.  We have by the definition of a median:  :$\displaystyle \map \Pr {X < M} = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x = \frac 1 2$ Evaluating this integral: {{begin-eqn}} {{eqn	| l = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x 	| r = \frac 1 \beta \sqbrk {-\beta e^{-\frac x \beta} }_0^M 	| c = Primitive of Exponential Function }} {{eqn	| r = 1 - e^{-\frac M \beta} }} {{end-eqn}} So:  :$1 - e^{-\frac M \beta} = \dfrac 1 2$ Then:  :$e^{-\frac M \beta} = \dfrac 1 2$ giving:  :$-\dfrac M \beta = \map \ln {\dfrac 1 2}$ So, by Logarithm of Reciprocal:  :$M = \beta \ln 2$ {{qed}} Category:Exponential Distribution j5t0e96gx4xvfgvj4qdzsdgyhjfg3cf
Let $L = \bigcap \mathbb L$. From Intersection of Ring Ideals is Ideal  $L$ is indeed an ideal of $R$. Let $L = \bigcap \mathbb L$. From Intersection of Subrings is Largest Subring Contained in all Subrings, we have that $L$ is the largest subring of $R$ contained in each member of $\mathbb L$. As $L$ is the largest subring of $R$ contained in each member of $\mathbb L$, and it is an ideal of $R$, there can be no larger ideal as it would then not be a subring. So $L$ is the largest ideal of $R$ contained in each member of $\mathbb L$. {{qed}}
Let $\displaystyle L = \bigcap \mathbb L$. From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$. By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {R, +}$ contained in each member of $\mathbb L$. By Intersection of Subsemigroups, $\struct {L, \circ}$ is the largest subsemigroup of $\struct {R, \circ}$ contained in each member of $\mathbb L$. Let $S$ be a subring of $R$ such that: :$\forall K \in \mathbb L: S \subseteq K$ By Intersection is Largest Subset, $S \subseteq L$. Thus $L$ is the largest subring of $R$ contained in each member of $\mathbb L$. {{qed}}
From the definition of the continuous uniform distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac 1 {b - a}$ Note that $f_X$ is non-zero, so the median is unique.  We have by the definition of a median:  :$\displaystyle \map \Pr {X < M} = \frac 1 {b - a} \int_a^M \rd x = \frac 1 2$ We have, by Primitive of Constant:  :$\dfrac {M - a} {b - a} = \dfrac 1 2$ So:  {{begin-eqn}} {{eqn	| l = M 	| r = a + \frac {b - a} 2 }} {{eqn	| r = \frac {b - a + 2 a} 2 }} {{eqn	| r = \frac {b + a} 2 }} {{end-eqn}} {{qed}} Category:Continuous Uniform Distribution joatsn7dvulzmoat6ha7uljktgmfme4
From Intersection of Subrings Containing Subset is Smallest, $L$ is the smallest subring of $R$ containing $S$. From Intersection of Ring Ideals is Ideal, $L$ is an ideal of $R$. As $L$ is the smallest subring of $R$ containing $S$, and it is an ideal of $R$, there can be no smaller ideal of $R$ containing $S$ as it would then not be a subring. So the intersection of the set of all ideals of $R$ containing $S$ is the smallest ideals of $R$ containing $S$.  {{qed}}
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$. Let $T$ be a subring of $R$ containing $S$. Let $x, y \in L$. By the Subring Test, we have that: {{begin-eqn}} {{eqn | l = x - y       | o = \in       | r = L }} {{eqn | l = x \circ y       | o = \in       | r = L }} {{end-eqn}} By Intersection is Largest Subset, it follows that $x, y \in T$. But $T$ is also a subring of $R$. So, by the Subring Test again, we have that: {{begin-eqn}} {{eqn | l = x - y       | o = \in       | r = T }} {{eqn | l = x \circ y       | o = \in       | r = T }} {{end-eqn}} So by definition of subset, $L \subseteq T$. {{qed}}
Let $L = \bigcap \mathbb K$. By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is a subgroup of $\struct {D, +}$. By the One-Step Subgroup Test: :$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$ By Intersection of Subgroups is Subgroup, $\struct {L, \circ}$ a subgroup of $\struct {D, \circ}$. By the Two-Step Subgroup Test: :$\forall x, y \in \struct {L, \circ}: x \circ y \in L$ :$\forall x \in \struct {L, \circ}: x^{-1} \in L$ By the Division Subring Test it follows that $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$. {{qed}} Category:Division Subrings 69ali8l00sj7xt1s2imaspbw72jzljt
Let $L = \bigcap \mathbb K$. A field is by definition also a division subring. From Intersection of Division Subrings is Division Subring, $L$ is itself a division subring of $F$. As $\struct {F, +, \circ}$ is a field, $\circ$ is commutative on $F$. By Restriction of Commutative Operation is Commutative, it follows that $\circ$ is also commutative on $L$. Thus $\struct {L, +, \circ}$ is a division ring where $\circ$ is commutative. Thus, by definition, of $\struct {L, +, \circ}$ is a field. Hence the result. {{qed}}
Let $L = \bigcap \mathbb K$. From Intersection of Division Subrings is Division Subring, $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$. By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {D, +}$ contained in each element of $\mathbb K$. By Intersection of Subgroups is Subgroup, $\struct {L, \circ}$ is the largest subgroup of $\struct {D, \circ}$ contained in each element of $\mathbb K$. Let $S$ be a subring of $D$ such that: :$\forall K \in \mathbb K: S \subseteq K$ By Intersection is Largest Subset, $S \subseteq L$. Thus $L$ is the largest division subring of $D$ contained in each element of $\mathbb L$. {{qed}}
From Intersection of Division Subrings is Division Subring, $L$ is indeed a division subring of $D$. Let $T$ be a division subring of $D$ containing $S$. Let $x, y \in L$. By the Division Subring Test, we have that: {{begin-eqn}} {{eqn | l = x - y       | o = \in       | r = L }} {{eqn | l = x \circ y       | o = \in       | r = L }} {{eqn | l = x^{-1} \circ y       | o = \in       | r = L }} {{end-eqn}} By Intersection is Largest Subset, it follows that $x, y \in T$. But $T$ is also a division subring of $D$. So, by the Division Subring Test again, we have that: {{begin-eqn}} {{eqn | l = x - y       | o = \in       | r = T }} {{eqn | l = x \circ y       | o = \in       | r = T }} {{eqn | l = x^{-1} \circ y       | o = \in       | r = T }} {{end-eqn}} So by definition of subset, $L \subseteq T$. {{qed}}