hcju/proofwikiqa huggingface dataset:
:600px Consider a sphere $S$ of radius $r$. Consider a cylinder $C$ whose bases are circles of radius $r$ and whose height is $2 r$. Let a double napped cone $K$ each of whose nappes has bases which coincide with the bases of $C$. Let $K$ be removed from $C$ to leave a solid figure $C'$ described as a cylinder with a conical hollow at either end. The bases of $C$ are circles of radius $r$. From Area of Circle, the area of each base is therefore $\pi r^2$. From Volume of Cylinder, $C$ has volume given by: {{begin-eqn}} {{eqn | l = V_C       | r = \pi r^2 \times 2 r       | c =  }} {{eqn | r = 2 \pi r^3       | c =  }} {{end-eqn}} $K$ has twice the volume of each of its nappes. Each of the nappes of $K$ is of height $r$ and has a base of area $\pi r^2$. Hence from Volume of Cone, $K$ has volume given by: {{begin-eqn}} {{eqn | l = V_K       | r = 2 \times \dfrac 1 3 \times \pi r^2 \times r       | c =  }} {{eqn | r = \dfrac 2 3 \pi r^3       | c =  }} {{end-eqn}} Thus the volume $V_{C'}$ of $C'$ is given by: {{begin-eqn}} {{eqn | l = V_{C'}       | r = 2 \pi r^3 - \dfrac 2 3 \pi r^3       | c =  }} {{eqn | r = \dfrac {\left({6 - 2}\right) \pi r^3} 3       | c =  }} {{eqn | r = \dfrac {4 \pi r^3} 3       | c =  }} {{end-eqn}} Let the sphere $S$ be positioned so as to be tangent to the plane containing the base of $C$ Consider a plane parallel to the base of $C$ intersecting both $S$ and $C$ at a distance $x$ from the center of $S$. Let $y$ be the radius of the circle which is the intersection of $S$ with that plane. From Pythagoras's Theorem, $y$ is given by: :$x^2 + y^2 = r^2$ and so: :$y^2 = r^2 - x^2$ Thus the area of this circle is given by: :$A = \pi \left({r^2 - x^2}\right)$ Similarly, consider the intersection of $C'$ with the same plane. This consists of: : the circle of radius $r$ which has an area equal to that of the base of $C$ with: : the circle of radius $x$ which is the intersection of the plane with the cone removed from it. So the area of the remaining circle with the inner circle removed is: :$A' = \pi r^2 - \pi x^2$ from Area of Circle. Thus the intersection of the plane with both $S$ and $C'$ are the same. Thus by Cavalieri's Principle $S$ and $C'$ have the same volume. Hence the result. {{qed}}
From Einstein's Law of Motion, we have: :$\mathbf F = \dfrac {m_0 \mathbf a} {\paren {1 - \dfrac {v^2} {c^2} }^{\tfrac 3 2} }$ where: :$\mathbf F$ is the force on the body :$\mathbf a$ is the acceleration induced on the body :$v$ is the magnitude of the velocity of the body :$c$ is the speed of light :$m_0$ is the rest mass of the body. {{WLOG}}, assume that the body is starting from rest at the origin of a cartesian plane. Assume the force $\mathbf F$ on the body is in the positive direction along the x-axis. To simplify the work, we consider the acceleration as a scalar quantity and write it $a$. Thus, from the Chain Rule for Derivatives: :$a = \dfrac {\d v} {\d t} = \dfrac {\d v} {\d x} \dfrac {\d x} {\d t} = v \dfrac {\d v} {\d x}$ Then from the definition of energy: :$\displaystyle E = \int_0^x F \rd x$ which leads us to: {{begin-eqn}} {{eqn | l = E       | r = m_0 \int_0^x \frac a {\paren {1 - v^2 / c^2}^{\tfrac 3 2} } \rd x       | c =  }} {{eqn | r = m_0 \int_0^v \frac v {\paren {1 - v^2 / c^2}^{\tfrac 3 2} } \rd v       | c =  }} {{eqn | r = m_0 \paren {- \frac {c^2} 2} \int_0^v \paren {1 - \frac {v^2} {c^2} }^{-\tfrac 3 2} \paren {- \frac {2 v \rd v} {c^2} }       | c =  }} {{eqn | r = \intlimits {m_0 c^2 \paren {1 - \frac {v^2} {c^2} }^{- \tfrac 1 2} } 0 v       | c =  }} {{eqn | r = m_0 c^2 \paren {\frac 1 {\sqrt {1 - \frac {v^2} {c^2} } } - 1}       | c =  }} {{eqn | r = c^2 \paren {\frac {m_0} {\sqrt {1 - \frac {v^2} {c^2} } } - m_0}       | c =  }} {{eqn | r = c^2 \paren {m - m_0}       | c = Einstein's Mass-Velocity Equation }} {{eqn | r = M c^2       | c =  }} {{end-eqn}} {{qed}}
