hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \laptrans {t^n}       | r = \int_0^\infty e^{-s t} t^n \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^\infty e^{-u} \paren {\dfrac u s}^n \rd \paren {\dfrac u s}       | c = Integration by Substitution: $u := s t$ where $s > 0$ is assumed }} {{eqn | r = \dfrac 1 {s^{n + 1} } \int_0^\infty u^n e^{-u} \rd u       | c =  }} {{eqn | r = \dfrac {\map \Gamma {n + 1} } {s^{n + 1} }       | c = {{Defof|Gamma Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\dfrac 1 {\sqrt t} }       | r = \laptrans {t^{-1 / 2} }       | c =  }} {{eqn | r = \dfrac {\map \Gamma {1 / 2} } {s^{1 / 2} }       | c = Laplace Transform of Real Power }} {{eqn | r = \dfrac {\sqrt \pi} {\sqrt s}       | c = Gamma Function of One Half }} {{eqn | r = \sqrt {\dfrac \pi s}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma {-\dfrac 3 2}       | r = \map \Gamma {\dfrac 1 2 + 1}       | c =  }} {{eqn | r = \dfrac 1 2 \map \Gamma {\dfrac 1 2}       | c = Gamma Difference Equation }} {{eqn | r = \dfrac {\sqrt \pi} 2       | c = Gamma Function of One Half }} {{end-eqn}} {{qed}} Category:Examples of Gamma Function Values pbardiulbf5exfhw7xs320tnft6zk9o
{{begin-eqn}} {{eqn | l = \map \Gamma {-\dfrac 1 2}       | r = -\dfrac 3 2 \map \Gamma {-\dfrac 3 2}       | c = Gamma Difference Equation }} {{eqn | ll= \leadsto       | l = \map \Gamma {-\dfrac 3 2}       | r = -\dfrac 2 3 \map \Gamma {-\dfrac 1 2}       | c =  }} {{eqn | r = -\dfrac 2 3 \paren {-2 \sqrt \pi}       | c = Gamma Function of Minus One Half }} {{eqn | r = \dfrac {4 \sqrt \pi} 3       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma {-\dfrac 3 2}       | r = -\dfrac 5 2 \map \Gamma {-\dfrac 5 2}       | c = Gamma Difference Equation }} {{eqn | ll= \leadsto       | l = \map \Gamma {-\dfrac 5 2}       | r = -\dfrac 2 5 \map \Gamma {-\dfrac 3 2}       | c =  }} {{eqn | r = -\dfrac 2 5 \paren {\dfrac {4 \sqrt \pi} 3}       | c = Gamma Function of Minus 3 over 2 }} {{eqn | r = -\dfrac {8 \sqrt \pi} {15}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma 1       | r = 0 \, \map \Gamma 0       | c = Gamma Difference Equation }} {{eqn | ll= \leadsto       | l = \map \Gamma 0       | r = \dfrac {\map \Gamma 1} 0       | c =  }} {{eqn | ll= \leadsto       | l = \map \Gamma 0       | r = \dfrac 1 0       | c = Gamma Function Extends Factorial }} {{end-eqn}} But $\dfrac 1 0$ is not defined. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma 0       | r = \paren {-1} \, \map \Gamma {-1}       | c = Gamma Difference Equation }} {{eqn | ll= \leadsto       | l = \map \Gamma {-1}       | r = \dfrac {\map \Gamma 0} {-1}       | c =  }} {{end-eqn}} But from Gamma Function of Zero, $\map \Gamma 0$ is not defined. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map \Gamma {-1}       | r = \paren {-2} \, \map \Gamma {-2}       | c = Gamma Difference Equation }} {{eqn | ll= \leadsto       | l = \map \Gamma {-2}       | r = \dfrac {\map \Gamma {-1} } {-2}       | c =  }} {{end-eqn}} But from Gamma Function of Minus 1, $\map \Gamma {-1}$ is not defined. Hence the result. {{qed}}
Using the technique of Evaluation of Integral using Laplace Transform: {{begin-eqn}} {{eqn | l = \int_0^\infty e^{-s t} \map {J_0} t \rd t       | r = \dfrac 1 {\sqrt {s^2 + 1} }       | c = Laplace Transform of Bessel Function of the First Kind of Order Zero }} {{eqn | ll= \leadsto       | l = \int_0^\infty \map {J_0} t \rd t       | r = 1       | c = letting $s \to 0^+$ }} {{end-eqn}} {{qed}}
Using the technique of Evaluation of Integral using Laplace Transform: {{begin-eqn}} {{eqn | l = \int_0^\infty e^{-s t} \erf \sqrt t \rd t       | r = \dfrac 1 {s \sqrt {s + 1} }       | c = Laplace Transform of Error Function of Root }} {{eqn | ll= \leadsto       | l = \int_0^\infty \erf \sqrt t \rd t       | r = \dfrac 1 {1 \sqrt {1 + 1} }       | c = letting $s \to 0^+$ }} {{eqn | r = \dfrac 1 {\sqrt 2}       | c = simplifying }} {{eqn | r = \dfrac {\sqrt 2} 2       | c =  }} {{end-eqn}} {{qed}}
