Algèbre

$\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$ natural, integer, rational, real, complex

Field$a+b=b+a$·$ab=ba$·$(a+b)+c=a+(b+c)$·$(ab)c=a(bc)$

$a+0=a$·$a\cdot 1=a$·$a+(-a)=0$·$a\cdot a^{-1}=1,\ a\neq 0$·$a(b+c)=ab+ac$

Orderexactly one of $a<b,\ a=b,\ a>b$·$a<b,\ b<c\Rightarrow a<c$

Completeevery bounded set in $\mathbb{R}$ has a supremum. $\mathbb{Q}$ does not.

Arithmetic$n=p_1^{k_1}\cdots p_r^{k_r}$ unique up to order·$\gcd\cdot\operatorname{lcm}=ab$

Bézout$ax+by=\gcd(a,b)$ solvable in $\mathbb{Z}$·Euclid: $\gcd(a,b)=\gcd(b,a\bmod b)$

Modulo$a\equiv b\ (m)\Leftrightarrow m\mid a-b$·Fermat $a^{p-1}\equiv 1\ (p)$, $p\nmid a$

Signs$(-a)(-b)=ab$·$-(a-b)=b-a$·$(-a)^{2k}=a^{2k}$·$(-a)^{2k+1}=-a^{2k+1}$

Fractions$\dfrac{a}{b}\pm\dfrac{c}{d}=\dfrac{ad\pm bc}{bd}$·$\dfrac{a}{b}\cdot\dfrac{c}{d}=\dfrac{ac}{bd}$·$\dfrac{a}{b}:\dfrac{c}{d}=\dfrac{ad}{bc}$

Proportion$\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc$·$\dfrac{ka}{kb}=\dfrac{a}{b},\ k\neq 0$

Zero$0\cdot a=0$·$ab=0\Leftrightarrow a=0$ or $b=0$·$a:0$ is undefined

Prioritybrackets, then powers and roots, then $\times\ :$, then $+\ -$; equal rank goes left to right

Modulus$|a|=a$ for $a\geqslant 0$, else $-a$·$|ab|=|a||b|$·$\left|\dfrac{a}{b}\right|=\dfrac{|a|}{|b|}$·$|a|^2=a^2$

Factorial$n!=1\cdot 2\cdots n$·$0!=1$·$n!=n\cdot(n-1)!$

$a^m a^n=a^{m+n}$·$\dfrac{a^m}{a^n}=a^{m-n}$·$(a^m)^n=a^{mn}$·$(ab)^n=a^n b^n$

$\left(\dfrac{a}{b}\right)^{\!n}=\dfrac{a^n}{b^n}$·$a^0=1\ (a\neq 0)$·$a^{-n}=\dfrac{1}{a^n}$·$a^{m/n}=\sqrt[n]{a^m}$

Roots$\sqrt[n]{ab}=\sqrt[n]{a}\,\sqrt[n]{b}$·$\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$·$\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$

$\sqrt{a^2}=|a|$, never $a$·$\left(\sqrt[n]{a}\right)^{n}=a$·$\sqrt[2k]{a}$ needs $a\geqslant 0$

Rationalise$\dfrac{1}{\sqrt{a}}=\dfrac{\sqrt{a}}{a}$·$\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{a-b}$

Compare$a>1:\ a^x$ grows·$0<a<1:\ a^x$ falls·$a^x=a^y\Leftrightarrow x=y$ for $a>0,a\neq1$

Sci.$x=m\cdot 10^k$, $1\leqslant m<10$

Def$\log_b x=y\iff b^{\,y}=x$, with $b>0$, $b\neq 1$, $x>0$

$\log_b(xy)=\log_b x+\log_b y$·$\log_b\dfrac{x}{y}=\log_b x-\log_b y$

$\log_b x^{\,n}=n\log_b x$·$\log_{b^k} x=\dfrac{1}{k}\log_b x$·$\log_b\sqrt[n]{x}=\dfrac{\log_b x}{n}$

Base$\log_b x=\dfrac{\log_c x}{\log_c b}$·$\log_b a=\dfrac{1}{\log_a b}$·$a^{\log_b c}=c^{\log_b a}$

Identity$b^{\log_b x}=x$·$\log_b b^{\,x}=x$·$\log_b b=1$·$\log_b 1=0$

Notation$\ln x=\log_e x$, $e=2.71828\ldots$·$\lg x=\log_{10}x$

Solving$\log_b f=\log_b g\iff f=g$ and $f>0$·check the domain first, always

$(a\pm b)^2=a^2\pm 2ab+b^2$·$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

$(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$·$a^2-b^2=(a-b)(a+b)$

$a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$·$a^n-b^n=(a-b)\!\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$

Binomial$(a+b)^n=\displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k}$, $\binom{n}{k}=\dfrac{n!}{k!\,(n-k)!}$

$\binom{n}{k}=\binom{n}{n-k}$·$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$·$\displaystyle\sum_k\binom{n}{k}=2^n$