From Newton's Law of Universal Gravitation: :$\mathbf F = \dfrac {G m_1 m_2 \mathbf r} {r^3}$ We have that: : The dimension of force is $M L T^{-2}$ : The dimension of displacement is $L$ : The dimension of mass is $M$. Let $x$ be the dimension of $G$. Then we have: :$M L T^{-2} = x \dfrac {M^2 L}{L^3}$ Hence, after algebra: :$x = M^{-1} L^3 T^{-2}$ {{qed}} Category:Dimensional Analysis 5e6cs4qtt18tlhzsszmt8pjekl6plv9
From Hooke's Law, we have: :$\mathbf F = -k \mathbf x$ where: :$\mathbf F$ is a force, of dimension $\mathsf {M L T}^{-2}$ :$k$ is the spring force constant :$\mathbf x$ is a displacement, of dimension $\mathsf L$. Let the dimension of $k$ be $D$. Then we have: :$\mathsf {M L T}^{-2} = D \mathsf L$ from which the result follows immediately: :$D = \mathsf {M T}^{-2}$ {{qed}} Category:Dimensional Analysis qeswg8tbvk3ce0qlt9rlqfjw1er197i
{{begin-eqn}} {{eqn | l = 1       | r = \sin \frac \pi 2       | c = Sine of Half-Integer Multiple of Pi }} {{eqn | r = 2 \sin \frac \pi 4 \cos \frac \pi 4       | c = Double Angle Formula for Sine }} {{eqn | r = 2 \paren {2 \sin \frac \pi 8 \cos \frac \pi 8} \cos \frac \pi 4       | c = Double Angle Formula for Sine }} {{eqn | r = 2 \paren {2 \paren {2 \sin \frac \pi {16} \cos \frac \pi {16} } \cos \frac \pi 8} \cos \frac \pi 4       | c = Double Angle Formula for Sine }} {{eqn | r = \cdots       | c =  }} {{eqn | r = 2^{n - 1} \sin \frac \pi {2^n} \cos \frac \pi {2^n} \cos \frac \pi {2^{n - 1} } \cdots \cos \frac \pi 8 \cos \frac \pi 4       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {2^{n - 1} \sin \frac \pi {2^n} }       | r = \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}       | c =  }} {{eqn | ll= \leadsto       | l = \frac 2 \pi \times \frac {\pi / 2^n} {\map \sin {\pi / 2^n} }       | r = \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}       | c =  }} {{end-eqn}} Then we have from the Half Angle Formula for Cosine that: {{begin-eqn}} {{eqn | l = \cos \frac \pi {2^{k} }       | r = \frac {\sqrt {2 + 2 \map \cos {\pi / 2^{k - 1} } } } 2       | c =  }} {{eqn | r = \frac {\sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 2} } } } } 2       | c =  }} {{eqn | r = \frac {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 3} } } } } } 2       | c =  }} {{end-eqn}} So we can replace all the instances of $\cos \dfrac \pi 4$, $\cos \dfrac \pi 8$, etc. with their expansions in square roots of $2$. Finally, we note that from Limit of Sine of X over X we have: :$\displaystyle \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$ As $n \to \infty$, then, we have that $\dfrac \pi {2^n} \to 0$, and so: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\map \sin {\pi / 2^n} } {\pi / 2^n} = 1$ The result follows after some algebra. {{qed}}
