Formula & theorem reference

B \subset A is false only when B contains some element that A lacks. Take B = \varnothing: it contains no elements at all, so there is nothing it could possibly contain that A doesn't also contain. The condition for B \subset A to fail can never be met, so \varnothing \subset A holds automatically — for every set A, including A = \varnothing itself.

Every natural number n has no fractional part and no negative sign, so it's also an integer: \mathbb{N} \subset \mathbb{Z}. Every integer n can be written as the ratio \frac{n}{1} of two integers, which is exactly the definition of rational: \mathbb{Z} \subset \mathbb{Q}. Every rational number is, by definition, a real number (the rationals are one of the two pieces — rational and irrational — that together make up \mathbb{R}): \mathbb{Q} \subset \mathbb{R}. Chaining all three containments gives \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}.

Every real number x satisfies -|x| \le x \le |x| directly from the definition of absolute value. Applying this to a and to b separately and adding the results gives -(|a|+|b|) \le a+b \le |a|+|b|. A number sandwiched between -k and k has absolute value at most k, so |a+b| \le |a|+|b|.

These five properties aren't derived from anything more basic here — they ARE the basics, the starting assumptions about + and \times that every other algebraic fact in this course is built from (including the Zero- Product Property and the Properties of Equality elsewhere in this registry, both proved by appealing directly to these). Asking "why is addition commutative?" inside algebra itself has no more basic answer to give; it's where the axioms start.

These four properties aren't derived from anything more basic in this course — they're part of what "=" means, the same way the Field Properties of the Real Numbers describe what "+" and "\times" mean. Every algebraic manipulation in this course — clearing a denominator, adding the same term to both sides, substituting one expression for an equal one — is really an appeal to one of these four properties, usually substitution. They're listed here explicitly because that appeal is normally silent.

Suppose ab = 0 and a \neq 0. By the Field Properties of the Real Numbers, a has a multiplicative inverse \frac1a with a\cdot\frac1a=1. Multiplying both sides of ab=0 by \frac1a gives \frac1a(ab) = \frac1a \cdot 0, \qquad \left(\frac1a \cdot a\right)b = 0, \qquad 1\cdot b = 0, \qquad b = 0, using associativity, then the inverse property, then the identity property, in turn. So whenever a \neq 0, b is forced to be 0 — which means a=0 or b=0 is unavoidable whenever ab=0.

For more than two factors, group and repeat: view a_1a_2\cdots a_n=0 as (a_1)(a_2\cdots a_n)=0. Either a_1=0, or a_1\neq0 and the argument above forces the shorter product a_2\cdots a_n to be 0 — to which the same splitting applies again, one factor shorter each time, until some single factor is forced to 0.

a < b means b - a > 0. Adding c to both sides of a<b is the claim (b+c)-(a+c) > 0, which is just b - a > 0 again — unchanged, so addition always preserves direction.

For multiplication, compare ac and bc by looking at their difference: bc - ac = (b-a)c. Since b - a > 0, the sign of this difference is exactly the sign of c. If c > 0, a positive times a positive is positive, so bc - ac > 0, i.e. ac < bc — direction preserved. If c < 0, a positive times a negative is negative, so bc - ac < 0, i.e. ac > bc — direction reversed. Dividing by c is multiplying by \frac{1}{c}, which carries the same sign as c, so the division rule follows immediately from the multiplication rule.

For u \ge 0, |u| = u, so |u| < a means u < a — combined with 0 \le u and a > 0, that already reads -a < u < a. For u < 0, |u| = -u, so |u| < a means -u < a, i.e. u > -a — combined with u < 0 < a, that's again -a < u < a. Either way, |u| < a \iff -a < u < a.

For |u| > a: if u \ge 0, |u| = u > a directly; if u < 0, |u| = -u > a means u < -a. So |u| > a holds exactly when u > a or u < -a — the two rays. The \le/\ge forms follow the identical case split with the strict inequality replaced throughout.

Suppose A \cdot B = 0 and A \neq 0. Since A \neq 0, it has a multiplicative inverse \frac{1}{A}. Multiply both sides by \frac{1}{A}: \frac{1}{A}(A \cdot B) = \frac{1}{A} \cdot 0 The left side simplifies to B (since \frac{1}{A} \cdot A = 1) and the right side is 0, so B = 0. By the same argument with the roles swapped, if B \neq 0 then A = 0. So at least one of the two factors must be 0.

Start from the general quadratic ax^2 + bx + c = 0 and divide every term by a (legal since a \neq 0): x^2 + \frac{b}{a}x + \frac{c}{a} = 0

Move the constant term to the right side: x^2 + \frac{b}{a}x = -\frac{c}{a}

Complete the square: take half of the x-coefficient, \frac{b}{2a}, square it, and add it to both sides. x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}

The left side is now a perfect-square trinomial: \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}

Take the square root of both sides — remembering a square root has two signs (strictly, \sqrt{4a^2} = 2|a| since a could itself be negative, but the very next \pm absorbs that sign either way, so writing 2a in the denominator costs nothing): x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}

Subtract \frac{b}{2a} from both sides: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The Quadratic Formula gives every solution of ax^2+bx+c=0 as x = \frac{-b \pm \sqrt{D}}{2a}, \qquad D = b^2 - 4ac, and the three cases fall directly out of what \sqrt{D} does under the \pm.

D > 0. \sqrt{D} is a positive real number, and +\sqrt D \ne -\sqrt D, so the formula produces two distinct real values. If D is a perfect square, \sqrt D is a whole number and the whole expression is a ratio of integers — rational. If D is not a perfect square, \sqrt D is irrational, and an irrational number added to the rational -b stays irrational, so both solutions are irrational.

D = 0. The \pm\sqrt D term vanishes entirely, collapsing the two branches into the single value x = -\frac{b}{2a} — one repeated solution.

D < 0. No real number squares to a negative, so \sqrt D has no real value and the formula names no real solution. Writing \sqrt D = i\sqrt{-D} instead produces the two complex values x = \frac{-b \pm i\sqrt{-D}}{2a}, which are conjugates of each other since they differ only in the sign in front of the imaginary part.

Write the roots from the quadratic formula, r_{1,2} = \dfrac{-b \pm \sqrt{D}}{2a} where D = b^2-4ac. Adding the two roots, the \pm\sqrt{D} terms cancel: r_1+r_2 = \frac{-b+\sqrt D}{2a} + \frac{-b-\sqrt D}{2a} = \frac{-2b}{2a} = -\frac{b}{a}. Multiplying uses a difference of squares: r_1 r_2 = \frac{(-b+\sqrt D)(-b-\sqrt D)}{4a^2} = \frac{b^2-D}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}.

This is the LCD method applied to the one shape where it collapses to a one-line shortcut. The LCD of \frac{a}{b} and \frac{c}{d} is bd; multiply both sides by it: \frac{a}{b} \cdot bd = \frac{c}{d} \cdot bd. On the left, the b in the denominator cancels the b just introduced, leaving ad; on the right, the d cancels, leaving bc. That gives ad = bc — the "multiply the diagonals" rule is nothing more than the LCD step with the cancellation already carried out. Since b \ne 0 and d \ne 0 (both are required for the original fractions to even be defined), multiplying by bd here is a genuinely reversible step — dividing back by bd recovers \frac{a}{b} = \frac{c}{d} from ad = bc — which is exactly why this one shape needs no extraneous-root check the way the general LCD method does. The moment a third term joins the equation, the LCD is no longer just bd shared cleanly between two single fractions, and this shortcut no longer applies.

The key modeling step, not just algebra: rates of the same kind of work add when the workers work simultaneously. If a job takes a hours alone, the worker completes \frac{1}{a} of the job every hour — that's the definition of "rate" as "amount of job per unit time." In t hours of working together, worker A contributes \frac{t}{a} of the job and worker B contributes \frac{t}{b}, and "together they finish the job" means their contributions sum to exactly one whole job: \frac{t}{a} + \frac{t}{b} = 1. Divide both sides by t (valid since t \ne 0 — finishing takes some positive amount of time): \frac{1}{a} + \frac{1}{b} = \frac{1}{t}. This is the same equation, just organized by rate instead of by amount completed — and it's why the pipes example above is set up as a sum of unit rates equal to \frac{1}{t} rather than derived from nothing: it's this identity, applied with a=4 and b=6.

It suffices to show squaring is not reversible: a = b does not follow from a^2 = b^2 alone. Move everything to one side and factor: a^2 = b^2 \iff a^2 - b^2 = 0 \iff (a-b)(a+b) = 0 \iff a = b \ \text{ or } \ a = -b. So a^2 = b^2 certifies only that a equals b or its negation — the squaring step cannot distinguish which. If the original equation was a = b, the transformed equation a^2 = b^2 therefore also admits every solution of the different, unrelated equation a = -b, with no flag marking which root belongs to which. Those extra roots are extraneous precisely when they solve a^2=b^2 but fail the original a=b — which is exactly what re-checking every candidate against the original equation catches.

The same argument, one level more general, is why multiplying both sides by an expression E(x) that can equal zero is also irreversible: P = Q implies PE = QE for any E, but PE = QE only implies P = Q when E \ne 0 (dividing both sides by E requires it). Wherever E(x) = 0, PE = QE collapses to the trivially true 0 = 0 regardless of whether P = Q actually held there — so the transformed equation can pick up every root of E(x) = 0 as a spurious solution, which is exactly why step 1 of the rational-equation method (find the excluded values first) exists.

