hcju/proofwikiqa huggingface dataset:
We start by stating Newton's Third Law of Motion in all its detail. We consider a collection of massive bodies denoted by the subscripts $1$ to $N$. These bodies interact with each other and exert forces on each other and these forces occur in equal and opposite pairs. The force $F_{i j}$ exerted by body $i$ on body $j$ is related to the force exerted by body $j$ on body $i$ by: :$(1): \quad \vec {F_{i j}} = -\vec {F_{j i}}$ The final part of Newton's Third Law of Motion is that these equal and opposite forces act through the line that connects the two bodies in question. This can be stated thus: :$(2): \quad \vec {F_{i j}} = a_{i j} \paren {\vec {r_j} - \vec {r_i} }$ where: :$\vec{r_i}$ is the position of body $i$ :$a_{i j} $ is the magnitude of the force. As a consequence of $(1)$: :$a_{j i} = a_{i j}$ Let the total torque $\vec {\tau_{\operatorname {total} } }$ on the system be measured about an origin located at $\vec {r_0}$. Thus: {{begin-eqn}} {{eqn | l = \vec {\tau_{\operatorname {total} } }       | r = \sum_i \vec{\tau_i}       | c =  }} {{eqn | r = \sum_i \paren {\paren {\vec {r_i} - \vec {r_0} } \times \sum_j \vec {F_{j i} } }       | c =  }} {{eqn | r = \sum_{i, j} \paren {\paren {\paren {\vec {r_j} - \vec {r_0} } - \paren {\vec {r_j} - \vec {r_i} } } \times \vec {F_{j i} } }       | c =  }} {{eqn | r = \sum_{i, j} \paren {\paren {\paren {\vec {r_j} - \vec {r_0} } - \paren {\vec {r_j} - \vec {r_i} } } \times a_{i j} \paren {\vec {r_j} - \vec {r_i} } }       | c =  }} {{eqn | r = \sum_{i, j} \paren {\vec {r_j} - \vec {r_0} } \times  a_{i j} \paren {\vec {r_j} - \vec {r_i} }       | c =  }} {{eqn | r = \sum_{i, j} \vec {r_j} \times  a_{i j} \paren {\vec {r_j} - \vec {r_i} } - \paren {\vec {r_0} \times \sum_{i, j} \vec {F_{i j} } }       | c =  }} {{end-eqn}} By hypothesis there is no external force. Thus the second term disappears, and: {{begin-eqn}} {{eqn | l = \vec {\tau_{\operatorname {total} } }       | r = \sum_{i, j} \vec {r_j} \times a_{i j} \paren {\vec {r_j} - \vec {r_i} }       | c =  }} {{eqn | r = \sum_{i \mathop > j} \vec {r_j} \times a_{i j} \paren {\vec {r_j} - \vec {r_i} } + \vec {r_i} \times  a_{j i} \paren {\vec {r_j} - \vec {r_i} }       | c =  }} {{eqn | r = \sum_{i \mathop > j} \vec {r_j} \times \vec {r_i} \paren {a_{j i} - a_{i j} }       | c =  }} {{eqn | r = \vec 0       | c =  }} {{end-eqn}} In summary, in a system of masses in which there is no external force, the total torque on the system is equal to $0$. This is because the pair of torque between two bodies must cancel out. Since the rate of change of angular momentum is proportional to the torque, the angular momentum of a system is conserved when no external force is applied. {{qed}} Category:Classical Mechanics tuykqogotz150r2cu0kldfktpohk7cb
It is straightforward to show that $\preccurlyeq$ is a total ordering on both $T_n$ and $T$. It remains to investigate the nature of $\preccurlyeq$ as to whether or not it is a well-ordering. === Proof for Finite Product === {{:Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering}} === Proof for Infinite Product === {{:Infinite Lexicographic Order on Well-Ordered Sets is not Well-Ordering}} Category:Lexicographic Order Category:Well-Orderings 6qy4ped8et7hakbkcqi35tt1gmjml8y