First we note that: :$(1): \quad \dfrac 1 {1 + t^2} = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \dfrac {t^{4 n + 2} } {1 + t^2}$ which is demonstrated here. Now consider the real number $x \in \R: 0 \le x \le 1$. We can integrate expression $(1)$ {{WRT|Integration}} $t$ from $0$ to $x$: :$\displaystyle (2): \quad \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots + \frac {x^{4 n + 1} } {4 n + 1} - \map {R_n} x$ where: :$\displaystyle \map {R_n} x = \int_0^x \frac {t^{4 n + 2} } {1 + t^2} \rd t$ From Square of Real Number is Non-Negative we have that: :$t^2 \ge 0$ and so: :$1 \le 1 + t^2$ From Relative Sizes of Definite Integrals, we have: :$\displaystyle 0 \le \map {R_n} x \le \int_0^x t^{4 n + 2} \rd t$ that is: :$\displaystyle 0 \le \map {R_n} x \le \frac {x^{4n + 3} } {4 n + 3}$ But as $0 \le x \le 1$ it is clear that: :$\dfrac {x^{4 n + 3} } {4 n + 3} \le \dfrac 1 {4 n + 3}$ So: :$0 \le \map {R_n} x \le \dfrac 1 {4 n + 3}$ From Basic Null Sequences and the Squeeze Theorem, $\dfrac 1 {4 n + 3} \to 0$ as $n \to \infty$. This leads us directly to: :$(3): \quad \displaystyle \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$ But from Derivative of Arctangent Function, we also have that: :$\dfrac \d {\d x} \arctan t = \dfrac 1 {1 + t^2}$ and thence from the Fundamental Theorem of Calculus we have: :$\displaystyle \arctan x = \int_0^x \frac {\d t} {1 + t^2}$ From $(3)$ it follows immediately that: :$(4): \quad \arctan x =  x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 \cdots$ Now all we need to do is plug $x = 1$ into $(4)$. {{qed}}
From Power Series Expansion for Real Arctangent Function, we obtain: :$\arctan x = x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 - \cdots$ Substituting $x = 1$ gives the required result. {{qed}}
:500px The area $OAT$ is a quarter-circle whose area is $\dfrac \pi 4$ by Area of Circle. Now consider the area $C$ of the segment $OPQT$, bounded by the arc $OT$ and the chord $OT$. Consider the area $OPQ$, bounded by the line segments $OP$ and $OQ$ and the arc $PQ$. As $P$ and $Q$ approach each other, the arc $PQ$ tends towards the straight line segment $\d s = PQ$. We can therefore consider the area $OPQ$ as a triangle. We extend the line segment $PQ$ and drop a perpendicular $OR$ to $O$. Using Area of Triangle in Terms of Side and Altitude, we see that the area $\d C$ of $\triangle OPQ$ is given by: :$\d C = \triangle OPQ = \dfrac {OR \cdot PQ} 2 = \dfrac {OR \cdot \d s} 2$ We also note from elementary Euclidean geometry that $\triangle ORS$ is similar to the small triangle on $PQ$. Thus: :$\dfrac {\d s} {\d x} = \dfrac {OS} {OR} \iff OR \cdot \d s = OS \cdot \d x$ Thus: :$\d C = \dfrac {OS \cdot \d x} 2 = \dfrac {y \rd x} 2$ where $y = OS$. We set the horizontal coordinate of $P$ as equal to $x$. Thus the total area $C$ is equal to the total of all the areas of these small triangles as $x$ increases from $0$ to $1$. So: :$\displaystyle C = \int \rd C = \frac 1 2 \int_0^1 y \rd x$ Now we use Integration by Parts to swap $x$ and $y$: :$\displaystyle C = \intlimits {\frac 1 2 x y} 0 1 - \frac 1 2 \int_0^1 x \rd y = \frac 1 2 - \frac 1 2 \int_0^1 x \rd y$ It can be seen that the limits on this new integral have to be $0$ and $1$ from the geometry of the situation. Now we note that: :$y = \tan \dfrac \phi 2$ :$x = 1 - \cos \phi = 2 \sin^2 \dfrac \phi 2$ (from Double Angle Formula for Cosine: Corollary 2) Thus: {{begin-eqn}} {{eqn | l = \tan^2 \frac \phi 2       | r = \frac {\sin^2 \frac \phi 2}{\cos^2 \frac \phi 2}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \sin^2 \frac \phi 2 \sec^2 \frac \phi 2       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \sin^2 \frac \phi 2 \paren {1 + \tan^2 \frac \phi 2}       | c = Difference of Squares of Secant and Tangent }} {{end-eqn}} and so: :$\dfrac x 2 = \dfrac {y^2} {1 + y^2}$ Using Sum of Geometric Sequence: :$\dfrac {y^2} {1 + y^2} = y^2 - y^4 + y^6 - y^8 + \cdots$ This gives us: {{begin-eqn}} {{eqn | l = C       | r = \frac 