Introduce a third point Q = (x_2, y_1) — same height as P_1, same sideways position as P_2. The segment P_1Q is horizontal with length |x_2 - x_1|, and the segment QP_2 is vertical with length |y_2 - y_1|. A horizontal segment meeting a vertical one always forms a right angle, so triangle P_1QP_2 is a right triangle with legs |x_2 - x_1| and |y_2 - y_1|, and its hypotenuse is exactly the segment P_1P_2 whose length d we want. The Pythagorean theorem gives d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2, and taking the (non-negative) square root of both sides gives the formula.

The step from P_1 to M is \left( \frac{x_1+x_2}{2} - x_1, \; \frac{y_1+y_2}{2} - y_1 \right) = \left( \frac{x_2-x_1}{2}, \; \frac{y_2-y_1}{2} \right), which is exactly half the step from P_1 to P_2. A point reached by taking half the step from P_1 toward P_2 sits exactly halfway along the segment — which is what "midpoint" means.

A point (x,y) lies on the circle exactly when its distance to the center (h,k) equals r. By the distance formula, \sqrt{(x-h)^2 + (y-k)^2} = r. Square both sides (both sides are non-negative, so this keeps the same solutions) to clear the radical: (x-h)^2 + (y-k)^2 = r^2.

Write the two lines in slope-intercept form, y = m_1x + b_1 and y = m_2x + b_2. A point lies on both lines exactly when m_1x + b_1 = m_2x + b_2 \quad\Longleftrightarrow\quad (m_1 - m_2)x = b_2 - b_1.

If m_1 \neq m_2, this last equation solves for a single value of x (divide both sides by m_1 - m_2), so the lines meet at exactly one point — they are not parallel.

If m_1 = m_2, the left side is 0 for every x, so the equation becomes 0 = b_2 - b_1. When b_1 \neq b_2 that's false for every x: no intersection at all, which is exactly what parallel means. When b_1 = b_2 the equation is true for every x: the two equations describe the same line, not two parallel ones — which is exactly why the theorem excludes that case by name.

Let the two lines meet at point P. Build the right triangle that defines line 1's slope with its right-angle corner at P: a horizontal leg of run a and a vertical leg of rise b, so m_1 = \frac{b}{a}.

Now rotate that triangle 90° about P. A 90° rotation is a rigid motion — it preserves every length and every angle — and it does one very specific thing to direction: it swaps horizontal and vertical, flipping the sign of one of them. The horizontal leg of length a becomes a vertical leg of length a; the vertical leg of length b becomes a horizontal leg of length b, now pointing the opposite way along the x-direction.

Because a 90° rotation about P is, by definition, what turns line 1 into the line through P perpendicular to it, the rotated triangle's hypotenuse lies exactly along line 2. Reading its slope from the rotated legs: m_2 = \frac{\text{rise}}{\text{run}} = \frac{a}{-b} = -\frac{a}{b}.

Multiply the two slopes: m_1 m_2 = \frac{b}{a} \cdot \left(-\frac{a}{b}\right) = -1.

Every step above is reversible — rotating by 90° the other way undoes it — so the same argument run backward proves the converse: if m_1 m_2 = -1, the lines meet at a right angle.

The vertical line x = a consists of exactly the points whose first coordinate is a — so the curve's intersections with that line are exactly the points (a, y) the curve assigns to input a.

If the curve is the graph of a function, input a has exactly one output, so x = a meets it in at most one point — for every a. Conversely, if some vertical line x = a met the curve twice, at (a, y_1) and (a, y_2) with y_1 \neq y_2, that would mean input a produced two different outputs, which is precisely what the definition of a function forbids. So the curve fails to be a function exactly when some vertical line crosses it more than once.

The horizontal line y = k meets the graph of f exactly at the points (x, k) with f(x) = k — one point for every input that produces output k.

If f is one-to-one, then f(a) = f(b) forces a = b, so no two distinct inputs share an output: every horizontal line meets the graph at most once. Conversely, suppose some horizontal line y = k met the graph twice, at (a, k) and (b, k) with a \neq b. Then f(a) = k = f(b) with a \neq b — two different inputs sharing an output, which is exactly the negation of one-to-one. So f fails to be one-to-one exactly when some horizontal line crosses its graph more than once.

Multiplicity k means f(x) = (x-c)^k g(x) for some polynomial g with g(c) \neq 0. A polynomial can only change sign by passing through a zero, and g has none at c, so on a small enough interval around c the sign of g(x) stays fixed at the sign of g(c). That means near x = c, the sign of f(x) is controlled entirely by the sign of (x-c)^k.

k odd. (x - c) itself changes sign at c — negative just to the left, positive just to the right. Raising to an odd power preserves that sign change: (x-c)^k is negative on one side and positive on the other, so f is too. The graph crosses.

k even. Raising (x-c) to an even power erases the sign entirely — (x-c)^k \geq 0 on both sides of c, with equality only at c itself. So f keeps the same sign on both sides (matching the sign of g(c)), touches zero at c, and turns back rather than crossing.

The flattening for k \geq 3 is a matter of scale rather than sign: for |x - c| < 1, higher powers of a small number shrink faster, so (x-c)^k (and hence f) sits closer to 0 near c the larger k is — the curve hugs the axis longer before pulling away.

Existence. This is exactly what the long-division procedure produces. At each step, the working dividend's leading term is cancelled by subtracting \frac{\text{lead}(f)}{\text{lead}(d)}x^{\,\deg f - \deg d} \cdot d(x) — a multiple of d chosen precisely so its leading term matches. That cancellation strictly lowers the degree of the working dividend at every step. Degrees are nonnegative integers, and a strictly decreasing sequence of nonnegative integers cannot continue forever, so the process must halt after finitely many steps. It halts exactly when the working dividend's degree drops below \deg d (or hits 0) — call what's left r(x), and call the accumulated multiples of d(x) subtracted along the way q(x). Adding everything back together gives f(x) = d(x)q(x) + r(x).

Uniqueness. Suppose two pairs both work: f = dq_1 + r_1 = dq_2 + r_2. Rearranging, d(q_1 - q_2) = r_2 - r_1. If q_1 \neq q_2, the left side is d times a nonzero polynomial, so it has degree at least \deg d. But the right side is a difference of two remainders, each either 0 or of degree less than \deg d — so \deg(r_2 - r_1) < \deg d, a contradiction. So q_1 = q_2, and then r_2 - r_1 = d(q_1-q_2) = 0 forces r_1 = r_2 as well.

Write f(x) = (x-c)q(x) + r with r a constant (the division algorithm guarantees \deg r < \deg(x-c) = 1, so r has degree 0). Substitute x = c: f(c) = (c-c)q(c) + r = 0 + r = r. So the remainder r equals f(c).

By the Remainder Theorem, dividing f(x) by x - c leaves remainder f(c). If f(c) = 0, the remainder is 0, so f(x) = (x-c)q(x) exactly — x - c divides evenly, i.e. it's a factor. Conversely, if x - c is a factor, f(x) = (x-c)q(x) for some q(x), so f(c) = (c-c)q(c) = 0.

Substitute x = \frac{p}{q} into f(x) = 0: a_n \left(\frac{p}{q}\right)^n + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + \cdots + a_1\left(\frac{p}{q}\right) + a_0 = 0. Multiply every term by q^n to clear denominators: a_np^n + a_{n-1}p^{n-1}q + \cdots + a_1pq^{n-1} + a_0q^n = 0.

Isolate the constant term: a_0q^n = -\left(a_np^n + a_{n-1}p^{n-1}q + \cdots + a_1pq^{n-1}\right). Every term on the right has a factor of p, so p divides a_0q^n. Because \frac{p}{q} is in lowest terms, p and q share no prime factor, so p shares no prime factor with q^n either — meaning p must divide a_0 on its own.

The same move on the leading term gives a_np^n = -\left(a_{n-1}p^{n-1}q + \cdots + a_0q^n\right), and every term on the right now has a factor of q, so q divides a_np^n. Again q shares no factor with p^n, so q must divide a_n.

Every known proof leans on machinery outside this course — complex analysis (Liouville's theorem: a bounded entire function is constant, applied to 1/f), topology (a degree-n map wraps the plane around itself n times, so it must hit 0), or Galois theory. None of those tools are available yet, so the result is taken on faith here, the way it is in every first course that uses it. What this course can do with it: combined with the Factor Theorem, it guarantees a degree-n polynomial factors completely into n linear factors over the complex numbers — that consequence, the Linear Factorization Theorem, is used freely and does have its own proof in the registry.

Apply the Fundamental Theorem of Algebra to find one zero c_1; by the Factor Theorem, f(x) = (x - c_1)q_1(x) for a degree-(n-1) quotient q_1. Apply the Fundamental Theorem again to q_1 to peel off c_2, and repeat. After n steps the quotient has degree 0 — a constant, which must be the leading coefficient a_n — leaving the full factorization f(x) = a_n(x-c_1)\cdots(x-c_n).

Conjugation respects addition and multiplication: for z = a+bi and w = c+di, direct expansion gives \overline{z+w} = \overline{z} + \overline{w} and \overline{zw} = \overline{z}\,\overline{w} (each side works out to the same real and imaginary parts). Since any polynomial expression is built from nothing but additions and multiplications, it follows that \overline{f(z)} = a_n\overline{z}^{\,n} + \cdots + a_1\overline{z} + a_0 provided every coefficient a_k is real — because \overline{a_k} = a_k exactly when a_k is real, which is precisely where the hypothesis gets spent.

Now suppose f(a+bi) = 0 for a real-coefficient f. Taking the conjugate of both sides: \overline{f(a+bi)} = f(\overline{a+bi}) = f(a - bi), and \overline{0} = 0, so f(a-bi) = 0 as well — the conjugate is also a zero.

