avewright/proofwiki-math huggingface dataset:
Theorem: Brahmagupta-Fibonacci Identity

Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$

Proof:
Let $z_j = a_j + i b_j$ for each $j = 1, 2, \ldots, n$.
Let $\ds c + i d = \prod_{j \mathop = 1}^n z_j$.
Then:
{{begin-eqn}}
{{eqn | l = c + i d
      | r = \prod_{j \mathop = 1}^n z_j
}}
{{eqn | r = \prod_{j \mathop = 1}^n \paren {a_j + i b_j}
}}
{{end-eqn}}
As $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are integers, so are $c$ and $d$.
Thus:
{{begin-eqn}}
{{eqn | l = c^2 + d^2
      | r = \cmod {c + i d}^2
      | c = {{Defof|Complex Modulus}}
}}
{{eqn | r = \cmod {\prod_{j \mathop = 1}^n z_j}^2
      | c = $\ds c + i d = \prod_{j \mathop = 1}^n z_j$
}}
{{eqn | r = \prod_{j \mathop = 1}^n \cmod {z_j}^2
      | c = Complex Modulus of Product of Complex Numbers: General Result
}}
{{eqn | r = \prod_{j \mathop = 1}^n \cmod {a_j + i b_j}^2
      | c = $z_j = a_j + i b_j$
}}
{{eqn | r = \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2}
      | c = {{Defof|Complex Modulus}}
}}
{{end-eqn}}
{{qed}}
Theorem: Brahmagupta-Fibonacci Identity

Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {a c + b d}^2 + \paren {a d - b c}^2
      | c = 
}}
{{eqn | r = \paren {\paren {a c}^2 + 2 \paren {a c} \paren {b d} + \paren {b d}^2} + \paren {\paren {a d}^2 - 2 \paren {a d} \paren {b c} + \paren {b c}^2}
      | c = Square of Sum, Square of Difference
}}
{{eqn | r = a^2 c^2 + 2 a b c d + b^2 d^2 + a^2 d^2 - 2 a b c d + b^2 c^2
      | c = multiplying out
}}
{{eqn | r = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2
      | c = simplifying
}}
{{eqn | r = \paren {a^2 + b^2} \paren {c^2 + d^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Brahmagupta-Fibonacci Identity

Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$

Proof:
From the more general version of Brahmagupta-Fibonacci Identity:
:$\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c + n b d}^2 + n \paren {a d - b c}^2$
The result follows by setting $n = 1$.
{{qed}}
Theorem: Brahmagupta-Fibonacci Identity

Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$

Proof:
Lagrange's Identity gives:
:$\ds \paren {\sum_{k \mathop = 1}^n {a_k}^2} \paren {\sum_{k \mathop = 1}^n {b_k}^2} - \paren {\sum_{k \mathop = 1}^n a_k b_k}^2 = \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \paren {a_i b_j - a_j b_i}^2$
Setting $n = 2$:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{k \mathop = 1}^2 {a_k}^2} \paren {\sum_{k \mathop = 1}^2 {b_k}^2} - \paren {\sum_{k \mathop = 1}^2 a_k b_k}^2
      | r = \sum_{i \mathop = 1}^1 \sum_{j \mathop = i + 1}^2 \paren {a_i b_j - a_j b_i}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren { {a_1}^2 + {a_2}^2} \paren { {b_1}^2 + {b_2}^2} - \paren {a_1 b_1 + a_2 b_2}^2
      | r = \paren {a_1 b_2 - a_2 b_1}^2
      | c = expanding out summations
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 + b^2} \paren {c^2 + d^2} - \paren {a c + b d}^2
      | r = \paren {a d - b c}^2
      | c = renaming $a_1 \to a, a_2 \to b, b_1 \to c, b_2 \to d$
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 + b^2} \paren {c^2 + d^2}
      | r = \paren {a c + b d}^2 + \paren {a d - b c}^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Brahmagupta-Fibonacci Identity

Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$

Proof:
Let $z_1 = a - b i, z_2 = c + d i$ be complex numbers.
Let $\cmod z$ denote the complex modulus of a given complex number $z \in \C$.
By definition of complex multiplication:
:$(1): \quad z_1 z_2 = \paren {a c + b d} + \paren {a d - b c} i$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_1 z_2} 
      | r = \cmod {z_1} \cmod{z_2}
      | c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | ll= \leadsto
      | l = \cmod {\paren {a - b i} \paren {c + d i} }^2
      | r = \cmod {a - b i}^2 \cmod {c + d i}^2
      | c = $z_1 = a - b i, z_2 = c + d i$
}}
{{eqn | ll= \leadsto
      | l = \paren {a c + b d}^2 + \paren {a d - b c}^2
      | r = \paren {a^2 + b^2} \paren {c^2 + d^2}
      | c = {{Defof|Complex Modulus}}, and from $(1)$
}}
{{end-eqn}}
{{qed}}
Theorem: Fermat's Two Squares Theorem

Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands.

Proof:
=== Uniqueness Lemma ===
{{:Fermat's Two Squares Theorem/Uniqueness Lemma}}{{qed|lemma}}
Theorem: Fermat's Two Squares Theorem

Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands.

Proof:
Suppose:
:$p = a^2 + b^2 = c^2 + d^2$
where $a, b, c, d \in \Z_{>0}$.
We have that:
{{begin-eqn}}
{{eqn | l = \paren {a c + b d} \paren {a d + b c}
      | r = \paren {a^2 + b^2} c d + \paren {c^2 + d^2} a b
      | c = multiplying out and gathering terms
}}
{{eqn | r = p \paren {a b + c d}
      | c = as $p = a^2 + b^2 = c^2 + d^2$
}}
{{end-eqn}}
$p$ is prime {{hypothesis}}.
From Euclid's Lemma for Prime Divisors:
:$p \divides \paren {a c + b d}$
or:
:$p \divides \paren {a d + b c}$
{{WLOG}}, suppose $p \divides \paren {a c + b d}$.
By {{Corollary|Absolute Value of Integer is not less than Divisors}}:
:$(1): \quad a c + b d \ge p$ 
We have:
{{begin-eqn}}
{{eqn | l = p^2
      | r = \paren {a^2 + b^2} \paren {c^2 + d^2}
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a c + b d}^2 + \paren {a d - b c}^2
      | c = Brahmagupta-Fibonacci Identity
}}
{{eqn | ll= \leadsto
      | l = p^2
      | o = \ge
      | r = p^2 + \paren {a d - b c}^2
      | c = from $(1)$: $p \le a c + b d$ and so $p^2 \le \paren {a c + b d}^2$
}}
{{eqn | ll= \leadsto
      | l = 0
      | o = \ge
      | r = \paren {a d - b c}^2
      | c = subtracting $p^2$ from both sides
}}
{{eqn | ll= \leadsto
      | l = a d - b c
      | r = 0
      | c = Square of Real Number is Non-Negative
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \dfrac c a 
      | r = \dfrac d b
      | c = dividing both sides by $a b$ and rearranging
}}
{{eqn | ll= \leadsto
      | l = \dfrac {c^2} {a^2} 
      | r = \dfrac {d^2} {b^2}
      | c = squaring both sides
}}
{{eqn | r = \dfrac {c^2 + d^2} {a^2 + b^2}
      | c = {{Corollary|Mediant is Between|2}}
}}
{{eqn | r = \dfrac p p
      | c = {{hypothesis}}
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Hence:
:$c = a$
and:
:$b = d$
{{qed}}
Theorem: Fermat's Two Squares Theorem

Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands.