From Laplace Transform of Power: :$\displaystyle \int_0^\infty e^{-s t} t^k \rd t = \dfrac {\map \Gamma {k + 1} } {s^{k + 1} }$ for $k > -1$. Differentiating {{WRT|Differentiation}} $k$: :$\displaystyle \int_0^\infty e^{-s t} t^k \ln t \rd t = \dfrac {\map {\Gamma'} {k + 1} - \map \Gamma {k + 1} \ln s} {s^{k + 1} }$ Setting $k = 0$: {{begin-eqn}} {{eqn | l = \int_0^\infty e^{-s t} \ln t \rd t       | r = \laptrans {\ln t}       | c =  }} {{eqn | r = \dfrac {\map {\Gamma'} 1 - \ln s}s       | c =  }} {{eqn | r = -\dfrac {\gamma + \ln s} s       | c = Derivative of Gamma Function at 1 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\d^2} {\d s^2} \laptrans {\map f t}       | r = \map {\frac \d {\d s} } {\dfrac \d {\d s} \laptrans {\map f t} }       | c = {{Defof|Second Derivative}} }} {{eqn | r = \map {\frac \d {\d s} } {-\laptrans {t \, \map f t} }       | c = Derivative of Laplace Transform }} {{eqn | r = -\frac \d {\d s} \laptrans {t \, \map f t}       | c =  }} {{eqn | r = -\paren {-\laptrans {t \paren {t \, \map f t} } }       | c = Derivative of Laplace Transform }} {{eqn | r = \laptrans {t^2 \, \map f t} }} {{end-eqn}} {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$ $\map P 0$ is the case: :$\map f u = \map F s$ which is the statement of the Laplace transform. Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: :$\displaystyle \laptrans {\int_0^t \map f u \rd u} = \dfrac {\map F s} s$ which is established in Laplace Transform of Integral Thus $\map P 1$ 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: :$\displaystyle \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k$ times} } \map f u \rd u^k} = \dfrac {\map F s} {s^k}$ from which it is to be shown that: :$\displaystyle \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k + 1$ times} } \map f u \rd u^{k + 1} } = \dfrac {\map F s} {s^{k + 1} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k + 1$ times} } \map f u \rd u^{k + 1} }       | r = \laptrans {\int_0^t \paren {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k$ times} } \map f u \rd u^k} \rd u}       | c =  }} {{eqn | r = \dfrac 1 s \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k$ times} } \map f u \rd u^k}       | c = Basis for the Induction }} {{eqn | r = \dfrac 1 s \paren {\dfrac {\map F s} {s^k} }       | c = Induction Hypothesis }} {{eqn | r = \dfrac {\map F s} {s^{k + 1} }       | c = simplification }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \displaystyle \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$ {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\int_0^t \map f u \map g {t - u} \rd u}       | r = \int_{t \mathop = 0}^\infty e^{-s t} \paren {\int_{u \mathop = 0}^t \map f u \map g {t - u} \rd u} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_{t \mathop = 0}^\infty \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t       | c =  }} {{eqn | r = \lim_{M \mathop \to \infty} \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t       | c =  }}  {{eqn | n = 1       | r = \lim_{M \mathop \to \infty} s_M       | c =  }}  {{end-eqn}} where $s_M$ is defined to be: :$\displaystyle \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$ The region in the plane over which $(1)$ is to be integrated is $\mathscr R_{t u}$ below: :400px Setting $t - u = v$, that is $t = u + v$, the shaded region above is transformed into the region $\mathscr R_{u v}$ the $u v$ plane: :400px Thus: {{begin-eqn}} {{eqn | l = s_M       | r = \iint_{\mathscr R_{t u} } e^{-s t} \map f u \map g {t - u} \rd u \rd t       | c = from $(1)$ above }} {{eqn | n = 2       | r = \iint_{\mathscr R_{u v} } e^{-s \paren {u + v} } \map f u \map g v \size {\dfrac {\map \partial {u, t} } {\map \partial {u, v} } } \rd u \rd v       | c =  }} {{end-eqn}} where $\dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ is the Jacobian of the transformation: {{begin-eqn}} {{eqn | l = J       | r = \dfrac {\map \partial {u, t} } {\map \partial {u, v} }       | c =  }} {{eqn | r = \begin{vmatrix} \dfrac {\partial u} {\partial u} & \dfrac {\partial u} {\partial v} \\ \dfrac {\partial t} {\partial u} & \dfrac {\partial t} {\partial v} \end{vmatrix}       | c =  }} {{eqn | r = \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}       | c =  }}  {{end-eqn}} Thus the {{RHS}} of $(2)$ is: :$(3): \quad \displaystyle s_M = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^{M - v} e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$ Let $\map K {u, v}$ be the function defined as: :$\map K {u, v} = \begin{cases} e^{-s \paren {u + v} } \map f u \map g v & : u + v \le M \\ 0 & : u + v > M \end{cases}$ This function is defined over the square region in the diagram below: :400px but is zero over the lighter shaded portion. Now we can write $(3)$ as: {{begin-eqn}} {{eqn | l = s_M       | r = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^M \map K {u, v} \rd u \rd v       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{M \mathop \to \infty} s_M       | r = \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty \map K {u, v} \rd u \rd v       | c =  }} {{eqn | r = \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v       | c =  }} {{eqn | r = \paren {\int_{v \mathop = 0}^\infty e^{-s u} \map f u \rd u} \paren {\int_{v \mathop = 0}^\infty e^{-s v} \map g u \rd v}       | c =  }} {{eqn | r = \map F s \map G s       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Harmonic Series is Divergent, we have that $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n$ diverges to $+\infty$. Thus from the Integral Test: :$\displaystyle \int_1^n \frac {\d x} x \to +\infty$ diverges. {{qed}}
From the definition of natural logarithm: :$\displaystyle \ln x = \int_1^x \dfrac 1 t \rd t$ The result follows from Logarithm Tends to Infinity. {{qed}}
Put $x = \dfrac 1 z$. Then: {{begin-eqn}} {{eqn | l = \int_\gamma^1 \frac {\d x} x       | r = \int_{1 / \gamma}^1 \frac {-z} {z^2} \rd z       | c = Integration by Substitution }} {{eqn | r = \int_1^{1 / \gamma} \frac {\d z} z       | c =  }} {{end-eqn}} But as $\gamma \to 0+$, we have that $\dfrac 1 \gamma \to +\infty$. Hence, from Integral of Reciprocal Unbounded Above is Divergent: :$\displaystyle \int_1^{1 / \gamma} \frac {\d z} z \to +\infty$ as $\dfrac 1 \gamma \to +\infty$. The result follows. {{qed}}
=== Lemma === {{:Convergence of P-Series/Lemma}} Since $x > 1$ it follows that $1 - x < 0$. Thus $P^{1 - x} \to 0$ as $P \to \infty$. Setting $x - 1 = \delta >0$, this limit is: :$\displaystyle -\frac 1 {\delta} \lim_{t \mathop \to \infty} \frac 1 {t^\delta} = 0$  Hence the integral is just $\dfrac 1 {1 - x}$ (that is, convergent) and so the sum converges as well. Since the terms of the sum were positive everywhere, it is absolutely convergent and hence so is: :$\displaystyle \sum_{n \mathop = 1}^\infty n^{-p}$ {{qed}}
=== Lemma === {{:Convergence of P-Series/Lemma}} Hence, the convergence of the $p$-series is dependent on the convergence of:  :$\displaystyle \lim_{t \mathop \to \infty} \frac{t^{1 - x} } {1 - x}$ Suppose $0 < x < 1$. Then: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to \infty} \frac{t^{1 - x} } {1 - x}       | r = \frac 1 {1 - x} \ \lim_{t \mathop \to \infty} \ t^{1 - x} }} {{eqn | o = \to       | r = +\infty       | c = Limit at Infinity of $x^n$ }} {{end-eqn}} The special case of $x = 1$ is covered by Integral of Reciprocal is Divergent. Hence the result from the Integral Test. {{qed}} Category:Convergence of P-Series nm6snv49i319ki80jsamztufjsjjzha