Every nonreal zero therefore drags its conjugate along as a second zero, so nonreal zeros come in pairs. By the Linear Factorization Theorem, a degree-n polynomial has exactly n zeros counted with multiplicity. If n is odd, the nonreal zeros can only ever be removed two at a time — leaving an odd number unaccounted for, and those leftover zeros can't be nonreal (they'd need an unpaired partner), so at least one must be real.

The idea is intuitive — a graph that starts below the x-axis and ends above it, without ever jumping, has to cross somewhere — but making "without ever jumping" precise is a statement about the real numbers themselves. A rigorous proof needs the completeness property of \mathbb R (typically via the least-upper-bound axiom, showing the crossing point is the supremum of \{x \in [a,b] : f(x) < 0\}), which belongs to a real analysis course, not this one. It's taken on faith here — a polynomial is continuous (its graph really is one unbroken curve), and continuity is exactly the property the sign-change test relies on.

Let d(x) = \big(x - (a+\sqrt b)\big)\big(x - (a-\sqrt b)\big) = (x-a)^2 - b = x^2 - 2ax + (a^2-b) — a quadratic whose coefficients are rational, since a and b are.

Divide f by d using the Division Algorithm: f(x) = d(x)q(x) + r(x) with \deg r \leq 1. Because d has rational coefficients and long division only ever adds, subtracts, multiplies, and divides the coefficients it's given, q and r come out with rational coefficients too — this is exactly where the rational-coefficient hypothesis earns its keep. Write r(x) = mx + n with m, n rational.

Evaluate at x = a + \sqrt b. Since d(a+\sqrt b) = 0 by construction, 0 = f(a+\sqrt b) = d(a+\sqrt b)q(a+\sqrt b) + r(a+\sqrt b) = r(a+\sqrt b) = (ma+n) + m\sqrt b. Both ma+n and m are rational, and \sqrt b is irrational. If m \neq 0, this equation could be solved for \sqrt b = -\frac{ma+n}{m}, a ratio of rational numbers — making \sqrt b rational, a contradiction. So m = 0, and then n = 0 follows immediately. That means r(x) \equiv 0, so f(x) = d(x)q(x) exactly.

Since a - \sqrt b is the other root of d, it makes d(x)q(x) = f(x) vanish too — it is also a zero of f. Repeating the same division on q shows each additional factor of d that divides f pulls another matched pair out together, so the two conjugate zeros always carry equal multiplicity.

Write p(x) = (x-c)^kP(x) and q(x) = (x-c)^nQ(x) with P(c) \neq 0 and Q(c) \neq 0. For x \neq c, cancel the shared factor: f(x) = \frac{(x-c)^kP(x)}{(x-c)^nQ(x)} = (x-c)^{k-n}\,\frac{P(x)}{Q(x)}.

k \geq n: the hole. The exponent k - n is \geq 0, so (x-c)^{k-n}\frac{P(x)}{Q(x)} is an ordinary polynomial-or-rational expression that's perfectly defined at x = c (recall Q(c) \neq 0). f agrees with this reduced expression at every x \neq c, and is simply undefined at the single point x = c where the original q(x) vanished — a removable gap, i.e. a hole.

k < n: the asymptote. Here f(x) = \dfrac{P(x)/Q(x)}{(x-c)^{n-k}} with n - k > 0. As x \to c, the numerator \frac{P(x)}{Q(x)} approaches the nonzero number \frac{P(c)}{Q(c)}, while the denominator (x-c)^{n-k} shrinks toward 0 — a nonzero number divided by something shrinking to 0 grows without bound, so |f(x)| \to \infty.

For the sign: (x-c)^{n-k} is positive on both sides of c when n-k is even (the sign-erasing behavior an even power always has), so f blows up toward the same infinity on both sides. When n-k is odd, (x-c)^{n-k} carries the sign of (x-c) itself — negative to the left of c, positive to the right — so f blows up toward opposite infinities on the two sides.

A real nth root \sqrt[n]{x} is, by definition, a real number y with y^n = x. When n is even, y^n \ge 0 for every real y — an even power is never negative — so x must be \ge 0 for such a y to exist. That's the bare-radical, n-even case. When n is odd, y^n ranges over all of \mathbb{R} as y does, since odd powers preserve sign and grow without bound in both directions, so every real x has a real nth root — the bare-radical, n-odd case.

The two denominator cases layer the ordinary "a denominator can't be zero" restriction on top of that: the boundary g(x)=0 is excluded whenever the radical sits under a fraction, regardless of n's parity, since \sqrt[n]{0}=0 either way.

By definition, x^{p/q} = \left(\sqrt[q]{x}\right)^p once q > 0. The Domain of a Radical Function rule applied to the inner \sqrt[q]{x} gives exactly the stated domain — all of \mathbb{R} if q is odd, [0,\infty) if q is even — and the outer whole-number power p restricts nothing further, since raising any real number to a whole-number power is always defined.

For the sign, write y = \sqrt[q]{x}, so x^{p/q} = y^p. If p is even, y^p \ge 0 for every real y, so the output is never negative no matter what x is. If p is odd, y^p carries the same sign as y — and y=\sqrt[q]{x} carries the same sign as x (an odd root preserves sign; an even root, defined only where x \ge 0, is itself \ge 0) — so the output matches the sign of x.

By definition, \log_b(u) is the exponent y satisfying b^y = u. For a base b > 0, b^y is strictly positive for every real y: a positive number raised to a positive-integer power stays positive, raised to a negative power becomes the reciprocal of a positive number (still positive), and any root of a positive number is again taken to be positive. So the equation b^y = u can only ever be satisfied when u itself is positive. There is no real exponent y that produces 0 (the powers of a positive base approach but never reach 0) or a negative value.

Whenever u \leq 0, the defining equation b^y = u has no solution, so \log_b(u) names nothing; it's undefined. Substituting u = g(x) gives the rule for an arbitrary argument: \log_b(g(x)) is defined exactly where g(x) > 0.

Both directions of each statement lean on the pair of inverse identities \log_b(b^x) = x and b^{\log_b x} = x.

Given b^m = b^n, apply \log_b to both sides: \log_b(b^m) = \log_b(b^n), and the inverse identity collapses each side to just the exponent, leaving m = n. The converse — equal inputs to b^{(\cdot)} give equal outputs — holds simply because b^{(\cdot)} is a function.

Given \log_b M = \log_b N, apply b^{(\cdot)} to both sides: b^{\log_b M} = b^{\log_b N}, and the other inverse identity collapses each side back to the original argument, leaving M = N. Again the converse holds because \log_b is a function.

The restriction b \neq 1 is exactly what makes this work: for 0 < b < 1 or b > 1, b^x is strictly increasing or strictly decreasing, so it never repeats an output and the inverse identities are valid. At b = 1 that fails completely — 1^m = 1^n = 1 for every m and n, so equal outputs say nothing at all about the exponents, which is precisely why b = 1 is excluded from the theorem.

Deriving this formula from scratch needs a differential equation (the rate T changes is proportional to the current gap T-T_a), which is beyond this course. What can be checked directly, with nothing but exponent rules, is that the formula behaves exactly the way that defining property demands.

Matches the starting temperature. At t=0, T(0) = T_a + (T_0-T_a)e^0 = T_a + (T_0-T_a) = T_0.

Approaches, never crosses, the ambient temperature. Since k>0, e^{-kt}\to0 as t\to\infty, so T(t)\to T_a+0=T_a — and e^{-kt} is never 0 for any finite t, so the gap T(t)-T_a=(T_0-T_a)e^{-kt} never actually reaches zero.

The gap shrinks by the same factor over any equal time step, which is the real content of "cooling slows down as the gap closes." For any time t and step \Delta t, T(t+\Delta t) - T_a = (T_0-T_a)e^{-k(t+\Delta t)} = \big[(T_0-T_a)e^{-kt}\big]\cdot e^{-k\Delta t} = \big(T(t)-T_a\big)\cdot e^{-k\Delta t}. The multiplier e^{-k\Delta t} depends only on \Delta t, not on t — so the gap after any additional 10 minutes, say, is always the same fraction of whatever the gap was 10 minutes earlier, whether that's the first 10 minutes or the fifty-first. That is exactly the "loses the same percentage of what's left" pattern every other exponential-decay model in this chapter shares, just applied to a gap instead of to the quantity itself.

Both scales measure the same full rotation, just in different units: 360^\circ = 2\pi \text{ rad}. Divide both sides by 360 to find what one degree equals in radians: 1^\circ = \frac{2\pi}{360} \text{ rad} = \frac{\pi}{180} \text{ rad}. Multiplying by \theta_{\text{deg}} scales this up to any angle: \theta_{\text{rad}} = \theta_{\text{deg}} \cdot \frac{\pi}{180}. The reverse conversion is the reciprocal factor, \frac{180}{\pi}.

By definition, one radian is the angle subtended by an arc equal in length to the radius — that is, an angle of 1 radian goes with an arc of length 1 \cdot r = r. Angles scale arc length proportionally: doubling the angle doubles the arc, tripling the angle triples it, and so on, because the arc is being swept out at a constant rate as \theta grows. So an angle of \theta radians — \theta copies of "one radius-length of arc" — subtends an arc of length \theta \cdot r = r\theta. The formula isn't really a new fact so much as radians' own definition, restated in symbols.

A sector with angle \theta is one particular fraction of the whole circle — specifically, the fraction \frac{\theta}{2\pi}, since 2\pi is the angle of the complete circle. The whole circle has area \pi r^2, so the sector's area is that same fraction of the total: A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{\theta \pi r^2}{2\pi} = \frac{1}{2}r^2\theta. The \pi in the numerator and denominator cancel, leaving a formula with no \pi in it at all — it's already baked into \theta being in radians.