Proof:
Suppose:
:$p = a^2 + b^2 = c^2 + d^2$
where $a > b > 0$ and $c > d > 0$.
We are going to show that $a = c$ and $b = d$.
From the two expressions for $p$, we have:
{{begin-eqn}}
{{eqn | l = \paren {a d - b c} \paren {a d + b c}
      | r = a^2 d^2 - b^2 c^2
      | c = Difference of Two Squares
}}
{{eqn | r = \paren {p - b^2} d^2 - b^2 \paren {p - d^2}
      | c = substituting for $a^2$ and $c^2$
}}
{{eqn | r = p d^2 - b^2 d^2 - p b^2 + b^2 d^2
      | c = 
}}
{{eqn | r = p \paren {d^2 - b^2}
      | c = 
}}
{{eqn | o = \equiv
      | r = 0
      | c = $\pmod p$
}}
{{end-eqn}}
So we have:
:$\paren {a d - b c} \paren {a d + b c} \equiv 0 \pmod p$
From Euclid's Lemma, that means:
:$p \divides \paren {a d - b c}$
or:
:$p \divides \paren {a d + b c}$
{{AimForCont}} $p \divides \paren {a d + b c}$.
Now, we have that each of $a^2, b^2, c^2, d^2$ must be less than $p$.
Hence $0 < a, b, c, d < \sqrt p$.
This implies $0 < a d + b c < 2 p$.
That must mean that $a d + b c = p$.
But then:
{{begin-eqn}}
{{eqn | l = p^2
      | r = \paren {a^2 + b^2} \paren {d^2 + c^2}
      | c = 
}}
{{eqn | r = \paren {a d + b c}^2 + \paren {a c - b d}^2
      | c = Brahmagupta-Fibonacci Identity
}}
{{eqn | r = p^2 + \paren {a c - b d}^2
      | c = 
}}
{{end-eqn}}
That means:
:$a c - b d = 0$
But since $a > b$ and $c > d$ we have:
:$a c > b d$
This contradiction shows that $a d + b c$ can not be divisible by $p$.
So this means:
:$p \divides \paren {a d - b c}$
Similarly, because $0 < a, b, c, d < \sqrt p$ we have:
:$-p < a d - b c < p$
This means:
:$a d = b c$
So:
:$a \divides b c$
But $a \perp b$ otherwise $a^2 + b^2$ has a common divisor greater than $1$ and less than $p$.
This cannot happen because $p$ is prime.
So by Euclid's Lemma:
:$a \divides c$
So we can put $c = k a$ and so $a d = b c$ becomes $d = k b$.
Hence:
:$p = c^2 + d^2 = k^2 \paren {a^2 + b^2} = k^2 p$
This means $k = 1$ which means $a = c$ and $b = d$ as we wanted to show.
{{qed}}
Theorem: Fermat's Two Squares Theorem

Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands.

Proof:
For $p = 2$ the only possibility is:
:$p = 1^2 + 1^2$
Assume $p \ge 3$.
Let $a \ge b$ and $c \ge d$ satisfy:
:$p = a^2 + b^2 = c^2 + d^2$
As $2 \nmid p$, we have:
:$(1): \quad a > b$ and $c > d$
Observe:
{{begin-eqn}}
{{eqn | l = \paren {a c + b d} \paren {a d + b c}
      | r = \paren {a^2 + b^2} c d + \paren {c^2 + d^2} a b
}}
{{eqn | r = p \paren {a b + c d}
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = p^2
      | r = \paren {a^2 + b^2} \paren {c^2 + d^2}
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a c - b d}^2 + \paren {a d + b c}^2
      | c = {{Corollary|Brahmagupta-Fibonacci Identity}}
}}
{{eqn | o = >
      | r = \paren {a d + b c}^2
      | c = as $a c - b d > 0$ by $(1)$ and Real Number Ordering is Compatible with Multiplication
}}
{{eqn | ll= \leadsto
      | l = p
      | o = \nmid
      | r = a d + b c
}}
{{end-eqn}}
Hence:
:$p \divides a c + b d$
In particular:
:$p \le a c + b d$
Thus:
{{begin-eqn}}
{{eqn | l = p^2
      | r = \paren {a^2 + b^2} \paren {c^2 + d^2}
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a c + b d}^2 + \paren {a d - b c}^2
      | c = Brahmagupta-Fibonacci Identity
      | n = 2
}}
{{eqn | o = \ge
      | r = p^2 + \paren {a d - b c}^2
}}
{{eqn | ll= \leadsto
      | l = a d - b c
      | r = 0
      | c = 
}}
{{end-eqn}}
Inserting this back into $(2)$, we obtain:
:$p^2 = \paren {a c + b d}^2$
That is:
:$p = a c + b d$
Hence:
{{begin-eqn}}
{{eqn | l = \paren {a - c}^2 + \paren {b - d}^2
      | r = a^2 + b^2 + c^2 + d^2 - 2 p
}}
{{eqn | r =  2 p - 2p
      | c = {{hypothesis}}
}}
{{eqn | r = 0
}}
{{end-eqn}}
Thus $a = c$ and $b = d$.
{{qed}}
Theorem: Integer as Sum of Two Squares