1 2 - \int_0^1 \paren {y^2 - y^4 + y^6 - y^8 + \cdots} \d y       | c =  }} {{eqn | r = \frac 1 2 - \intlimits {\frac {y^3} 3 - \frac {y^5} 5 + \frac {y^7} 7 - \frac {y^9} 9 + \cdots} 0 1       | c =  }} {{eqn | r = \frac 1 2 - \paren {\frac 1 3 - \frac 1 5 + \frac 1 7 - \frac 1 9 + \cdots}       | c =  }} {{eqn | r = \frac 1 2 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots       | c =  }} {{end-eqn}} Remember, $C$ is the area of the segment $OPQT$. Now we add to it the area of $\triangle OTA$, which trivially equals $\dfrac 1 2$, to get the area of the quarter circle which we know as equal to $\dfrac \pi 4$. Putting it all together, this gives us: :$\dfrac \pi 4 = 1 - \dfrac 1 3 + \dfrac 1 5 - \dfrac 1 7 + \dfrac 1 9 - \cdots$ {{qed}}
From Continuously Differentiable Curve has Finite Arc Length, $s$ exists and is given by: {{begin-eqn}} {{eqn | l = s       | r = \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} }^2} \rd u }} {{eqn | ll= \leadsto       | l = \frac {\d s} {\d x}       | r = \frac {\d} {\d x} \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} } ^2} \rd u       | c = differentiating both sides {{WRT|Differentiation}} $x$ }} {{eqn | r = \sqrt {1 + \paren {\frac {\d y} {\d x} }^2}       | c = Fundamental Theorem of Calculus: First Part }} {{end-eqn}} {{qed}}
Consider a length $\d s$ of $C$, short enough for it to be approximated to a straight line segment: :250px By Pythagoras's Theorem, we have: :$\d s^2 = \d x^2 + \d y^2$ Dividing by $\d x^2$ we have: {{begin-eqn}} {{eqn | l = \paren {\frac {\d s} {\d x} }^2       | r = \paren {\frac {\d x} {\d x} }^2 + \paren {\frac {\d y} {\d x} }^2       | c =  }} {{eqn | r = 1 + \paren {\frac {\d y} {\d x} }^2       | c =  }} {{end-eqn}} Hence the result, by taking the principal square root of both sides. {{qed}}
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $+$ is commutative on $\Q$. {{qed}}
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $+$ is associative on $\Q$. {{qed}}
=== Natural Numbers === {{:Subtraction on Numbers is Anticommutative/Natural Numbers}} === Integers, Rationals, Reals, Complex Numbers === {{:Subtraction on Numbers is Anticommutative/Integral Domains}}
From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers. From Integers under Addition form Abelian Group, the integers under addition form an abelian group, from which commutativity follows a priori. {{qed}}
Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$. Then: {{begin-eqn}} {{eqn | l = x + y       | r = \eqclass {a, b} {} + \eqclass {c, d} {}       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{eqn | r = \eqclass {a + c, b + d} {}       | c = {{Defof|Integer Addition}} }} {{eqn | r = \eqclass {c + a, d + b} {}       | c = Natural Number Addition is Commutative }} {{eqn | r = \eqclass {c, d} {} + \eqclass {a, b} {}       | c = {{Defof|Integer Addition}} }} {{eqn | r = y + x       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{end-eqn}} {{qed}}
From the formal definition of integers, $\left[\!\left[{a, b}\right]\!\right]$ is an equivalence class of ordered pairs of natural numbers. From Integers under Addition form Abelian Group, the integers under addition form a group, from which associativity follows a priori. {{qed}}