Start from the arc-length formula, s = r\theta (with \theta in radians). Both sides describe the same motion over the same time interval t, so divide both sides by t: \frac{s}{t} = \frac{r\theta}{t} = r \cdot \frac{\theta}{t} The left side, \frac{s}{t}, is exactly the definition of linear speed v; the factor \frac{\theta}{t} on the right is exactly the definition of angular speed \omega. Substituting both gives v = r\omega.

Take four identical copies of the right triangle (legs a, b, hypotenuse c) and arrange them inside a square of side length a+b, with each triangle's hypotenuse forming one side of a smaller, tilted square in the middle. That tilted inner square has side length c, so its area is c^2.

The big square's area can be computed two ways. Directly, it's (a+b)^2. Piece by piece, it's the tilted inner square plus the four triangles: c^2 + 4 \cdot \frac{1}{2}ab = c^2 + 2ab. Both expressions describe the same region, so they're equal: (a+b)^2 = c^2 + 2ab a^2 + 2ab + b^2 = c^2 + 2ab a^2 + b^2 = c^2 A fully general treatment — proving the dissection actually tiles the square with no gaps or overlaps for any right triangle — is geometry-course territory; the rearrangement above is the intuition that makes the result believable, not just memorizable.

Every entry is just opposite/hypotenuse, adjacent/hypotenuse, or opposite/adjacent, read straight off the two triangles derived above.

From the 1 : \sqrt3 : 2 triangle, at the 30° angle the opposite leg is 1, the adjacent leg is \sqrt3, and the hypotenuse is 2: \sin 30° = \frac{1}{2}, \qquad \cos 30° = \frac{\sqrt3}{2}, \qquad \tan 30° = \frac{1}{\sqrt3} = \frac{\sqrt3}{3} At the 60° angle in the same triangle, opposite and adjacent swap (the side that was adjacent to 30° is opposite 60°, and vice versa): \sin 60° = \frac{\sqrt3}{2}, \qquad \cos 60° = \frac{1}{2}, \qquad \tan 60° = \sqrt3 From the 1 : 1 : \sqrt2 triangle, at either 45° angle opposite and adjacent are both 1: \sin 45° = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \cos 45° = \frac{\sqrt2}{2}, \qquad \tan 45° = \frac{1}{1} = 1

The horizontal line at A's feet and the horizontal line at B's altitude are parallel (both are just "horizontal"), and the line of sight connecting A and B is a single straight line crossing both. A line crossing two parallel lines creates equal alternate interior angles wherever it cuts them — a geometry fact independent of trigonometry — and the elevation/depression pair is exactly that alternate-interior-angle pair.

Take an acute angle \theta, so its terminal side crosses the unit circle at a point (x,y) in the first quadrant. Drop a perpendicular segment straight down from (x,y) to the x-axis. That segment, the x-axis, and the terminal side form a right triangle with:

• hypotenuse = 1 (the radius of the unit circle),
• the side adjacent to \theta running along the x-axis, with length x,
• the side opposite \theta standing vertically, with length y.

By the right-triangle definition already in hand, \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{1} = x, \qquad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{1} = y. So for every acute angle, the coordinate definition and the right-triangle definition hand back the exact same numbers — the unit circle isn't replacing the old definition, it's the same idea with the hypotenuse fixed at 1 so "adjacent" and "opposite" collapse into plain x and y.

For \theta in any other quadrant, drop the same perpendicular from (x,y) to the x-axis and let \theta' be the resulting reference angle — the acute angle between the terminal side and the x-axis. That perpendicular still forms a right triangle with hypotenuse 1, legs |x| and |y|, and angle \theta', so the acute case already proven gives \cos\theta' = |x| and \sin\theta' = |y|. The sign is supplied automatically, not separately argued: x and y are the point's actual (signed) coordinates, negative exactly in the quadrants where the terminal side puts them, so \cos\theta = x and \sin\theta = y hold with their correct sign built in — nothing about the coordinate definition ever stops applying, only the right-triangle language ("opposite," "adjacent") stops making sense once \theta passes 90°.

Scale (x, y) down to the unit circle by dividing by r: the point \left(\frac{x}{r}, \frac{y}{r}\right) is on the terminal side of the same angle \theta and satisfies \left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = \frac{x^2+y^2}{r^2} = 1, so it's the actual unit-circle point for \theta. By the unit circle definition, \cos\theta = \frac{x}{r} and \sin\theta = \frac{y}{r}; the other four follow from their usual quotient and reciprocal relationships.

One full cycle of the parent \sin\theta occurs exactly as \theta runs through an interval of length 2\pi — that's the defining fact about the period of sine.

Here \theta has been replaced by B(x - C). Setting up the same "one full sweep" condition: the cycle completes as B(x-C) increases by 2\pi.

Solve for how much x itself must increase. If B(x-C) increases by 2\pi over an x-interval of length T, then B \cdot T = 2\pi, so T = \frac{2\pi}{B}.

Taking |B| handles a negative B (which reflects the wave horizontally but can't produce a negative period, since a period is a length): T = \frac{2\pi}{|B|}.

\tan\theta=\sin\theta/\cos\theta and \sec\theta=1/\cos\theta are both undefined exactly where \cos\theta=0 — at \theta=\frac{\pi}{2}+k\pi, the zeros of cosine established in the unit-circle chapter. \cot\theta=\cos\theta/\sin\theta and \csc\theta=1/\sin\theta are both undefined exactly where \sin\theta=0 — at \theta=k\pi, sine's zeros. Every one of the four asymptote locations is nothing more than "where does the function's own denominator vanish," read straight off sine and cosine's already-known zeros.

The terminal side of \theta meets the unit circle at the point (\cos\theta, \sin\theta), and every point on the unit circle satisfies x^2 + y^2 = 1 by definition of the circle. Substitute x = \cos\theta and y = \sin\theta directly: \cos^2\theta + \sin^2\theta = 1. That's the whole proof. No triangle, no case split for which quadrant \theta lands in — the unit-circle equation itself doesn't care, so neither does this identity.

Place two points on the unit circle: P = (\cos\alpha, \sin\alpha) at angle \alpha from standard position, and Q = (\cos\beta, \sin\beta) at angle \beta. The angle between segments OP and OQ (where O is the origin) has measure \alpha - \beta. We'll compute the straight-line distance PQ two different ways and set the results equal.

Way 1 — the distance formula. PQ^2 = (\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2 Expand both squares: PQ^2 = \cos^2\alpha - 2\cos\alpha\cos\beta + \cos^2\beta + \sin^2\alpha - 2\sin\alpha\sin\beta + \sin^2\beta Regroup by point, and apply the Pythagorean identity to each point separately — \cos^2\alpha+\sin^2\alpha=1 since P is on the unit circle, and likewise for Q: PQ^2 = \underbrace{(\cos^2\alpha+\sin^2\alpha)}_{=\,1} + \underbrace{(\cos^2\beta+\sin^2\beta)}_{=\,1} - 2(\cos\alpha\cos\beta+\sin\alpha\sin\beta) = 2 - 2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)

Way 2 — the Law of Cosines. Triangle OPQ has two sides of length 1 (both P and Q are on the unit circle, so OP = OQ = 1) enclosing an angle of measure \alpha-\beta. The Law of Cosines gives PQ^2 = OP^2 + OQ^2 - 2\cdot OP\cdot OQ\cdot\cos(\alpha-\beta) = 1 + 1 - 2\cos(\alpha-\beta) = 2 - 2\cos(\alpha-\beta) This route treats \alpha-\beta as an actual interior angle of a triangle, which only makes literal sense for 0<\alpha-\beta<\pi — Way 1 makes no such assumption at all (it's pure coordinate algebra, valid for every real \alpha,\beta), which is what actually earns the theorem's "for any angles" claim; Way 2 is included because the Law-of-Cosines route is the more memorable picture, not because it alone would suffice.

Set the two expressions for PQ^2 equal — they measure the same segment: 2 - 2(\cos\alpha\cos\beta+\sin\alpha\sin\beta) = 2 - 2\cos(\alpha-\beta) Subtract 2 from both sides, then divide by -2: \cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos(\alpha-\beta) which is the identity, read right to left. \blacksquare

A nonvertical line's slope is the tangent of its inclination angle, the angle it makes with the positive x-axis. Let \theta_1, \theta_2 be the two lines' inclination angles, so m_1=\tan\theta_1 and m_2=\tan\theta_2. One angle between the two lines is \theta_2-\theta_1, and the tangent difference formula gives \tan(\theta_2-\theta_1) = \frac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_1\tan\theta_2} = \frac{m_2-m_1}{1+m_1m_2}. Two intersecting lines actually form two pairs of equal vertical angles — one acute (or right), one obtuse (or right) — and \theta_2-\theta_1 might land on either one depending on which line is labeled "1" and which is "2." Taking the absolute value of the ratio always selects the acute angle: if the raw ratio is negative, \theta_2-\theta_1 named the obtuse angle, and its supplement (the acute one) has the same tangent magnitude with the opposite sign, which the absolute value produces either way, without needing to know in advance which of the two named angles was acute.

1+m_1m_2=0 making \tan\theta undefined recovers the perpendicular test from Chapter 4 exactly: m_1m_2=-1 is precisely the negative- reciprocal condition, and a tangent that blows up to infinity signals an angle of 90°, consistent with that earlier, purely algebraic rule.