Let $n$ be a positive integer.
Then:
: $n$ can be expressed as the sum of two squares
{{iff}}:
:each of its prime divisors of the form $4 k + 3$ (if any) occur to an even power.

Proof:
Let us extract the largest square divisor of $n$, and write:
:$n = m^2 r$
where $r$ is square-free.
Theorem: Square Modulo 8

Let $x \in \Z$ be an integer.
: If $x$ is even then:
::: $x^2 \equiv 0 \pmod 8$ or $x^2 \equiv 4 \pmod 8$
: If $x$ is odd then:
::: $x^2 \equiv 1 \pmod 8$

Proof:
=== Proof for Even Integer ===
Let $x \in \Z$ be even.
Then from Square Modulo 4:
: $x^2 \equiv 0 \pmod 4$
Hence there are two possibilities for $x^2$:
: $x^2 \equiv 0 \pmod 8$
: $x^2 \equiv 4 \pmod 8$
The fact that there do exist such squares can be demonstrated by example:
: $2^2 = 4 \equiv 4 \pmod 8$
: $4^2 = 16 \equiv 0 \pmod 8$
{{qed|lemma}}
Theorem: Integer as Sum of Three Squares

Let $r$ be a positive integer.
Then $r$ can be expressed as the sum of three squares {{iff}} it is not of the form:
:$4^n \paren {8 m + 7}$
for some $m, n \in \Z_{\ge 0}$.

Proof:
=== Sufficient Condition ===
Suppose $r$ is not of the form $4^n \paren {8 m + 7}$.
Then we need to show that it can always be expressed as the sum of three squares.
{{ProofWanted}}
Theorem: Product of Sums of Four Squares

Let $a, b, c, d, w, x, y, z$ be numbers.
Then:
{{begin-eqn}}
{{eqn | o = 
      | r = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)
      | c = 
}}
{{eqn | l = =
      | o = 
      | r = \left({a w + b x + c y + d z}\right)^2
      | c = 
}}
{{eqn | o = +
      | r = \left({a x - b w + c z - d y}\right)^2
      | c = 
}}
{{eqn | o = +
      | r = \left({a y - b z - c w + d x}\right)^2
      | c = 
}}
{{eqn | o = +
      | r = \left({a z + b y - c x - d w}\right)^2
      | c = 
}}
{{end-eqn}}