We use a Proof by Contraposition. Thus, we show that if a basis expansion of a (real) number terminates or recurs, then that number is rational. Suppose $x \in \R$ were to terminate in some number base $b$. Then (using the notation of that definition): :$\exists k \in \N: f_k = 0$ and so we can express $x$ precisely as: :$x = \left[{s . d_1 d_2 d_3 \ldots d_{k - 1} }\right]_b$ This means: :$x = s + \dfrac {d_1} b + \dfrac {d_2} {b^2} + \dfrac {d_3} {b^3} + \cdots + \dfrac {d_{k - 1} } {b^{k - 1} }$ This is the same as: :$x = \dfrac {s b^{k - 1} + d_1 b^{k - 2} + d_2 b^{k - 3} + \cdots + d_{k - 1} } {b^{k - 1} }$ Both numerator and denominator are integers and so $x$ is a rational number. Now suppose $x \in \R$ were to recur in some number base $b$. Let $y$ be the recurring part of $x$. That is, $y$ is obtained from $x$ by first subtracting $\floor x$ and then subtracting the non-recurring part of the basis expansion (if any). Then we can express $y$ precisely as: :$y = \sqbrk {0 . 0 0 \ldots 0 d_1 d_2 \ldots d_{k - 1} d_1 d_2 \ldots d_{k - 1} \ldots}_b$ for some number $m$ (possibly $m = 0$) of zeroes following the point. We define: :$z:= \sqbrk {0 . d_1 d_2 \ldots d_{k - 1} }_b = \dfrac {d_1} b + \dfrac {d_2} {b^2} + \dfrac {d_3} {b^3} + \cdots + \dfrac {d_{k - 1} } {b^{k - 1} }$ Then $z$ is a rational number. Furthermore: :$y = \dfrac z {b^m} + \dfrac z {b^{m + k} } + \dfrac z {b^{m + 2 k} } + \cdots = \dfrac z {b^m} \paren {1 + \dfrac 1 {b^k} + \dfrac 1 {b^{2 k } } + \cdots} = \dfrac z {b^m} \cdot \dfrac 1 {1 - \dfrac 1 {b^k} }$ so $y$ is a rational number as well. Thus as $x - y$ is rational, $x$ is also rational. {{Qed}} Category:Arithmetic Category:Proofs by Contraposition Category:Basis Expansions 16apnzaix4hlgyx904es6ug7zb69rbo
Take the determinant $\det \left({C_n}\right)$: :$\det \left({C_n}\right) = \begin{vmatrix} x + y & y & y & \cdots & y \\ y & x + y & y & \cdots & y \\ y & y & x + y & \cdots & y \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y & y & y & \cdots & x + y \end{vmatrix}$ Subtract column $1$ from columns $2$ to $n$. From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant: :$\det \left({C_n}\right) = \begin{vmatrix} x + y & -x & -x & \cdots & -x \\ y & x & 0 & \cdots & 0 \\ y & 0 & x & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y & 0 & 0 & \cdots & x \end{vmatrix}$ Add rows $2$ to $n$ to row $1$. Again, from Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant: :$\det \left({C_n}\right) = \begin{vmatrix} x + n y & 0 & 0 & \cdots & 0 \\ y & x & 0 & \cdots & 0 \\ y & 0 & x & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y & 0 & 0 & \cdots & x \end{vmatrix}$ This is now the determinant of a (lower) triangular matrix. From Determinant of Triangular Matrix, it follows immediately that: :$\det \left({C_n}\right) = x^{n-1} \left ({x + n y}\right)$ {{qed}}
Let $A, B \in \powerset S$. Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$. We also have $A \cup B \subseteq S \iff A \subseteq S \land B \subseteq S$ from Union is Smallest Superset. Thus $A \cup B \in \powerset S$, and closure is proved. {{qed}}