Start from the target shape and expand it with the sine sum formula proved earlier in this chapter: R\sin(\theta+\varphi) = R\sin\theta\cos\varphi + R\cos\theta\sin\varphi = (R\cos\varphi)\sin\theta + (R\sin\varphi)\cos\theta

For this to equal a\sin\theta+b\cos\theta for every value of \theta, the coefficients of \sin\theta and \cos\theta must match independently (two different, linearly independent functions of \theta can only agree everywhere if their coefficients agree term by term): R\cos\varphi = a \qquad\text{and}\qquad R\sin\varphi = b

Square both equations and add them. The left side collapses via the Pythagorean identity: R^2\cos^2\varphi + R^2\sin^2\varphi = a^2+b^2 \implies R^2\underbrace{(\cos^2\varphi+\sin^2\varphi)}_{=\,1} = a^2+b^2 \implies R = \sqrt{a^2+b^2}

With R known, divide the two original equations by R to isolate \varphi: \cos\varphi = \frac aR, \qquad \sin\varphi = \frac bR \qquad\blacksquare

Substitute \alpha=\beta=\theta into the sine sum formula \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta: \sin(2\theta) = \sin(\theta+\theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta. The two middle terms are identical, so they combine by ordinary like-terms addition — not a new trig fact, just x + x = 2x wearing a trig costume.

Set \alpha=\beta=\theta in the cosine sum formula \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta: \cos(2\theta) = \cos(\theta+\theta) = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta. That's the first form, and no further work is required — it's already a true statement about \cos(2\theta).

The other two forms come from a single substitution each, using the Pythagorean identity \sin^2\theta+\cos^2\theta=1. Replace \sin^2\theta with 1-\cos^2\theta: \cos(2\theta) = \cos^2\theta - \sin^2\theta = \cos^2\theta - (1-\cos^2\theta) = 2\cos^2\theta - 1. Or replace \cos^2\theta with 1-\sin^2\theta instead: \cos(2\theta) = \cos^2\theta - \sin^2\theta = (1-\sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta. All three expressions equal \cos(2\theta); they're the same quantity written three ways, not three different facts.

By the quotient identity, \tan(2\theta) = \dfrac{\sin(2\theta)}{\cos(2\theta)}. Substitute the sine and cosine double-angle forms just proved: \tan(2\theta) = \frac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta} Divide every term, top and bottom, by \cos^2\theta — valid only once \cos\theta\neq0, i.e. \theta\neq\frac{\pi}{2}+k\pi, since dividing by \cos^2\theta is what introduces \tan\theta in the first place: \tan(2\theta) = \frac{\dfrac{2\sin\theta\cos\theta}{\cos^2\theta}}{\dfrac{\cos^2\theta}{\cos^2\theta} - \dfrac{\sin^2\theta}{\cos^2\theta}} = \frac{2\tan\theta}{1-\tan^2\theta} Two separate restrictions survive into the final form, for two different reasons. \theta \neq \frac{\pi}{2}+k\pi is carried over from the division step above — \tan\theta itself doesn't exist there, so the right-hand side can't even be evaluated, even though \tan(2\theta) on the left is perfectly fine (e.g. \theta=\frac{\pi}{2} gives \tan(2\theta)=\tan\pi=0). Separately, \tan\theta \neq \pm 1 keeps the denominator 1-\tan^2\theta away from 0 — and that condition is exactly \theta = \frac{\pi}{4}+k\frac{\pi}{2}, precisely where 2\theta itself has no tangent, so both sides genuinely agree there is no value to assign.

Start from \cos(2\theta) = 1 - 2\sin^2\theta. Subtract 1 from both sides, then multiply by -1: \cos(2\theta) - 1 = -2\sin^2\theta \quad\Longrightarrow\quad 1 - \cos(2\theta) = 2\sin^2\theta Divide both sides by 2: \sin^2\theta = \frac{1-\cos(2\theta)}{2} Start instead from \cos(2\theta) = 2\cos^2\theta - 1. Add 1 to both sides: \cos(2\theta) + 1 = 2\cos^2\theta Divide both sides by 2: \cos^2\theta = \frac{1+\cos(2\theta)}{2}

Substitute \theta \to \dfrac{\theta}{2} into both power-reduction identities (so every 2\theta becomes \theta): \sin^2\frac{\theta}{2} = \frac{1-\cos\theta}{2}, \qquad \cos^2\frac{\theta}{2} = \frac{1+\cos\theta}{2} Square-rooting a squared quantity always produces a \pm — the square root property from Chapter 3 applies here exactly as it did to any equation of the form u^2 = k: \sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}, \qquad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}} Dividing the two gives the tangent version, and it also simplifies to two radical-free forms, \tan\dfrac{\theta}{2} = \dfrac{1-\cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1+\cos\theta}, which carry their own sign automatically — but the radical form above is the one that matches sine and cosine's pattern, and the one this course uses by default.

Write the cosine difference and sum formulas together: \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta, \qquad \cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta Add them — the \sin\alpha\sin\beta terms cancel, leaving \cos\alpha\cos\beta doubled: \cos(\alpha-\beta)+\cos(\alpha+\beta) = 2\cos\alpha\cos\beta \implies \cos\alpha\cos\beta = \tfrac12\big[\cos(\alpha-\beta)+\cos(\alpha+\beta)\big] Subtract them instead — the \cos\alpha\cos\beta terms cancel, leaving \sin\alpha\sin\beta doubled: \cos(\alpha-\beta)-\cos(\alpha+\beta) = 2\sin\alpha\sin\beta \implies \sin\alpha\sin\beta = \tfrac12\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big] The same add-and-cancel move on the sine sum and difference formulas, \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta and \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta, isolates the third identity above.

A function has an inverse only where it's one-to-one — otherwise "the" angle with \sin y = x isn't unique, and \arcsin x couldn't return a single answer. Sine, cosine, and tangent are each periodic, so none of them is one-to-one on their full natural domain: infinitely many angles share the same sine (or cosine, or tangent).

Restricting to one piece where the function is strictly monotonic (always increasing, or always decreasing) fixes that, since a monotonic function never repeats a value. \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] is exactly one full monotonic piece of sine (increasing from -1 to 1); [0,\pi] is one full monotonic piece of cosine (decreasing from 1 to -1); \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) is one full monotonic piece of tangent (increasing across every real value). Each restricted function is now one-to-one on its piece, so it has a genuine inverse — and that inverse's domain is exactly the parent's restricted range, while its range is exactly the parent's restricted domain, which is precisely what's stated above.

Other monotonic pieces exist too (sine is also monotonic on \left[\frac{\pi}{2}, \frac{3\pi}{2}\right], for instance) — the pieces above are simply the standard convention, chosen for sitting closest to the origin.

This is periodicity, not a new fact: \sin(\theta + 2\pi) = \sin\theta for every \theta, since a full revolution returns the terminal side to the same point. So if \theta_1 solves \sin\theta = k, then \theta_1 + 2\pi, \theta_1 + 4\pi, \theta_1 - 2\pi, and so on all solve it too — and because 2\pi is sine's smallest period, no solutions are skipped by stepping in increments of anything less. The same argument with period \pi handles tangent, where the extra shortcut — \theta_1 and \theta_1 + \pi always landing in opposite, sign-matching quadrants — is exactly why one family, not two, exhausts every solution.

Take triangle ABC and drop the altitude from vertex C straight down to side AB, meeting it (or its extension) at point D, with length h. This slices the triangle into two right triangles, ACD and BCD, both sharing that same altitude h.

In right triangle ACD, angle A sits opposite the altitude, and the hypotenuse of that right triangle is side b (side AC, opposite vertex B). So \sin A = \frac{h}{b} \quad\Longrightarrow\quad h = b\sin A.

In right triangle BCD, angle B sits opposite the same altitude, and the hypotenuse of that right triangle is side a (side BC, opposite vertex A). So \sin B = \frac{h}{a} \quad\Longrightarrow\quad h = a\sin B.

Both expressions equal the same h, so b\sin A = a\sin B \quad\Longrightarrow\quad \frac{a}{\sin A} = \frac{b}{\sin B}.

Drop a second altitude — say from vertex A to side BC — and the identical argument relates the other pair: \frac{b}{\sin B} = \frac{c}{\sin C}.

Chaining the two gives \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. (If angle C happens to be obtuse, the altitude from C lands outside segment AB rather than on it — but the identity \sin(180° - \theta) = \sin\theta makes the same two equations hold regardless, so the result survives untouched.)

Place triangle ABC with angle C at vertex C, flanked by sides CA = b and CB = a; the side opposite C is AB = c. Drop an altitude from vertex B down to line CA, meeting it at point D, with height h = BD.

Triangle BDC is a right triangle with hypotenuse CB = a and angle C at vertex C, so CD = a\cos C, \qquad h = a\sin C. (If C is obtuse, \cos C is negative and D falls past C, outside segment CA — the signed length CD = a\cos C still comes out correct, so nothing below needs to change for that case.)

That leaves AD = CA - CD = b - a\cos C. Triangle ABD has its right angle at D and hypotenuse AB = c, so the Pythagorean theorem gives c^2 = AD^2 + h^2 = (b - a\cos C)^2 + (a\sin C)^2. Expand and group the a^2 terms together: c^2 = b^2 - 2ab\cos C + a^2\cos^2 C + a^2\sin^2 C = b^2 - 2ab\cos C + a^2\left(\cos^2 C + \sin^2 C\right). The Pythagorean identity collapses \cos^2 C + \sin^2 C to 1: c^2 = a^2 + b^2 - 2ab\cos C. Relabeling the same argument around the other two vertices gives the A- and B-rotations.

Sanity check: set C = 90°. Then \cos C = 0, the whole correction term vanishes, and what's left is c^2 = a^2 + b^2 — the Pythagorean theorem, exactly as it must be, since a right triangle is just the C = 90° special case of every triangle.