Proof:
Taking each of the squares on the {{RHS}} and multiplying them out in turn:
{{begin-eqn}}
{{eqn | l = \paren {a w + b x + c y + d z}^2
      | r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a x - b w + c z - d y}^2
      | r = a^2 x^2 + b^2 w^2 + c^2 z^2 + d^2 y^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {-a b w x + a c x z - a d x y - b c w z + b d w y - c d y z}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a y - b z - c w + d x}^2
      | r = a^2 y^2 + b^2 z^2 + c^2 w^2 + d^2 x^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {-a b y z - a c w y + a d x y + b c w z - b d x z - c d w x}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a z + b y - c x - d w}^2
      | r = a^2 z^2 + b^2 y^2 + c^2 x^2 + d^2 w^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {a b y z - a c x z - a d w z - b c x y - b d w y + c d w x}
      | c = 
}}
{{end-eqn}}
All the non-square terms cancel out with each other, leaving:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {a w + b x + c y + d z}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a x - b w + c z - d y}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a y - b z - c w + d x}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a z + b y - c x - d w}^2
      | c = 
}}
{{eqn | r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = a^2 x^2 + b^2 w^2 + c^2 z^2 + d^2 y^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = a^2 y^2 + b^2 z^2 + c^2 w^2 + d^2 x^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = a^2 z^2 + b^2 y^2 + c^2 x^2 + d^2 w^2
      | c = 
}}
{{eqn | r = \paren {a^2 + b^2 + c^2 + d^2} \paren {w^2 + x^2 + y^2 + z^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Sums of Four Squares

Let $a, b, c, d, w, x, y, z$ be numbers.
Then:
{{begin-eqn}}
{{eqn | o = 
      | r = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)
      | c = 
}}
{{eqn | l = =
      | o = 
      | r = \left({a w + b x + c y + d z}\right)^2
      | c = 
}}
{{eqn | o = +
      | r = \left({a x - b w + c z - d y}\right)^2
      | c = 
}}
{{eqn | o = +
      | r = \left({a y - b z - c w + d x}\right)^2
      | c = 
}}
{{eqn | o = +
      | r = \left({a z + b y - c x - d w}\right)^2
      | c = 
}}
{{end-eqn}}

Proof:
Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two quaternions.
Then:
{{begin-eqn}}
{{eqn | l = \size {\mathbf m} \size {\mathbf n}
      | r = \size {\mathbf m \mathbf n}
      | c = Quaternion Modulus of Product of Quaternions
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 + b^2 + c^2 + d^2} \paren {\paren {-w}^2 + x^2 + y^2 + z^2}
      | r = \size {\mathbf m \mathbf n}
      | c = {{Defof|Quaternion Modulus}}
}}
{{eqn | r = \paren {-a w - b x - c y - d z}^2
      | c = {{Defof|Quaternion Multiplication}}
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a x - b w + c z - d y}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a y - b z - c w + d x}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a z + b y - c x - d w}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 + b^2 + c^2 + d^2} \paren {w^2 + x^2 + y^2 + z^2}
      | r = \paren {a w + b x + c y + d z}^2
      | c = simplifying
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a x - b w + c z - d y}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a y - b z - c w + d x}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a z + b y - c x - d w}^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Necessary Condition for Existence of BIBD

Let there exist a BIBD with parameters $v, b, r, k, \lambda$.
Then the following are true:
:$(1): \quad b k = r v$
:$(2): \quad \lambda \paren {v - 1} = r \paren {k - 1}$
:$(3): \quad b \dbinom k 2 = \lambda \dbinom v 2$
:$(4): \quad k < v$
:$(5): \quad r > \lambda$
All of $v, b, r, k, \lambda$ are integers.
Some sources prefer to report $(3)$ as:
:$b = \dfrac {\dbinom v 2} {\dbinom k 2} \lambda$
which is less appealing visually, and typographically horrendous.