Pascal1·1 1·1 2 1·1 3 3 1·1 4 6 4 1·1 5 10 10 5 1

Complete sq.$ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^{\!2}+c-\dfrac{b^2}{4a}$

Linear$ax+b=0\Rightarrow x=-\dfrac{b}{a}$·$a=0,b\neq0$: none·$a=b=0$: all $x$

Quadratic$ax^2+bx+c=0$, $D=b^2-4ac$, $x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}$

$D>0$: two real·$D=0$: one double $x=-\dfrac{b}{2a}$·$D<0$: two conjugate complex

Viète$x_1+x_2=-\dfrac{b}{a}$·$x_1x_2=\dfrac{c}{a}$·$ax^2+bx+c=a(x-x_1)(x-x_2)$

Cubic$t^3+pt+q=0$: $t=\sqrt[3]{-\dfrac{q}{2}+\sqrt{Q}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{Q}}$, $Q=\dfrac{q^2}{4}+\dfrac{p^3}{27}$ (Cardano)

Abel–Ruffinino general formula in radicals for degree $\geqslant 5$. See panels 10 and 14.

Typesrational: clear denominators, keep $\text{denom}\neq0$·radical: isolate, raise, verify roots

exponential: same base or take $\log$·$|f|=g\iff g\geqslant 0$ and $f=\pm g$

Rulesadd anything to both sides·multiply by $k>0$: sign stays·by $k<0$: sign flips

$a<b,\ c<d\Rightarrow a+c<b+d$·$0<a<b\Rightarrow \dfrac{1}{a}>\dfrac{1}{b}$·$0<a<b\Rightarrow a^n<b^n$

Intervalsfactor, mark roots on the line, read signs of each factor between them

Modulus$|f|<g\iff -g<f<g$·$|f|>g\iff f>g$ or $f<-g$

Triangle$|a+b|\leqslant|a|+|b|$·$\bigl||a|-|b|\bigr|\leqslant|a-b|$

Means$\dfrac{a+b}{2}\geqslant\sqrt{ab}\geqslant\dfrac{2ab}{a+b}$, $a,b>0$, equality iff $a=b$

Cauchy$\left(\textstyle\sum a_ib_i\right)^2\leqslant\sum a_i^2\sum b_i^2$·Bernoulli$(1+x)^n\geqslant 1+nx$, $x\geqslant-1$

$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, $a_n\neq 0$, degree $n$

Division$P=Q\cdot D+R$ with $\deg R<\deg D$·use long division or Horner

Remainder$P(x)=(x-c)Q(x)+P(c)$·Factor$P(c)=0\iff (x-c)\mid P$

Rational rootif $\dfrac{p}{q}$ is a root of an integer polynomial then $p\mid a_0$ and $q\mid a_n$

Viète, general$\displaystyle\sum x_i=-\frac{a_{n-1}}{a_n}$·$\displaystyle\prod x_i=(-1)^n\frac{a_0}{a_n}$·$e_k=(-1)^k\dfrac{a_{n-k}}{a_n}$

Multiplicity$c$ has multiplicity $m$ iff $(x-c)^m\mid P$ and $(x-c)^{m+1}\nmid P$

repeated roots are exactly the roots of $\gcd(P,P')$·$\deg(PQ)=\deg P+\deg Q$

Endsfor large $|x|$ the term $a_nx^n$ decides the sign and the shape

Def$f:X\to Y$ assigns one $y$ to each $x$·domain $D(f)$, range $E(f)$

Kindsinjective: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$·surjective: hits all of $Y$·bijective: both

Inverseexists iff bijective·$f^{-1}(f(x))=x$·graph mirrored in $y=x$

Compose$(f\circ g)(x)=f(g(x))$, order matters·$(f\circ g)^{-1}=g^{-1}\circ f^{-1}$

Parityeven $f(-x)=f(x)$·odd $f(-x)=-f(x)$·period $f(x+T)=f(x)$

Shifts$f(x)+a$ up·$f(x+a)$ left·$kf(x)$ stretch·$f(kx)$ squeeze·$-f(x)$, $f(-x)$ reflect

Parabolavertex $x=-\dfrac{b}{2a}$, form $y=a(x-h)^2+k$·Rationalzeros of denominator give vertical asymptotes

$i^2=-1$·$z=a+bi$·$\bar z=a-bi$·$|z|=\sqrt{a^2+b^2}$·$z\bar z=|z|^2$

$(a+bi)(c+di)=(ac-bd)+(ad+bc)i$·$\dfrac{1}{z}=\dfrac{\bar z}{|z|^2}$·$\overline{z+w}=\bar z+\bar w$

Polar$z=r(\cos\varphi+i\sin\varphi)=re^{i\varphi}$, $r=|z|$, $\varphi=\arg z$

multiply: moduli multiply, arguments add·divide: moduli divide, arguments subtract