Treat side a as the base. Drop the altitude from the vertex opposite a down to the line containing a; inside the right triangle it forms with side b and angle C, that altitude has length h = b\sin C. Substitute into the familiar \text{Area} = \frac12 \times \text{base} \times \text{height}: \text{Area} = \frac12 \cdot a \cdot (b\sin C) = \frac12 ab\sin C. When C=90°, \sin C = 1 and this collapses to the ordinary \text{Area} = \frac12 ab — because at a right angle, side b is the height, standing straight up from the base with nothing left to project.

A full derivation starts from the SAS area formula \text{Area}=\frac12 ab\sin C, converts \sin C to \cos C via the Pythagorean identity, then substitutes the Law of Cosines' expression for \cos C in terms of a, b, c — the same two tools from earlier in this chapter, combined. The algebra that follows is a long chain of factoring (it eventually needs the difference-of-squares pattern twice over), so rather than grinding through every line here, take away the important fact: Heron's formula isn't a new idea, it's the SAS area formula and the Law of Cosines fused into one side-only shortcut.

Dividing a vector by the positive scalar |\mathbf{v}| is scalar multiplication by k = \frac{1}{|\mathbf{v}|} > 0, which — as established above — scales magnitude by |k| and leaves direction unchanged. So \hat{\mathbf{v}} points the same way as \mathbf{v}, and its magnitude is

Draw \mathbf{u} and \mathbf{v} from a common starting point O, with tips at points U and V. The three points O, U, V form a triangle: two sides have lengths |\mathbf{u}| and |\mathbf{v}|, meeting at angle \theta; the third side, from U to V, has length |\mathbf{u}-\mathbf{v}|. The Law of Cosines applied to this triangle gives

Now expand the left side directly from components. Writing \mathbf{u}=\langle u_1,u_2\rangle and \mathbf{v}=\langle v_1,v_2\rangle, so \mathbf{u}-\mathbf{v} = \langle u_1-v_1,\ u_2-v_2\rangle:

The first four terms are exactly |\mathbf{u}|^2+|\mathbf{v}|^2, and the last piece is 2(\mathbf{u}\cdot\mathbf{v}) by the algebraic definition, so

Both boxed expressions equal |\mathbf{u}-\mathbf{v}|^2, so set them equal to each other and cancel the shared |\mathbf{u}|^2+|\mathbf{v}|^2 from both sides:

Divide both sides by -2:

Drop a perpendicular from the tip of \mathbf{v} down onto the line through \mathbf{u}; this forms a right triangle with hypotenuse |\mathbf{v}| and angle \theta between \mathbf{u} and \mathbf{v}. Exactly as in Chapter 8, the leg lying along \mathbf{u}'s line — the scalar projection — has length |\mathbf{v}|\cos\theta (adjacent equals hypotenuse times cosine). From the theorem above, \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}, so the scalar projection simplifies to

The vector projection points along \mathbf{u}'s own direction, so multiply this scalar length by the unit vector \frac{\mathbf{u}}{|\mathbf{u}|}:

This is nothing but the polar-to-rectangular conversion from the polar coordinate system, relabeled. There, a point at distance r and angle \theta has coordinates x = r\cos\theta and y = r\sin\theta. Here, the real axis is standing in for x and the imaginary axis for y, so the same two equations read a = r\cos\theta, \qquad b = r\sin\theta. Substitute directly into z = a + bi: z = a + bi = r\cos\theta + i(r\sin\theta) = r(\cos\theta + i\sin\theta). No new geometry was needed — the polar coordinate system already proved the conversion; this is that exact result wearing an i.

Multiply the two polar forms out like any product of binomials, treating \cos\alpha + i\sin\alpha and \cos\beta+i\sin\beta as the two factors: z_1z_2 = r_1r_2(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta) = r_1r_2\big[\cos\alpha\cos\beta + i\cos\alpha\sin\beta + i\sin\alpha\cos\beta + i^2\sin\alpha\sin\beta\big] +++ Apply i^2 = -1 to the last term, then separate the real terms from the i terms: = r_1r_2\Big[\underbrace{(\cos\alpha\cos\beta - \sin\alpha\sin\beta)}_{\text{real part}} + i\underbrace{(\sin\alpha\cos\beta+\cos\alpha\sin\beta)}_{\text{imaginary part}}\Big] +++ Both underbraced pieces are exactly the sum formulas proved in the identities chapters: \cos\alpha\cos\beta-\sin\alpha\sin\beta = \cos(\alpha+\beta), \qquad \sin\alpha\cos\beta+\cos\alpha\sin\beta = \sin(\alpha+\beta) +++ Substitute both in: z_1z_2 = r_1r_2\big[\cos(\alpha+\beta) + i\sin(\alpha+\beta)\big] \qquad \blacksquare

This is mathematical induction, a proof technique that gets a full treatment of its own in Chapter 14 — worth previewing here in full, rather than waiting, since it's the only way to certify a pattern for every positive integer n at once rather than checking cases one at a time. Applied to the statement P(n): "[r\operatorname{cis}\theta]^n = r^n\operatorname{cis}(n\theta)."

Base case (n=1): [r\operatorname{cis}\theta]^1 = r\operatorname{cis}\theta = r^1\operatorname{cis}(1\cdot\theta). True. +++ Inductive hypothesis: assume P(k) holds for some integer k \ge 1 — that is, assume [r\operatorname{cis}\theta]^k = r^k\operatorname{cis}(k\theta). +++ Inductive step: show P(k+1) follows. Split off one more factor of z = r\operatorname{cis}\theta: [r\operatorname{cis}\theta]^{k+1} = [r\operatorname{cis}\theta]^k \cdot [r\operatorname{cis}\theta] +++ Apply the inductive hypothesis to the first factor, then multiply the two polar numbers using the product rule just proved — moduli multiply, arguments add: = \big[r^k\operatorname{cis}(k\theta)\big]\cdot\big[r\operatorname{cis}\theta\big] = r^k \cdot r\;\operatorname{cis}\big(k\theta+\theta\big) = r^{k+1}\operatorname{cis}\big((k+1)\theta\big) This is exactly P(k+1). +++ Base case and inductive step both hold, so by the Principle of Mathematical Induction, [r\operatorname{cis}\theta]^n = r^n\operatorname{cis}(n\theta) for every integer n \ge 1. \blacksquare

Each row falls straight out of the standard forms from the last two sections — the general equation is exactly what you get if you multiply a standard form out and shove every term to one side.

Circle. The standard form is (x-h)^2+(y-k)^2=r^2. Expand it and the coefficients of x^2 and y^2 are both exactly 1 (or whatever common factor the whole equation gets multiplied by) — identical numbers, same sign, by construction. A=C is not a coincidence; it's the algebraic fingerprint of "the two directions are treated identically," which is precisely what makes a circle round instead of oval.

Ellipse. The standard form \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 has different denominators whenever a \neq b. Multiplying through by a^2b^2 to clear denominators gives leading coefficients b^2 and a^2 — both positive (same sign), but unequal whenever a \neq b. An ellipse is a circle that got stretched unevenly, and the inequality A \neq C is the algebra recording that unevenness.

Hyperbola. The standard form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 has a minus sign built into the definition itself — clearing denominators gives leading coefficients b^2 and -a^2, opposite signs no matter what a and b are. That minus sign isn't a computational accident; it's the entire geometric difference between a closed loop (ellipse) and two runaway branches (hyperbola).

Parabola. The standard forms y^2=4px and x^2=4py each contain only ONE squared variable. There's no term to write for the other one, which is the same as writing its coefficient as 0. A parabola isn't a squashed circle or a lopsided ellipse — it's a fundamentally different shape, with only one distance rule (focus and directrix) instead of two foci, and the missing squared term is exactly where that difference shows up in the algebra.

Substitute x=x'\cos\theta-y'\sin\theta and y=x'\sin\theta+y'\cos\theta into Ax^2+Bxy+Cy^2 and expand. Collecting only the x'y' terms from each piece: Ax^2 \to -2A\cos\theta\sin\theta\;x'y' + \cdots, \qquad Cy^2 \to 2C\sin\theta\cos\theta\;x'y' + \cdots Bxy \to B(\cos^2\theta-\sin^2\theta)\;x'y' + \cdots +++ Add the three x'y' coefficients together: \big[-2A\cos\theta\sin\theta + 2C\sin\theta\cos\theta + B(\cos^2\theta-\sin^2\theta)\big]x'y' Apply the double-angle identities 2\sin\theta\cos\theta=\sin2\theta and \cos^2\theta-\sin^2\theta=\cos2\theta from earlier in this course: = \big[(C-A)\sin2\theta + B\cos2\theta\big]x'y' +++ Set this bracket to 0 (the whole point: no x'y' term survives) and solve for \theta: (C-A)\sin2\theta = -B\cos2\theta \implies \frac{\cos2\theta}{\sin2\theta} = \frac{A-C}{B} \implies \cot2\theta = \frac{A-C}{B} \qquad\blacksquare

A polar pair (r, \theta) names a point by facing the direction \theta from the origin and walking a signed distance r along that ray — positive r forward, negative r backward along the same line.

Adding 2\pi k. Turning through a full extra 2\pi (or several) returns to facing exactly the same direction — \theta and \theta+2\pi k are coterminal angles, sharing one terminal ray. Walking the same signed distance r along the same ray lands on the same point.

Flipping the sign and adding \pi. Turning through an extra \pi reverses the facing direction to point exactly the opposite way. Walking distance -r backward along that reversed ray retraces the same path as walking distance r forward along the original ray — same point, reached from the opposite facing.