Proof:
:$(1)$: We have by definition of balanced incomplete block design that:
:each treatment is in exactly $r$ blocks
:each block is of size $k$.
We have that $b k$ is the number of blocks times the size of each block.
We also have that $r v$ is the number of treatments times the number of blocks each treatment is in.
The two must clearly be equal.
:$(2)$: Comparing the {{LHS}} and {{RHS}} of the equation we can see that:
{{LHS}}: An arbitrary treatment must be paired with $v - 1$ other treatments.
If $\lambda > 1$ then every treatment is paired $\lambda \paren {v - 1}$ times.
{{RHS}}: An arbitrary treatment is paired with $k - 1$ other treatments for each of the $r$ blocks it is in.
Therefore it is paired $r \paren {k - 1}$ times.
Both values give the number of times an arbitrary treatment is paired, therefore the {{LHS}} equals the {{RHS}}.
:$(3)$: From equation $(1)$, we have that $r = \dfrac {b k} v$
From $(2)$ we have that:
:$r = \dfrac {v - 1} {k - 1} \lambda$
So:
{{begin-eqn}}
{{eqn | l = \frac {b k} v
      | r = \frac {v - 1} {k - 1} \lambda
      | c = substituting for $r$
}}
{{eqn | ll= \leadsto
      | l = b k \paren {k - 1}
      | r = \lambda v \paren {v - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \binom k 2
      | r = \lambda \binom v 2
      | c = Binomial Coefficient with Two
}}
{{end-eqn}}
{{qed}}
{{proof wanted|Yes I know the other ones are obvious but there's other things I'd rather do at the moment.}}
Category:Balanced Incomplete Block Designs
bs2im9moj4xhypnk07wgqnzx69q6vz6
Theorem: Mapping from Singleton is Injection

Let $f: S \to T$ be a mapping.
Let $S$ be a singleton.
Then $f$ is an injection.

Proof:
Let $S = \set s$.
For $f$ to be an injection, all we need to do is show:
:$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$
But as $S$ is a singleton, it follows that:
:$x_1 = x_2 = s$
Hence the result.
{{qed}}
Category:Injections
Category:Singletons
8uicojvbyk3yq3v7gffgk7xd1qriew8
Theorem: Mapping to Singleton is Surjection

Let $S$ be a non-empty set.
Let $f: S \to T$ be a mapping.
Let $T$ be a singleton.
Then $f$ is a surjection.

Proof:
Let $T = \set t$.
For $f$ to be a surjection, all we need to do is show:
:$\forall y \in T: \exists x \in S: \map f x = y$.
As $S \ne \O$, $\exists s \in S$.
As $f: S \to T$ is a mapping, it follows that $\map f s \in T$.
So as $\map f s \in T$ it follows that $t = \map f s$.
As $T = \set t$, it follows that $\forall y \in T: \exists x \in S: y = \map f x$.
Hence the result.
{{qed}}
Category:Surjections
Category:Singletons
4vvbsh8xrbo595u9i52y03m8m5rcxic
Theorem: Generating Function for Constant Sequence

Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N: a_n = r$
for some $r \in \R$.
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac r {1 - z}$ for $\size z < 1$

Proof:
{{begin-eqn}}
{{eqn | l = \map G z
      | r = \sum_{n \mathop = 0}^\infty r z^n
      | c = {{Defof|Generating Function}}
}}
{{eqn | r = r \sum_{n \mathop = 0}^\infty z^n
      | c = 
}}
{{eqn | r = \frac r {1 - z}
      | c = Sum of Infinite Geometric Sequence
}}
{{end-eqn}}
for $\size z < 1$.
{{qed}}
Theorem: Area of Circle

The area $A$ of a circle is given by:
:$A = \pi r^2$
where $r$ is the radius of the circle.