Let $p = x + i y$. Then: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \size {n^{-p} }       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x n^{i y} } }       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x e^{-i y \, \map \log n} } }       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x} \size {e^{-i y \, \map \log n} } }       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x} }       | c =  }} {{end-eqn}} by Euler's Formula. Now since $x > 0$, and all $n \ge 1$, all terms are positive and we may do away with the absolute values. Then by the Integral Test: :$\displaystyle \sum_{n \mathop = 1}^{\to \infty} \frac 1 {n^x}$ converges {{iff}} $\displaystyle \int_1^\infty \frac {\d t} {t^x}$ converges. First let $x \ne 1$: {{begin-eqn}} {{eqn | l = \int_1^{\to \infty} \frac {\d t} {t^x}       | r = \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} {t^x}       | c =  }} {{eqn | r = \lim_{P \mathop \to \infty} \intlimits {\dfrac {t^{1 - x} } {1 - x} } 1 P       | c = Primitive of Power }} {{eqn | r = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} - \dfrac {1^{1 - x} } {1 - x} }       | c =  }} {{eqn | r = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} } - \dfrac {1^{1 - x} } {1 - x}       | c =  }} {{end-eqn}} Hence: :$\displaystyle \int_1^{\to \infty} \frac {\d t} {t^x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} } - \dfrac {1^{1 - x} } {1 - x}$ {{iff}}: :$\displaystyle \int_1^{\to \infty} \frac {\d t} {t^x} + \dfrac {1^{1 - x} } {1 - x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} }$ and so: :$\displaystyle \sum_{n \mathop = 1}^\infty \size {n^{-p} }$ converges {{iff}}: :$\displaystyle \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges. For $x = 1$: {{begin-eqn}} {{eqn | l = \int_1^{\to \infty} \frac {\d t} t       | r = \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} t       | c =  }} {{eqn | r = \lim_{P \mathop \to \infty} \bigintlimits {\ln x} 1 P       | c = Primitive of Reciprocal }} {{eqn | r = \lim_{P \mathop \to \infty} \ln P       | c = Logarithm of 1 is 0 }} {{end-eqn}} which diverges. {{qed}}
First let $x \ne 1$. Then: {{begin-eqn}} {{eqn | l = \int_1^\infty \dfrac {\d t} {t^x}       | r = \lim_{P \mathop \to \infty} \int_1^P t^{-x} \rd t       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{P \mathop \to \infty} \intlimits {\dfrac {t^{-x + 1} } {-x + 1} } 1 P       | c = Primitive of Power }} {{eqn | r = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} - \dfrac {1^{1 - x} } {1 - x} }       | c =  }} {{eqn | r = \lim_{P \mathop \to \infty} \dfrac 1 {x - 1} \paren {1 - \dfrac 1 {P^{x - 1} } }       | c = simplifying }} {{end-eqn}} If $x > 1$, then $x - 1 > 0$. Hence from Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 {P^{x - 1} } \to 0$ as $P \to +\infty$. If $x < 1$, then $x - 1 < 0$. Hence $P^{x - 1} \to 0$ as $P \to +\infty$. Then from Reciprocal of Null Sequence it follows that $\dfrac 1 {P^{x - 1} } \to \infty$ as $P \to +\infty$. Finally we have that from Integral of Reciprocal is Divergent: :$\displaystyle \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} t \to \infty$ All cases are covered, and the result follows. {{qed}}
By the Integral Test: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges {{iff}} the improper integral $\displaystyle \int_1^\infty \frac {\d t} {t^x}$ exists. The result follows from Integral to Infinity of Reciprocal of Power of x. {{qed}}