This last claim can be checked algebraically using the reduction formulas \cos(\theta+\pi) = -\cos\theta and \sin(\theta+\pi) = -\sin\theta: rectangular coordinates built from (-r, \theta+\pi) come out to (-r)\cos(\theta+\pi) = (-r)(-\cos\theta) = r\cos\theta, \qquad (-r)\sin(\theta+\pi) = (-r)(-\sin\theta) = r\sin\theta, exactly matching the coordinates built from (r, \theta) — confirming both name the same point.

Since either move alone leaves the point unchanged, applying both in any combination still leaves it unchanged, which is why the two rules combine freely.

This is nothing but the unit-circle definitions of cosine and sine — \cos\theta and \sin\theta are the x- and y-coordinates on a circle of radius 1 — stretched from radius 1 out to radius r.

r = \sqrt{x^2+y^2} is just the distance formula from the origin to (x,y). Dividing the two conversion equations above, \frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \tan\theta, whenever x \neq 0.

Take r=a\cos(n\theta) (the sine version is the same curve rotated, so the petal count is identical). r is zero exactly when n\theta = \frac\pi2+k\pi, i.e. at \theta = \frac{\pi}{2n}+\frac{k\pi}{n} — consecutive zeros spaced \frac{\pi}{n} apart. Between any two consecutive zeros, r sweeps from 0 out to an extreme of \pm a and back to 0: one petal-shaped arch, of length |a|. A full revolution \theta\in[0,2\pi) contains 2\pi \div \frac{\pi}{n} = 2n such arches — the question is how many of those 2n arches are genuinely distinct petals versus retraces of one already drawn.

Compare the arch starting at \theta to the one starting at \theta+\pi: r(\theta+\pi) = a\cos(n\theta+n\pi) = a\cos(n\theta)\cos(n\pi) = (-1)^n\, r(\theta), since \sin(n\pi)=0 for every integer n.

• n odd: (-1)^n=-1, so r(\theta+\pi) = -r(\theta). A point plotted with a negative radius at angle \theta+\pi lands at the exact same location as a positive radius at angle \theta — that's the standard (-r,\alpha)=(r,\alpha+\pi) equivalence, applied here to give (-r(\theta),\ \theta+\pi) = (r(\theta),\ \theta+2\pi) = (r(\theta),\ \theta). So the arch at \theta+\pi exactly retraces the arch at \theta, meaning the second half of the revolution (\theta\in[\pi,2\pi)) duplicates the first half — only n of the 2n arches are distinct.
• n even: (-1)^n=1, so r(\theta+\pi) = r(\theta): the SAME radius at the OPPOSITE angle, which is a different point (unless r=0). No duplication occurs, so all 2n arches are distinct petals.

Take r = a+b\cos\theta (the sine version is the same curve rotated). The loop/cardioid/no-loop boundary is a question of whether r ever reaches 0: solving a+b\cos\theta=0 gives \cos\theta = -a/b, which has a real solution exactly when -1 \le -a/b \le 1, i.e. a/b \le 1 (using a,b>0).

• a/b < 1: \cos\theta=-a/b has two solutions in [0,2\pi), so r crosses zero twice — once going negative, once coming back — tracing a small inner loop while r<0.
• a/b = 1: \cos\theta = -1 has exactly one solution, \theta=\pi, where r touches 0 without crossing into negative values anywhere else — the loop degenerates to a single point, the cardioid's cusp.
• a/b > 1: \cos\theta=-a/b has no real solution at all (the right side is outside [-1,1]), so r never reaches 0 and the curve has no loop and no cusp.

The finer dimpled-versus-convex split at a/b=2, among the no-loop curves, isn't a question of where r is zero — both shapes have r>0 everywhere. It turns on where the curve's boundary is locally concave instead of convex, which is a statement about curvature (built from first AND second derivatives of r with respect to \theta) rather than about r's value alone, so pinning down a/b=2 exactly needs more calculus than this course builds. It's taken on faith here.

Start from the time of flight found by factoring y(t)=0: v_0\sin\theta \cdot t - \tfrac12 gt^2 = t\left(v_0\sin\theta - \tfrac12gt\right) = 0, so the nonzero root is T = \dfrac{2v_0\sin\theta}{g}. +++ Substitute T into x(t) = v_0\cos\theta \cdot t: R = x(T) = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g} = \frac{2v_0^2\sin\theta\cos\theta}{g}. +++ The double-angle identity \sin(2\theta) = 2\sin\theta\cos\theta collapses the numerator to a single term: R = \frac{v_0^2 \sin(2\theta)}{g}.

2\times2 swap. \det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc. Swapping the rows gives \det\begin{pmatrix}c&d\\a&b\end{pmatrix}=cb-da= -(ad-bc) — exactly the negative of the original.

2\times2 scaling. Scaling row 1 by k gives \det\begin{pmatrix}ka&kb\\c&d\end{pmatrix}=(ka)d-(kb)c=k(ad-bc) — exactly k times the original. Scaling row 2 (or either column) works identically, by the same factoring.

3\times3 scaling. Expanding along the scaled row, each cofactor term carries exactly one factor from that row, so scaling the row by k scales every term of the expansion — and therefore the whole sum — by k. This argument uses only that cofactor expansion is a sum of terms each linear in the entries of the expansion row, so it applies to any row (or, by the analogous column expansion, any column).

3\times3 swap. Swap rows 1 and 2, and expand the determinant along row 3 (the row NOT touched by the swap). Each cofactor in this expansion is the determinant of a 2\times2 minor formed by deleting row 3 and one column — and that minor's two remaining rows are exactly rows 1 and 2, which the swap exchanged. By the 2\times2 swap property just proved, each such minor negates. Row 3's entries (the expansion's coefficients) are untouched by the swap, so every term in the sum negates, and therefore the whole determinant negates. Swapping columns instead of rows follows the identical argument, expanding along a column instead of a row.

Zero row or column. Take k=0 in the scaling property: a row (or column) of all zeros scales the determinant by 0, so the determinant is 0 — regardless of what the matrix's other entries are.

Let \vec u = (x_2-x_1,\ y_2-y_1) and \vec v = (x_3-x_1,\ y_3-y_1) be the two vectors from vertex 1 to the other two vertices — the triangle's two side vectors. The 2\times2 determinant of two vectors is the signed area of the parallelogram they span: \det(\vec u, \vec v) = (x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1). The diagonal from (x_1,y_1) to the parallelogram's far corner splits that parallelogram into two congruent triangles, and the triangle with vertices (x_1,y_1), (x_2,y_2), (x_3,y_3) is exactly one of them. Its area is therefore half the parallelogram's, and taking the absolute value discards the sign the determinant carries for orientation: \text{Area} = \frac12\,|\det(\vec u, \vec v)|.

For the corollary: \vec u and \vec v point along the same line through the origin exactly when one is a scalar multiple of the other — the proportional-rows condition under which a 2\times2 determinant is 0. Geometrically, a triangle whose area has collapsed to 0 has been flattened into a line segment, and a triangle flattens into a line segment exactly when its three vertices already lay on one line to begin with.

Write q(x) = (x-r)^k m(x) with m(r) \neq 0 (every other factor of q). The cover-up value A_1 = \dfrac{p(r)}{m(r)} is well-defined since m(r)\neq0, and \frac{p(x)}{q(x)} - \frac{A_1}{(x-r)^k} = \frac{p(x) - A_1\,m(x)}{(x-r)^k m(x)}. The new numerator vanishes at x=r, since p(r) - A_1 m(r) = p(r) - p(r) = 0 by exactly how A_1 was chosen — so by the Factor Theorem, (x-r) divides it: p(x) - A_1 m(x) = (x-r)\,p_1(x) for some polynomial p_1. Substituting back, \frac{p(x)}{q(x)} - \frac{A_1}{(x-r)^k} = \frac{(x-r)\,p_1(x)}{(x-r)^k m(x)} = \frac{p_1(x)}{(x-r)^{k-1}m(x)}, a rational function with the SAME shape as the one just started with, except the power of (x-r) in the denominator dropped from k to k-1. Repeat the identical step on \frac{p_1(x)}{(x-r)^{k-1}m(x)} to peel off A_2 and drop to k-2, and so on. After k repetitions every power of (x-r) has been stripped out, leaving exactly \frac{A_1}{x-r}+\cdots+\frac{A_k}{(x-r)^k} pulled out and a remaining proper fraction over m(x) — the other factors of q(x), handled the same way by the rest of the Partial Fraction Decomposition Recipe.

Clearing every denominator turns the claimed decomposition into a single polynomial identity: p(x) on one side, and a polynomial built from the unknown constants on the other. Both sides have degree less than \deg q(x) = n, and two such polynomials are equal exactly when all n of their coefficients match — so the identity is really a system of n linear equations. It has exactly n unknown constants too: a linear factor (x-r)^k contributes k of them (matching its degree-k contribution to q), an irreducible quadratic factor repeated k times contributes 2k of them (matching ITS degree-2k contribution), and these degrees sum to n by construction. A system with as many equations as unknowns has a unique solution whenever its coefficient matrix is invertible; that this particular system's matrix always is, for any proper p(x)/q(x), needs more linear algebra than this course builds, so that part is taken on faith here.

What can be verified in full is the piece that does the real explanatory work — how a repeated linear factor's own k terms are actually produced, one at a time. See the Partial Fractions: Repeated Linear Factor card for that argument, proved directly from the Factor Theorem with no linear algebra at all; every other case in this recipe (distinct factors, irreducible quadratics) follows the identical peeling idea.

Assume (a,b)\neq(0,0) (otherwise z is the same constant everywhere and every point, vertices included, is trivially optimal). Suppose z attains its maximum at some feasible point P.