Proof:
Let the circle of radius $r$ be divided into many sectors:
:400px
If they are made small enough, they can be approximated to triangles whose heights are all $r$.
Let the bases of these triangles be denoted:
:$b_1, b_2, b_3, \ldots$
From Area of Triangle in Terms of Side and Altitude, their areas are:
:$\dfrac {r b_1} 2, \dfrac {r b_2} 2, \dfrac {r b_3} 2, \ldots$
The area $\AA$ of the circle is given by the sum of the areas of each of these triangles:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \dfrac {r b_1} 2 + \dfrac {r b_2} 2 + \dfrac {r b_3} 2 + \cdots
      | c = 
}}
{{eqn | r = \dfrac r 2 \paren {b_1 + b_2 + b_3 + \cdots}
      | c = 
}}
{{end-eqn}}
But $b_1 + b_2 + b_3 + \cdots$ is the length of the circumference of the circle.
From Perimeter of Circle:
:$b_1 + b_2 + b_3 + \cdots = 2 \pi r$
Hence:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \dfrac r 2 \paren {b_1 + b_2 + b_3 + \cdots}
      | c = 
}}
{{eqn | r = \dfrac r 2 \paren {2 \pi r}
      | c = 
}}
{{eqn | r = \pi r^2
      | c = 
}}
{{end-eqn}}
It needs to be noted that this proof is intuitive and non-rigorous.
{{qed}}
Theorem: Area of Circle

The area $A$ of a circle is given by:
:$A = \pi r^2$
where $r$ is the radius of the circle.

Proof:
From Equation of Circle:
:$x^2 + y^2 = r^2$
Thus $y = \pm \sqrt {r^2 - x^2}$.
It follows that from the geometric interpretation of the definite integral:
{{begin-eqn}}
{{eqn | l = A
      | r = \int_{-r}^r \paren {\sqrt {r^2 - x^2} - \paren {-\sqrt {r^2 - x^2} } } \rd x
}}
{{eqn | r = \int_{-r}^r 2 \sqrt {r^2 - x^2} \rd x
}}
{{eqn | r = \int_{-r}^r 2 r \sqrt {1 - \frac {x^2} {r^2} } \rd x
}}
{{end-eqn}}
Let $x = r \sin \theta$ (note that we can do this because $-r \le x \le r$).
Thus $\theta = \map \arcsin {\dfrac x r}$ and $\rd x = r \cos \theta \rd \theta$.
{{begin-eqn}}
{{eqn | l = A
      | r = \int_{\map \arcsin {\frac {-r} r} }^{\map \arcsin {\frac r r} } 2 r^2 \sqrt {1 - \frac {\paren {r \sin \theta}^2} {r^2} } \cos \theta \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta
}}
{{eqn | r = \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {\cos^2 \theta} \cos \theta \rd \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = r^2 \int_{-\frac \pi 2}^{\frac \pi 2} 2 \cos^2 \theta \rd \theta
}}
{{eqn | r = r^2 \int_{-\frac \pi 2}^{\frac \pi 2} \paren {1 + \map \cos {2 \theta} } \rd \theta
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | r = r^2 \intlimits {\theta + \frac 1 2 \map \sin {2 \theta} } {-\frac \pi 2} {\frac \pi 2}
      | c = from Definite Integral of Constant and Primitive of Cosine Function
}}
{{eqn | r = r^2 \paren {\frac \pi 2 + \frac 1 2 \map \sin {2 \cdot \frac \pi 2} - \frac {-\pi} 2 - \frac 1 2 \map \sin {2 \cdot \frac {-\pi} 2} }
}}
{{eqn | r = r^2 \paren {2 \cdot \frac \pi 2 + 2 \cdot \frac 1 2 \cdot 0}
}}
{{eqn | r = \pi r^2
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Circle

The area $A$ of a circle is given by:
:$A = \pi r^2$
where $r$ is the radius of the circle.

Proof:
Proof by shell integration:
The circle can be divided into a set of infinitesimally thin rings, each of which has area $2 \pi t \rd t$, since the ring has length $2 \pi t$ and thickness $\rd t$.
{{Handwaving|The fact that the above is a valid approximation needs to be established.}}
{{begin-eqn}}
{{eqn | l = A
      | r = \int_0^r 2 \pi t \rd t
      | c = Perimeter of Circle
}}
{{eqn | r = \bigintlimits {\pi t^2} 0 r
}}
{{eqn | r = \pi r^2
}}
{{end-eqn}}
{{Qed}}