Let $\mathbf A = \paren {a_{i j} }$. Let $A_{i j}$ denote the cofactor of $a_{i j} \in \mathbf A$. === Right Multiplication === We show that $\mathbf A \cdot \adj {\mathbf A} = \map \det {\mathbf A} \cdot \mathbf I_n$. Let $i, j \in \set {1, \ldots, n}$.  If $i = j$, expanding $\map \det {\mathbf A}$ along row $i$ shows that: :$\displaystyle \map \det {\mathbf A} = \sum_{k \mathop = 1}^n a_{i k} A_{i k}$ If $i \ne j$, define $\mathbf A'$ as the matrix obtained by replacing row $j$ of $\mathbf A$ with row $i$ of $\mathbf A$. Then $\mathbf A' = \begin{bmatrix} a' \end{bmatrix}_n$ has two identical rows, so: {{begin-eqn}} {{eqn | l = 0_R       | r = \map \det {\mathbf A'}       | c = Square Matrix with Duplicate Rows has Zero Determinant }} {{eqn | r = \sum_{k \mathop = 1}^n a'_{j k} A'_{j k}       | c = expanding $\map \det {\mathbf A'}$ along row $j$ }} {{eqn | r = \sum_{k \mathop = 1}^n a_{i k} A_{j k} }} {{end-eqn}} By definition of matrix product, element $\tuple {i, j}$ of $\mathbf A \cdot \adj {\mathbf A}$ is: :$\displaystyle \sum_{k \mathop = 1}^n a_{i k} A_{j k} = \begin{cases} 0_R & \text {for} &  i \ne j \\ \map \det {\mathbf A} & \text {for} &  i = j \end{cases}$ Hence: :$\mathbf A \cdot \adj {\mathbf A} = \map \det {\mathbf A} \cdot \mathbf I_n$ {{qed|lemma}} === Left Multiplication === We show that $\adj {\mathbf A} \cdot \mathbf A = \map \det {\mathbf A} \cdot \mathbf I_n$. Let $i, j \in \set {1, \ldots, n}$. If $i = j$, expanding $\map \det {\mathbf A}$ along column $j$ shows that: :$\displaystyle \map \det {\mathbf A} = \sum_{k \mathop = 1}^n a_{k j} A_{k j}$ If $i \ne j$, define $\mathbf A'$ as the matrix obtained by replacing column $i$ of $\mathbf A$ with column $j$ of $\mathbf A$. Then $\mathbf A' = \begin{bmatrix} a' \end{bmatrix}_n$ has two identical columns, so: {{begin-eqn}} {{eqn | l = 0_R       | r = \map \det {\mathbf A'}       | c = Square Matrix with Duplicate Columns has Zero Determinant }} {{eqn | r = \sum_{k \mathop = 1}^n a'_{k i} A'_{k i}       | c = expanding $\map \det {\mathbf A'}$ along column $i$ }} {{eqn | r = \sum_{k \mathop = 1}^n a_{k j} A_{k i} }} {{end-eqn}} By definition of matrix product, element $\tuple {i, j}$ of $\adj {\mathbf A} \cdot \mathbf A$ is: :$\displaystyle \sum_{k \mathop = 1}^n A_{k i} a_{k j} = \begin{cases} 0_R & \text {for} &  i \ne j \\ \map \det {\mathbf A} & \text {for} &  i = j \end{cases}$ Hence: :$\adj {\mathbf A} \cdot \mathbf A = \map \det {\mathbf A} \cdot \mathbf I_n$ {{qed}}
Follows directly from the definition of matrix multiplication: :$\displaystyle \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$ In this case, $m = n$ and $a_{i k} = b_{k j} = 1$. Hence: :$\displaystyle c_{i j} = \sum_{k \mathop = 1}^n 1 \times 1 = n$ {{qed}} Category:Matrix Algebra 4o70eshdycot8iszwvk6jr2n1p8qn7j
From the definition of the combinatorial matrix: :$C_n = x \mathbf I_n + y \mathbf J_n$ where: :$\mathbf I_n$ is the unit matrix of order $n$ :$\mathbf J_n$ is the square ones matrix of order $n$. From Square of Ones Matrix we have $\mathbf J_n^2 = n \mathbf J_n$. Hence: {{begin-eqn}} {{eqn | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {x \mathbf I_n - y \mathbf J_n}       | r = x^2 \mathbf I_n - n y^2 \mathbf J_n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {x \mathbf I_n - y \mathbf J_n} + n y^2 \mathbf J_n       | r = x^2 \mathbf I_n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {x \mathbf I_n - y \mathbf J_n} + n y \mathbf I_n y \mathbf J_n       | r = x^2 \mathbf I_n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {x \mathbf I_n - y \mathbf J_n} + x n y \mathbf I_n + n y \mathbf I_n y \mathbf J_n       | r = x^2 \mathbf I_n + x n y \mathbf