Let $x = \left[\!\left[{a, b}\right]\!\right]$, $y = \left[\!\left[{c, d}\right]\!\right]$ and $z = \left[\!\left[{e, f}\right]\!\right]$ for some $x, y, z \in \Z$. Then: {{begin-eqn}} {{eqn | l = x + \left({y + z}\right)       | r = \left[\!\left[{a, b}\right]\!\right] + \left({\left[\!\left[{c, d}\right]\!\right] + \left[\!\left[{e, f}\right]\!\right]}\right)       | c = Definition of Integers }} {{eqn | r = \left[\!\left[{a, b}\right]\!\right] + \left[\!\left[{c + e, d + f}\right]\!\right]       | c = Definition of Integer Addition }} {{eqn | r = \left[\!\left[{a + \left({c + e}\right), b + \left({d + f}\right)}\right]\!\right]       | c = Definition of Integer Addition }} {{eqn | r = \left[\!\left[{\left({a + c}\right) + e, \left({b + d}\right) + f}\right]\!\right]       | c = Natural Number Addition is Associative }} {{eqn | r = \left[\!\left[{a + c, b + d}\right]\!\right] + \left[\!\left[{e, f}\right]\!\right]       | c = Definition of Integer Addition }} {{eqn | r = \left({\left[\!\left[{a, b}\right]\!\right] + \left[\!\left[{c, d}\right]\!\right]}\right) + \left[\!\left[{e, f}\right]\!\right]       | c = Definition of Integer Addition }} {{eqn | r = \left({x + y}\right) + z       | c = Definition of Integers }} {{end-eqn}} {{qed}}
Proof by induction. Consider the natural numbers $\N$ defined as the elements of the minimal infinite successor set $\omega$. From the definition of addition in $\omega$, we have that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m + 0       | r = m }} {{eqn | l = m + n^+       | r = \paren {m + n}^+ }} {{end-eqn}} For all $n \in \N$, let $\map P n$ be the proposition: :$\forall m \in \N: m + n = n + m$ === Basis for the Induction === From Natural Number Addition Commutes with Zero, we have: :$\forall m \in \N: m + 0 = m = 0 + m$ Thus $\map P 0$ is seen to be true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true. So this is our induction hypothesis $\map P k$: :$\forall m \in \N: m + k = k + m$ Then we need to show that $\map P {k^+}$ follows directly from $\map P k$: :$\forall m \in \N: m + k^+ = k^+ + m$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = k^+ + m       | r = \paren {k + m}^+       | c = Natural Number Addition Commutativity with Successor }} {{eqn | r = \paren {m + k}^+       | c = from the induction hypothesis }} {{eqn | r = m + k^+       | c = by definition }} {{end-eqn}} So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n \in \N: m + n = n + m$ {{qed}}
Consider the natural numbers defined as a naturally ordered semigroup. By definition, the operation in a naturally ordered semigroup is commutative. Hence the result. {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Let $x \in \N_{> 0}$ be arbitrary. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$x + n = n + x$ === Basis for the Induction === From Natural Number Commutes with 1 under Addition, we have that: :$\forall x \in \N_{> 0}: x + 1 = 1 + x$ and so $P \left({1}\right)$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$x + k = k + x$ Then we need to show: :$x + \left({k + 1}\right) = \left({k + 1}\right) + x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = x + \left({k + 1}\right)       | r = \left({x + k}\right) + 1       | c = Natural Number Addition is Associative }} {{eqn | r = \left({k + x}\right) + 1       | c = Induction Hypothesis }} {{eqn | r = k + \left({x + 1}\right)       | c = Natural Number Addition is Associative }} {{eqn | r = k + \left({1 + x}\right)       | c = Basis for the Induction }} {{eqn | r = \left({k + 1}\right) + x       | c = Natural Number Addition is Associative }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. {{qed}}