P can't be an interior point. If P is interior, every point close enough to P in every direction is still feasible. Moving a tiny step from P in the direction (a,b) changes z by a positive amount (that's what (a,b)\neq(0,0) buys), producing a nearby feasible point with a strictly larger z — contradicting that P was already maximal.

So P sits on the boundary — either at a vertex (done), or in the interior of an edge joining two vertices V_1 and V_2. Along that edge, every point is (1-t)V_1+tV_2 for some t\in[0,1], and because z is linear, z\big((1-t)V_1+tV_2\big) = (1-t)\,z(V_1) + t\,z(V_2), an affine function of t alone. An affine function on [0,1] reaches its largest value at t=0 or t=1 — the two endpoints — unless it's constant on the whole interval, in which case z(V_1)=z(V_2)=z(P) too. Either way, some vertex achieves a value at least as large as z(P)'s, and since P was already maximal, that vertex ties or matches it.

The minimum case is identical with every inequality reversed.

Inserting k means between a and b builds a sequence of k+2 terms total: a itself, the k inserted means, and b — so b sits k+1 steps after a.

Arithmetic case. By the nth-term formula, the term k+2 steps into the sequence (i.e. b, which is term number k+2 counting a as term 1) satisfies b = a + (k+1)d. Solving for d gives d=\dfrac{b-a}{k+1}. For k=1, d=\dfrac{b-a}{2}, and the single inserted mean is a+d = a+\dfrac{b-a}{2} = \dfrac{a+b}{2}, the ordinary average.

Geometric case. By the nth-term formula for a geometric sequence, b = a\cdot r^{k+1}. Solving for r gives r=\sqrt[k+1]{b/a} (the sign condition on a,b keeps b/a positive, so the root is a real number). For k=1, r=\sqrt{b/a}, and the single inserted mean is a\cdot r = a\sqrt{b/a} = \sqrt{a^2\cdot(b/a)} = \sqrt{ab}, the geometric mean.

Row 1 of Pascal's triangle already matches the claim: (a+b)^1 = a+b = \binom{1}{0}a + \binom{1}{1}b.

Suppose the formula holds for some row n: (a+b)^n = \sum_{k=0}^n \binom{n}{k}a^{n-k}b^k. Multiply both sides by (a+b): (a+b)^{n+1} = \sum_{k=0}^n \binom{n}{k}a^{n+1-k}b^k \ + \ \sum_{k=0}^n \binom{n}{k}a^{n-k}b^{k+1}. Reindex the second sum by k \to k-1 and combine it with the first: the coefficient of a^{n+1-k}b^k in the total is \binom{n}{k}+\binom{n}{k-1}, which is exactly \binom{n+1}{k} by the same row-building identity that produces Pascal's triangle. So (a+b)^{n+1} = \sum_{k=0}^{n+1}\binom{n+1}{k}a^{n+1-k}b^k, the identical formula one row later. That row-to-row step is precisely how each row of Pascal's triangle is generated from the one above it, so the formula holds for every row, starting from row 1.

This isn't really a theorem to prove so much as a fact to see. Lay the first choice's m options down the rows of a grid and the second choice's n options across the columns. Every cell of that grid is exactly one valid pairing — one option from the first choice matched with one option from the second — and every valid pairing lands in exactly one cell. A grid with m rows and n columns has m \times n cells. Counting the pairings and counting the cells are the same question, so they have the same answer. Chaining a third independent choice onto the pair just replaces each cell of that grid with a fresh n_3-row stack, and multiplication absorbs each new stage the same way.

This is just the counting principle, run r times in a row. The first position in the arrangement can be filled by any of the n items. The second position can be filled by any of the n - 1 items that remain (one has been used). The third by any of the n-2 remaining, and so on, down to the rth position, which has n - (r-1) = n - r + 1 choices left. Multiply across all r positions: P(n,r) = n \cdot (n-1) \cdot (n-2) \cdots (n - r + 1). That product has exactly r factors, starting at n and counting down. To see it equals \frac{n!}{(n-r)!}, write n! in full and notice the tail end cancels: n! = \bigl[n(n-1)\cdots(n-r+1)\bigr] \cdot (n-r)!, so dividing both sides by (n-r)! leaves exactly the product above.

Start from P(n,r), which counts every arrangement of r chosen items. Each unordered group of r items corresponds to many of those arrangements — specifically, every one of the r! ways to reorder that same group is a different permutation but the exact same combination. So P(n,r) overcounts each combination by a factor of r!, and dividing it out gives the true count: C(n,r) = \frac{P(n,r)}{r!} = \frac{n!/(n-r)!}{r!} = \frac{n!}{r!\,(n-r)!}.

Start from n!, the count if every one of the n objects were secretly distinct (as if each carried its own invisible label). Within that count, every group of n_1 identical objects can be relabeled among themselves in n_1! ways without changing the arrangement anyone can actually see — so n! overcounts the first repeated group by a factor of n_1!. The same overcounting happens independently for the second group (factor n_2!), the third, and so on. Dividing n! by n_1!\, n_2! \cdots n_k! removes every one of those redundant relabelings at once, since the groups' internal shufflings are independent of each other.

This is a weighted average, and the weighting is what makes it meaningful over the long run. Imagine repeating the underlying random process N times, for very large N: outcome x_i will occur approximately N p_i times (that's what "probability p_i" means in the long run). The total payout across all N trials is then approximately \sum_i x_i \cdot (N p_i), and dividing by N trials to get the average payout per trial leaves exactly \sum_i x_i p_i — the probabilities' role as relative frequencies is what turns a plain average into a probability-weighted one.

This principle isn't derived from something more basic in this course — it's taken as one of the foundational properties of the integers themselves (equivalent to the well-ordering principle: every nonempty set of integers bounded below has a least element, a fact from set theory this course doesn't build). The domino-chain intuition is a genuine guide to why it's believable — P(n_0) is true, and the inductive step means each true case knocks over the next one — but making that intuition airtight for infinitely many dominoes is exactly what the well-ordering principle supplies, and that belongs to a foundations-of-mathematics course, not this one. It's used freely here because every course that reaches integers uses it freely.

Rather than taking this as a second, separate axiom alongside the Principle of Mathematical Induction, it can be built from that ordinary principle directly — the only genuinely new idea is what statement to induct on.

Given P(n) satisfying strong induction's two hypotheses, define a new statement Q(n): "P(n_0), P(n_0+1), \ldots, P(n) are all true." Q(n_0) holds because that's exactly the base case. And if Q(k) holds — every one of P(n_0) through P(k) is true — then the strong inductive step says P(k+1) is true too, so Q(k+1) (everything through P(k+1)) holds as well. That's the base case and inductive step of the ordinary Principle of Mathematical Induction, applied to Q instead of P — so Q(n) is true for every n \ge n_0. Since Q(n) includes P(n) as one of its conjuncts, P(n) is true for every n \ge n_0 too.

The Conditional Probability formula, applied to P(A\mid B) and to P(B\mid A) separately, gives two ways to write the same joint probability: P(A\text{ and }B) = P(A\mid B)\,P(B), \qquad P(A\text{ and }B) = P(B\mid A)\,P(A). Both right-hand sides equal the same quantity, so they equal each other: P(A\mid B)\,P(B) = P(B\mid A)\,P(A). Dividing both sides by P(B) (nonzero by hypothesis) isolates P(A\mid B): P(A\mid B) = \frac{P(B\mid A)\,P(A)}{P(B)}.

Fix one particular arrangement of k successes and n-k failures — say, successes on the first k trials and failures on the rest. Independence lets the trials' probabilities multiply straight through: \underbrace{p \cdot p \cdots p}_{k}\; \cdot\; \underbrace{(1-p)\cdot(1-p)\cdots(1-p)}_{n-k} = p^k(1-p)^{n-k}. Every OTHER arrangement of k successes among the n trials — successes scattered in different positions — multiplies the exact same k factors of p and n-k factors of 1-p, just in a different order, so it has this same probability p^k(1-p)^{n-k} too. The number of such arrangements is exactly the number of ways to choose which k of the n trials are the successes: \binom{n}{k}. These arrangements are mutually exclusive (a specific sequence of results can't match two different success-position patterns at once), so their probabilities add: P(k\text{ successes}) = \binom{n}{k}\,p^k(1-p)^{n-k}.

The deposit made at the end of period k (for k=1,\dots,n) sits in the account for n-k remaining periods and grows to PMT(1+i)^{n-k}. Summing over every deposit, FV = \sum_{k=1}^{n} PMT(1+i)^{n-k} = PMT\sum_{j=0}^{n-1}(1+i)^{j}, reindexing with j=n-k. The remaining sum is a finite geometric series with first term 1, common ratio (1+i), and n terms, so the geometric partial-sum formula gives \sum_{j=0}^{n-1}(1+i)^{j} = \frac{(1+i)^{n}-1}{(1+i)-1} = \frac{(1+i)^{n}-1}{i}, which produces the boxed FV formula.

Present value is defined as the lump sum today that grows to the same FV after n periods at rate i, i.e. PV(1+i)^n=FV. Solving for PV and substituting the FV formula: PV = \frac{FV}{(1+i)^n} = \frac{PMT\big[(1+i)^n-1\big]}{i(1+i)^n} = PMT\cdot\frac{1-(1+i)^{-n}}{i}, splitting the fraction and simplifying \frac{(1+i)^n}{i(1+i)^n}=\frac1i and \frac{1}{i(1+i)^n}=\frac{(1+i)^{-n}}{i}.

The amortization payment follows immediately by solving the PV equation for PMT: multiply both sides by i and divide by 1-(1+i)^{-n}.