De Moivre$z^n=r^n(\cos n\varphi+i\sin n\varphi)$

Roots$\sqrt[n]{z}=\sqrt[n]{r}\left(\cos\dfrac{\varphi+2\pi k}{n}+i\sin\dfrac{\varphi+2\pi k}{n}\right)$, $k=0,\ldots,n-1$

the $n$ values sit on a circle of radius $\sqrt[n]{r}$, evenly spaced·$\varepsilon_k=e^{2\pi ik/n}$, $\sum\varepsilon_k=0$

Euler$e^{i\varphi}=\cos\varphi+i\sin\varphi$·$\hi{e^{i\pi}+1=0}$

Arithmetic$a_n=a_1+(n-1)d$·$d=a_{n+1}-a_n$·$S_n=\dfrac{a_1+a_n}{2}\,n$

$S_n=\dfrac{2a_1+(n-1)d}{2}\,n$·$a_n=\dfrac{a_{n-1}+a_{n+1}}{2}$

Geometric$a_n=a_1q^{\,n-1}$·$q=\dfrac{a_{n+1}}{a_n}$·$S_n=a_1\dfrac{1-q^{\,n}}{1-q}$, $q\neq1$

$a_n^2=a_{n-1}a_{n+1}$·infinite: $S=\dfrac{a_1}{1-q}$ when $|q|<1$

Sums$\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$·$\displaystyle\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$

$\displaystyle\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^{\!2}$·$\displaystyle\sum_{k=0}^{n}q^k=\frac{q^{n+1}-1}{q-1}$

Telescope$\displaystyle\sum\left(b_k-b_{k+1}\right)=b_1-b_{n+1}$·$\dfrac{1}{k(k+1)}=\dfrac1k-\dfrac1{k+1}$

System$A\mathbf{x}=\mathbf{b}$·solve by substitution, elimination, or Gauss to row echelon form

Determinant$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$·$3\times3$ by cofactor expansion or Sarrus

Cramer$x_i=\dfrac{\Delta_i}{\Delta}$ when $\Delta=\det A\neq 0$·$\det(AB)=\det A\cdot\det B$

Productrow by column, $(AB)_{ij}=\sum_k a_{ik}b_{kj}$·$AB\neq BA$ in general·$AI=A$

Inverse$A^{-1}=\dfrac{1}{\det A}\operatorname{adj}A$, exists iff $\det A\neq 0$·$(AB)^{-1}=B^{-1}A^{-1}$

RankKronecker–Capelli: solvable iff $\operatorname{rank}A=\operatorname{rank}[A|\mathbf{b}]$; unique iff that equals $n$

Spacebasis, dimension, independence·rank–nullity $\operatorname{rank}A+\dim\ker A=n$

Basis$P$ holds the new basis in its columns·$\mathbf{x}_{\text{old}}=P\mathbf{x}_{\text{new}}$·$A'=P^{-1}AP$

Orthogonal$\langle\mathbf{u},\mathbf{v}\rangle=0$·projection $\dfrac{\langle\mathbf{u},\mathbf{v}\rangle}{\langle\mathbf{u},\mathbf{u}\rangle}\mathbf{u}$·Gram–Schmidt subtracts it away

Spectrum$\det(A-\lambda I)=0$ gives eigenvalues, $A\mathbf{v}=\lambda\mathbf{v}$ the eigenvectors·l’atelier →

Diagonal$A=PDP^{-1}$ when there are enough independent eigenvectors·then $A^n=PD^nP^{-1}$

Group$(G,*)$: closed, associative, has $e$, has $a^{-1}$·abelian if $a*b=b*a$

order $|G|$·cyclic $G=\langle g\rangle$·$S_n$ permutations, $|S_n|=n!$

Lagrangefor finite $G$ and $H\leqslant G$, $|H|$ divides $|G|$·so $a^{|G|}=e$

Morphism$\varphi(ab)=\varphi(a)\varphi(b)$·$\ker\varphi\trianglelefteq G$·$G/\ker\varphi\cong\operatorname{im}\varphi$

Ringtwo operations, $+$ abelian, $\cdot$ associative, distributive·ideal $I$: $rI\subseteq I$

Fielda commutative ring where every $a\neq0$ is invertible: $\mathbb{Q},\mathbb{R},\mathbb{C},\mathbb{F}_p$

Galoiseach equation carries a group of symmetries of its roots

solvable in radicals $\iff$ that group is solvable. $S_n$ is not solvable for $n\geqslant5$, which is exactly why the quintic has no formula.

Implication$A\Rightarrow B$·contrapositive $\lnot B\Rightarrow\lnot A$, always equivalent

Equivalence$A\iff B$ means both directions·a solution chain must preserve $\iff$, or be checked

Inductionprove $P(1)$, then $P(n)\Rightarrow P(n+1)$·Contradictionassume $\lnot A$, derive nonsense

Pièges$\sqrt{x^2}=|x|$, not $x$·dividing by an expression can lose roots, factor instead

squaring can create false roots, verify every one·$(a+b)^2\neq a^2+b^2$

multiplying an inequality by a negative flips it·$\log$ and even roots need their domain checked first

$AB\neq BA$·$a:0$ never·a double root still counts twice·$\sqrt{a}\sqrt{b}=\sqrt{ab}$ fails for two negatives