I_n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {x \mathbf I_n - y \mathbf J_n} + n y \mathbf I_n \paren {x \mathbf I_n + y \mathbf J_n}       | r = x^2 \mathbf I_n + x n y \mathbf I_n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {\paren {x + n y} \mathbf I_n - y \mathbf J_n}       | r = x \paren {x + n y} \mathbf I_n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \mathbf I_n + y \mathbf J_n} \paren {\frac {\paren {x + n y} } {x \paren {x + n y} } \mathbf I_n - \frac y {x \paren {x + n y} } \mathbf J_n}       | r = \mathbf I_n       | c =  }} {{end-eqn}} So we have found a specification for the matrix which, when multiplied by $C_n$, yields $\mathbf I_n$. By using the identities $\mathbf I_n = \sqbrk {\delta_{i j} }_n$ and $\mathbf J_n = \sqbrk 1_n$ we obtain the stated result: :$b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$ {{qed}}
Let $y = \floor x$. By Floor Function is Integer, $y$ is an integer. Then from Real Number is Integer iff equals Floor: :$\floor y = y$ So: :$\floor {\floor x} = \floor x$ {{qed}}
Let $ABC$ be a circle, and $D$ a point in it such that $DA = DB = DC$. Then $D$ can be proved to be the center of circle $ABC$ as follows. :300px Join $AB$ and $BC$, and bisect them at $E$ and $F$. Join $ED$ and $FD$, and produce them in both directions to $GK$ and $HL$. We have that $AE = EB$, $AD = BD$ and $ED$ is common. From Triangle Side-Side-Side Equality, $\triangle AED = \triangle BED$ and so $\angle AED = \angle BED$. So from {{EuclidDefLink|I|10|Right Angle}}, each of $\angle AED$ and $\angle BED$ is a right angle. Hence from the porism to Finding Center of Circle, the center of circle $ABC$ is on $GK$. The same applies to $\triangle BFD = \triangle CFD$, and so by the same construction, the center of circle $ABC$ is also on $HL$. The straight lines $GK$ and $HL$ have only point $D$ in common. Hence the result. {{qed}} {{Euclid Note|9|III|As pointed out by {{AuthorRef|Augustus De Morgan}}, this theorem is logically equivalent to Relative Lengths of Lines Inside Circle.}}
Let $S_k$ denote the partial sums of the $x$s. Using Summation by Parts: :$\displaystyle \frac 1 {b_n} \sum_{k \mathop = 1}^n b_k x_k = S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{n - 1} \paren {b_{k + 1} - b_k} S_k$ Now, pick any $\epsilon > 0$. Choose $N$ such that $S_k$ is $\epsilon$-close to $s$ for $k > N$. This can be done, as the sequence $S_k$ converges to $s$. Then the {{RHS}} is: {{begin-eqn}} {{eqn | o =       | r = S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{N - 1} \paren {b_{k + 1} - b_k} S_k - \frac 1 {b_n} \sum_{k \mathop = N}^{n - 1} \paren {b_{k + 1} - b_k} S_k       | c =  }} {{eqn | r = S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{N - 1} \paren {b_{k + 1} - b_k} S_k - \frac 1 {b_n} \sum_{k \mathop = N}^{n - 1} \paren {b_{k + 1} - b_k} s - \frac 1 {b_n} \sum_{k \mathop = N}^{n - 1} \paren {b_{k + 1} - b_k} \paren {S_k - s}       | c =  }} {{eqn | r = S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{N - 1} \paren {b_{k + 1} - b_k} S_k - \frac {b_n - b_N} {b_n} s - \frac 1 {b_n} \sum_{k \mathop = N}^{n - 1} \paren {b_{k + 1} - b_k} \paren {S_k - s}       | c =  }} {{end-eqn}} Now, let $n \to \infty$. The first term goes to $s$, which cancels with the third term. The second term goes to zero (as the sum is a fixed value). Since the $b$ sequence is increasing, the last term is bounded by $\epsilon \dfrac {b_n - b_N} {b_n} \le \epsilon$. {{qed}} {{MissingLinks|Needs tightening up a bit, linking to existing results, some definitions need doing.}} {{Namedfor|Leopold Kronecker|cat = Kronecker}}