Let $p = 1$. Then from Harmonic Series is Divergent the $p$-series diverges. So let $p > 1$. We note that the sequence of partial sums is increasing. Hence it is sufficient to show that they are bounded above. Let: :$s_{2^N} := 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dotsb + \dfrac 1 {N^p}$ Then: {{begin-eqn}} {{eqn | l = s_N       | o = \le       | r = S_{2^N - 1}       | c =  }} {{eqn | r = 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dfrac 1 {4^p} + \dotsb + \dfrac 1 {\paren {2^N - 1}^p}       | c =  }} {{eqn | r = 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {3^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {5^p} + \dfrac 1 {6^p} + \dfrac 1 {7^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {\paren {2^N - 1}^p} }       | c =  }} {{eqn | o = \le       | r = 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {2^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {2^{\paren {N - 1} p} } }       | c =  }} {{eqn | r = 1 + \dfrac 2 {2^p} + \dfrac 4 {4^p} + \dotsb + \dfrac {2^{N - 1} } {2^{\paren {N - 1} p} }       | c =  }} {{eqn | r = 1 + \dfrac 1 {2^{p - 1} } + \paren {\dfrac 1 {2^{p - 1} } }^2 + \dotsb + \paren {\dfrac 1 {2^{p - 1} } }^{n - 1}       | c =  }} {{eqn | r = \dfrac {1 - \paren {1 / 2^{p - 1} }^N} {1 - \paren {1 / 2^{p - 1} } }       | c =  }} {{eqn | o = \le       | r = \dfrac 1 {1 - \paren {1 / 2^{p - 1} } }       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain. From Ordered Integral Domain is Totally Ordered Ring, the usual ordering $\le$ is a total ordering. From Relation Induced by Strict Positivity Property is Transitive it follows that $<$ is transitive. {{qed}}
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain. Thus: {{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = b - a       | o = >       | r = 0       | c = {{Defof|Positivity Property}} }} {{eqn | ll= \leadsto       | l = c \times \paren {b - a}       | o = >       | r = 0       | c = {{Defof|Ordered Integral Domain|index = 1}}: $(\text P 1)$: Closure under Ring Addition }} {{eqn | ll= \leadsto       | l = b \times c - a \times c       | o = >       | r = 0       | c = {{Ring-axiom|D}} }} {{eqn | ll= \leadsto       | l = a \times c       | o = <       | r = b \times c       | c = {{Defof|Positivity Property}} }} {{end-eqn}} {{Qed}}
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain. Thus: {{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = b - a       | o = >       | r = 0       | c = {{Defof|Positivity Property}} }} {{eqn | ll= \leadsto       | l = c \times \paren {b - a}       | o = <       | r = 0       | c = Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative }} {{eqn | ll= \leadsto       | l = b \times c - a \times c       | o = <       | r = 0       | c = {{Ring-axiom|D}} }} {{eqn | ll= \leadsto       | l = a \times c       | o = >       | r = b \times c       | c = {{Defof|Positivity Property}} }} {{end-eqn}} {{Qed}}
Because $b^2 - 4 a c > 0$, we have from Solution to Quadratic Equation with Real Coefficients that the roots of $\map Q x$ are real and unequal. This demonstrates the existence of $\alpha$ and $\beta$, where by hypothesis we state that $\alpha < \beta$. We can express $\map Q x$ as: :$\map Q x = a \paren {x - \alpha} \paren {x - \beta}$ When $\alpha < x < \beta$ we have that: :$x - \alpha > 0$ :$x - \beta < 0$ and so: :$\map Q x = a \paren {x - \alpha} \paren {x - \beta} < 0$ {{qed|lemma}} When $x < \alpha$ we have that: :$x - \alpha < 0$ :$x - \beta < 0$ and so: :$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$ {{qed|lemma}} When $x > \beta$ we have that: :$x - \alpha > 0$ :$x - \beta > 0$ and so: :$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$ {{qed|lemma}} Hence the result. {{qed}}
{{AimForCont}} $t = r + s \sqrt 2$ be rational. Then: :$\sqrt 2 = \dfrac {t - r} s$ is also rational. This contradicts the fact that Square Root of 2 is Irrational. Hence the result by Proof by Contradiction. {{qed}}