Consider the von Neumann construction of natural numbers $\N$, as elements of the minimal infinite successor set $\omega$. We are to show that: :$\paren {x + y} + n = x + \paren {y + n}$ for all $x, y, n \in \N$. From the definition of addition, we have that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m + 0       | r = m       | c =  }} {{eqn | l = m + n^+       | r = \paren {m + n}^+       | c =  }} {{end-eqn}} Let $x, y \in \N$ be arbitrary. For all $n \in \N$, let $\map P n$ be the proposition: :$\paren {x + y} + n = x + \paren {y + n}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \paren {x + y} + 0       | r = x + y       | c =  }} {{eqn | r = x + \paren {y + 0}       | c =  }} {{end-eqn}} and so $\map P 0$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k^+}$ is true. So this is our induction hypothesis: :$\paren {x + y} + k = x + \paren {y + k}$ Then we need to show: :$\paren {x + y} + \paren {k^+} = x + \paren {y + \paren {k^+} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {x + y} + k^+       | r = \paren {\paren {x + y} + k}^+       | c = {{Defof|Addition in Minimal Infinite Successor Set}} }} {{eqn | r = \paren {x + \paren {y + k} }^+       | c = induction Hypothesis }} {{eqn | r = x + \paren {\paren {y + k}^+}       | c = {{Defof|Addition in Minimal Infinite Successor Set}} }} {{eqn | r = x + \paren {y + k^+}       | c = {{Defof|Addition in Minimal Infinite Successor Set}} }} {{end-eqn}} So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
Consider the natural numbers defined as a naturally ordered semigroup. By  definition, the operation in a semigroup is associative. Hence the result. {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Let $x, y \in \N_{> 0}$ be arbitrary. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\paren {x + y} + n = x + \paren {y + n}$ === Basis for the Induction === From Axiom $\text C$, we have by definition that: :$\forall x, y \in \N_{> 0}: \paren {x + y} + 1 = x + \paren {y + 1}$ and so $\map P 1$ holds. 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: :$\paren {x + y} + k = x + \paren {y + k}$ Then we need to show: :$\paren {x + y} + \paren {k + 1} = x + \paren {y + \paren {k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {x + y} + \paren {k + 1}       | r = \paren {\paren {x + y} + k} + 1       | c = Basis for the Induction }} {{eqn | r = \paren {x + \paren {y + k} } + 1       | c = Induction Hypothesis }} {{eqn | r = x + \paren {\paren {y + k} + 1}       | c = Basis for the Induction }} {{eqn | r = x + \paren {y + \paren {k + 1} }       | c = Basis for the Induction }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $\times$ is commutative on $\Q$. {{qed}}
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $\times$ is associative on $\Q$. {{qed}}
Let $A, B$ be two (natural) numbers, and let $A$ by multiplying $B$ make $C$, and $B$ by multiplying $A$ make $D$. We need to show that $C = D$. :350px We have that $A \times B = C$. So $B$ measures $C$ according to the units of $A$. But the unit $E$ also measures $A$ according to the units in it. So $E$ measures $A$ the same number of times that $B$ measures $C$. Therefore from {{EuclidPropLink|book=VII|prop=15|title=Alternate Ratios of Multiples}} $E$ measures $B$ the same number of times that $A$ measures $C$. We also have that $A$ measures $D$ according to the units of $B$. But the unit $E$ also measures $B$ according to the units in it. Therefore from {{EuclidPropLink|book=VII|prop=15|title=Alternate Ratios of Multiples}} $E$ measures $B$ the same number of times that $A$ measures $D$. But we also have that $E$ measures $B$ the same number of times that $A$ measures $C$. So $A$ measures $C$ and $D$ the same number of times. Therefore $C = D$. {{qed}}