We pick the elements of $S$ in any arbitrary order. There are $n$ elements of $S$, so there are $n$ options for the first element. Then there are $n - 1$ elements left in $S$ that we haven't picked, so there are $n-1$ options for the second element. Then there are $n - 2$ elements left, so there are $n - 2$ options for the third element. And so on, to the $r$th element of our selection: we now have $n - \paren {r - 1}$ possible choices. Each mapping is independent of the choices made for all the other mappings, so by the Product Rule for Counting, the total number of ordered selections from $S$: {{begin-eqn}} {{eqn | l = {}^r P_n       | r = n \paren {n - 1} \paren {n - 2} \cdots \paren {n - r + 1} }} {{eqn | r = n \paren {n - 1} \paren {n - 2} \ldots \paren {n - r + 1} \dfrac {\paren {n - r}!} {\paren {n - r}!}        | c = multiplying top and bottom by $\paren {n - r}!$ }} {{eqn | r = \dfrac {n!} {\paren {n - r}!}       | c = simplifying the numerator }} {{end-eqn}} {{qed}}
From the definition, an $r$-permutation of $S$ is an ordered selection of $r$ elements of $S$. It can be seen that an $r$-permutation is an injection from a subset of $S$ into $S$. From Cardinality of Set of Injections, we see that the number of $r$-permutations ${}^r P_n$ on a set of $n$ elements is given by: :${}^r P_n = \dfrac {n!} {\left({n-r}\right)!}$ From this definition, it can be seen that a bijection $f: S \to S$ is an $n$-permutation. Hence the number of $n$-permutations on a set of $n$ elements is: :${}^n P_n = \dfrac {n!} {\left({n - n}\right)!} = n!$ {{Qed}}
{{begin-eqn}} {{eqn | l = {}^{n - 1} P_n       | r = \dfrac {n!} {\paren {n - \paren {n - 1} }!}       | c = Number of Permutations }} {{eqn | r = \dfrac {n!} {1!}       | c =  }} {{eqn | r = n!       | c =  }} {{eqn | r = {}^n P_n       | c = Number of Permutations }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = {}^{n - 1} P_n       | r = n^{\underline {n - 1} }       | c = Number of Permutations: $n^{\underline {n - 1} }$ denotes Falling Factorial }} {{eqn | r = n!       | c = Integer to Power of Itself Less One Falling is Factorial }} {{eqn | r = {}^n P_n       | c = Number of Permutations }} {{end-eqn}} {{qed}}
The following is an inductive method of creating all the permutations of $n$ objects. === Base Case === There is clearly one way to arrange one object in order. === Inductive Hypothesis === We assume that we have constructed all $n!$ permutations of $n$ objects. === Induction Step === {{WLOG}}, let a set $S_n$ of $n$ objects be $\left\{{1, 2, \ldots, n}\right\}$. Take a permutation of $S_n$: :$a_1 \, a_2 \, a_3 \, \ldots \, a_n$ Now we take the number $n + 1$. We can form $n + 1$ permutations from this one by putting $n + 1$ in all places possible: :$a_{n + 1} \, a_1 \, a_2 \, a_3 \, \ldots \, a_n, \quad a_1 \, a_{n + 1} \, a_2 \, a_3 \, \ldots \, a_n, \quad a_1 \, a_2 \, a_{n + 1} \, a_3 \, \ldots \, a_n, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, a_{n + 1}$ It is clear that all permutations of $n + 1$ objects can be obtained in this manner, and no permutation is obtained more than once. As there are ${}^n P_n$ permutations on $n$ objects, there are $\left({n + 1}\right) \times {}^n P_n$ permutations on $n + 1$ objects. Hence by induction, and the recursive definition of the factorial: :${}^n P_n = n!$ {{qed}}