We have that: {{begin-eqn}} {{eqn | l = a \paren {r + s \sqrt 2}^2 + b \paren {r + s \sqrt 2} + c       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a r^2 + 2 a s + br + c} + \paren {2 a + b} s \sqrt 2       | r = 0       | c =  }} {{end-eqn}} Because $a$, $b$, $c$, $r$ and $s$ are rational, it must be that $\paren {2 a + b} s = 0$. Hence: {{begin-eqn}} {{eqn | l = a \paren {r - s \sqrt 2}^2 + b \paren {r - s \sqrt 2} + c       | r = \paren {a r^2 + 2 a s + br + c} - \paren {2 a + b} s \sqrt 2       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} and so $\beta$ is also a root of $(1)$. {{qed}}
Proof by Counterexample: Consider the subset $T$ of the set of real numbers $\R$: :$T := \set {x \in \R: 1 \le x < 2}$ The number $2$ cannot be the greatest element of $T$ as $2 \notin T$. However, $2$ is the supremum of $T$ in $S$. Indeed, by definition: :$\forall x \in T: x < 2$ So, let $x < 2$. Then consider $y = \dfrac x 1 + \dfrac 2 1 = \dfrac {x + 2} 2$. We have that $x < y < 2$ by Mediant is Between. Thus $y \in T$ but $y > x$ and so $x$ cannot be the greatest element of $T$. Neither can $y$ be the supremum of $T$ in $S$. The conclusion is that there is no greatest element of $T$. Hence the result. {{qed}}
Let $V$ be the subset of the real numbers $\R$ defined as: :$V := \set {x \in \R: x > 0}$ From Infimum of Subset of Real Numbers: Example 3, $V$ admits an infimum: :$\inf V = 0$ But $V$ has no smallest element, as follows. We note that $\inf V = 0 \notin V$. {{AimForCont}} $x \in V$ is the smallest element of $V$. Then $\dfrac x 2$ is also in $V$ as $\dfrac x 2 > 0$. But $\dfrac x 2 < x$ which contradicts $x$ as being the smallest element of $V$. Hence there can be no smallest element of $V$. {{qed}}
{{begin-eqn}} {{eqn | l = \size {\xi - x}       | o = \le       | r = \delta       | c = }} {{eqn | ll= \leadstoandfrom       | l = -\delta       | o = \le       | r = \xi - x \le \delta       | c = Negative of Absolute Value: Corollary 2 }} {{eqn | ll= \leadstoandfrom       | l = \delta       | o = \le       | r = x - \xi \le -\delta       | c = }} {{eqn | ll= \leadstoandfrom       | l = \xi + \delta       | o = \le       | r = x \le \xi - \delta       | c = }} {{end-eqn}} But: :$\closedint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta \le x \le \xi + \delta}$ {{qed}}
{{begin-eqn}} {{eqn | l = \size {\xi - x}       | o = <       | r = \delta       | c = }} {{eqn | ll= \leadstoandfrom       | l = -\delta       | o = <       | r = \xi - x < \delta       | c = Negative of Absolute Value: Corollary 1 }} {{eqn | ll= \leadstoandfrom       | l = \delta       | o = >       | r = x - \xi > -\delta       | c = }} {{eqn | ll= \leadstoandfrom       | l = \xi + \delta       | o = >       | r = x > \xi - \delta       | c = }} {{end-eqn}} But: :$\openint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta < x < \xi + \delta}$ {{qed}}
{{AimForCont}} $\R_{>0}$ has a smallest element. Let $m$ be that smallest element. Then we have that: :$0 < \dfrac m 2 < m$ But as $0 < \dfrac m 2$ it follows that $\dfrac m 2 \in \R_{>0}$. This contradicts our assertion that $m$ is the smallest element of $\R_{>0}$. Hence the result by Proof by Contradiction. {{qed}}
From the definition of distance: :$\forall x, y \in \R: \map d {x, y} = \size {x - y}$ Thus: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$ Because $I$ is an interval, if $x \notin I$ then $x$ is either an upper bound or a lower bound for $I$. Suppose $x$ is an upper bound for $I$. Let $B$ be the supremum of $I$. Then because $I$ is closed, $B \in I$. So: {{begin-eqn}} {{eqn | ll= \forall y \in I:       | l = \size {x - y}       | r = x - y       | c =  }} {{eqn | r = x - B + B - y       | c =  }} {{eqn | r = x - B + \size {B - y}       | c =  }} {{end-eqn}} Now from Infimum Plus Constant: :$\inf_{y \mathop \in S} \size {x - y} = x - B + \inf_{y \mathop \in S} \size {B - y}$ But we also have: :$x - B \ge 0$ :$\map d {B, S} \ge 0$ :$\map d {x, S} = 0$ So it follows that $x = B$ and so $x \in I$. A similar argument applies if $x$ is a lower bound for $I$. {{Qed}}