The following is an inductive method of creating all the permutations of $n$ objects. === Base Case === There is clearly one way to arrange one object in order. === Inductive Hypothesis === We assume that we have constructed all $n!$ permutations of $n$ objects. === Induction Step === {{WLOG}}, let a set $S_n$ of $n$ objects be $\left\{{1, 2, \ldots, n}\right\}$. Take a permutation of $S_n$: :$a_1 \, a_2 \, a_3 \, \ldots \, a_n$ Construct the array: :$a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 1 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 3 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 5 2, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \left({n + \dfrac 1 2}\right)$ This will contain $n + 1$ objects. Now rename the elements of each permutation using the numbers $\left\{ {1, 2, \ldots, n}\right\}$ preserving order. It is clear that all permutations of $n + 1$ objects can be obtained in this manner, and no permutation is obtained more than once. As there are ${}^n P_n$ permutations on $n$ objects, there are $\left({n + 1}\right) \times {}^n P_n$ permutations on $n + 1$ objects. Hence by induction, and the recursive definition of the factorial: :${}^n P_n = n!$ {{qed}}
Note that although the summation given in the statement of the theorem is given as an infinite sum, in fact it terminates after a finite number of terms (because when $p^k > n$ we have $0 < n/p^k < 1$). From Number of Multiples less than Given Number, we have that $\floor{\dfrac n {p^k} }$ is the number of integers $m$ such that $0 < m \le n$ which are multiples of $p^k$. We look more closely at $n!$: :$n! = 1 \times 2 \times \ldots \times \paren {n - 1} \times n$ We see that any integer $m$ such that $0 < m \le n$ which is divisible by $p^j$ and not $p^{j + 1}$ must be counted exactly $j$ times. That is: :once in $\floor {\dfrac n p}$ :once in $\floor {\dfrac n {p^2} }$ $\ldots$ :once in $\floor {\dfrac n {p^j} }$ And that is all the occurrences of $p$ as a factor of $n!$. Thus: :$\mu = \floor {\dfrac n p} + \floor {\dfrac n {p^2} } + \dotsb + \floor {\dfrac n {p^j} }$ Hence the result. {{qed}}
For all $n \in \N$, let $P \left({n}\right)$ be the proposition: : $\displaystyle \sum_{i \mathop = 0}^n \binom n i = 2^n$ $P(0)$ is true, as this just says $\dbinom 0 0 = 1$. This holds by definition. === Basis for the Induction === $P(1)$ is true, as this just says $\dbinom 1 0 + \dbinom 1 1 = 2$. This holds by Binomial Coefficient with Zero and Binomial Coefficient with One (or Binomial Coefficient with Self). 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: :$\displaystyle \sum_{i \mathop = 0}^k \binom k i = 2^k$ Then we need to show: :$\displaystyle \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i = 2^{k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i       | r = \binom {k + 1} 0 + \sum_{i \mathop = 1}^k \binom {k + 1} i + \binom {k + 1} {k + 1}       | c =  }} {{eqn | r = \binom {k + 1} 0 + \sum_{i \mathop = 1}^k \left({\binom k {i - 1} + \binom k i}\right) + \binom {k + 1} {k + 1}       | c = Pascal's Rule }} {{eqn | r = \left({\sum_{i \mathop = 0}^{k - 1} \binom k i + \binom {k + 1} {k + 1} }\right) + \left({\binom {k + 1} 0 + \sum_{i \mathop = 1}^k \binom k i}\right)       | c = Translation of Index Variable of Summation }} {{eqn | r = \left({\sum_{i \mathop = 0}^{k - 1} \binom k i + \binom k k}\right) + \left({\binom k 0 + \sum_{i \mathop = 1}^k \binom k i}\right)       | c = as $\displaystyle \binom {k + 1} {k + 1} = \binom k k = 1$ and $\displaystyle \binom {k + 1} 0 = \binom k 0 = 1$ }} {{eqn | r = \sum_{i \mathop = 0}^k \binom k i + \sum_{i \mathop = 0}^k \binom k i       | c = Translation of Index Variable of Summation }} {{eqn | r = 2^k + 2^k       | c =  }} {{eqn | r = 2^{k + 1}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i = 2^n$ {{qed}}
Let $S$ be a set with $n$ elements. From the definition of $r$-combination, $\displaystyle \sum_{i \mathop = 0}^n \binom n i$ is the total number of subsets of $S$. Hence $\displaystyle \sum_{i \mathop = 0}^n \binom n i$ is equal to the cardinality of the power set of $S$. Hence the result. {{qed}}
From the Binomial Theorem, we have that: :$\displaystyle \forall n \in \Z_{\ge 0}: \left({x + y}\right)^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i}y^i$ Putting $x = y = 1$ we get: {{begin-eqn}} {{eqn | l = 2^n       | r = \left({1 + 1}\right)^n       | c =  }} {{eqn | r = \sum_{i \mathop = 0}^n \binom n i 1^{n-i} \ 1^i       | c =  }} {{eqn | r = \sum_{i \mathop = 0}^n \binom n i       | c =  }} {{end-eqn}} {{qed}}
We have by definition of vacuous summation that: :$\displaystyle \forall n \in \Z: n < 0: \sum_{i \mathop = 0}^n \binom n 1 = 0$ Then from Zero Choose Zero: :$\displaystyle \sum_{i \mathop = 0}^0 \binom 0 0 = 1$ For $n > 0$: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^n \left({-1}\right)^i \binom n i       | r = \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \left({-1}\right)^i \binom n i + \left({-1}\right)^n \binom n n       | c =  }} {{eqn | r = \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \left({-1}\right)^i \left({\binom {n - 1} {i - 1} + \binom {n - 1} i}\right) + \left({-1}\right)^n \binom n n       | c = Pascal's Rule }} {{eqn | r = \binom n 0 - \binom {n - 1} 0 + \left({-1}\right)^{n - 1} \binom {n - 1} {n - 1} + \left({-1}\right)^n \binom n n       | c = as the series telescopes }} {{end-eqn}} {{explain|Expand the working to demonstrate how the series telescopes}} We note: :$\dbinom n 0 = \dbinom {n - 1} 0 = 1$ so: :$\dbinom n 0 - \dbinom {n - 1} 0 = 0$ :$\left({-1}\right)^{n - 1} \dbinom {n - 1} {n - 1} = - \left({-1}\right)^n \dbinom n n = - \left({-1}\right)^n$ so: :$\left({-1}\right)^{n - 1} \dbinom {n - 1} {n - 1} + \left({-1}\right)^n \dbinom n n = 0$ Hence the result. {{qed}}
We have by definition of vacuous summation that: :$\displaystyle \forall n \in \Z: n < 0: \sum_{i \mathop = 0}^n \binom n 1 = 0$ Then from Zero Choose Zero: :$\displaystyle \sum_{i \mathop = 0}^0 \binom 0 0 = 1$ Let $n > 0$. From the Binomial Theorem, we have that: :$\displaystyle \forall n \in \Z_{\ge 0}: \left({x + y}\right)^n = \sum_{i \mathop = 0}^n  \binom n i x^{n - i} y^i$ Putting $x = 1, y = -1$, we get: {{begin-eqn}} {{eqn | l = 0       | r = \left({1 - 1}\right)^n       | c = which holds for all $n > 0$ }} {{eqn | r = \sum_{i \mathop = 0}^n \binom n i 1^{n - i} \left({-1}\right)^i       | c =  }} {{eqn | r = \sum_{i \mathop = 0}^n \left({-1}\right)^i \binom n i       | c =  }} {{end-eqn}} {{qed}}