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If $A_n$ are countable, choose enumerations and map each element of ${\bigcup }_n{A}_n$ to the first pair $(n,k)$ at which it occurs. Since ${\mathbb{N}}^2$ is countable, the union is countable.

A periodic function of period $k$ is determined by its $k$ values on a complete residue system. Thus all periodic functions form a countable union over $k$ of the countable sets ${\mathbb{N}}^k$, and are countable.

The set of bijections $\mathbb{N}\to \mathbb{N}$ is uncountable. Indeed, each binary sequence determines a bijection that swaps $2j-1$ and $2j$ exactly when its $j$th bit is one. This is an injection from the uncountable set of binary sequences.

For prime $p$, the nonzero residues pair with their distinct inverses except for $1$ and $-1$, so Wilson theorem givesIf composite $n>4$ has a factorization $n=ab$ with $1<a<b<n$, both factors occur in $(n-1)!$. If $n=m^2$, the distinct factors $m$ and $2m$ occur and their product is divisible by $n$; the excluded case $m=2$ is exactly $n=4$. Thus $(n-1)!\equiv 0(\mathrm{mod}n)$.

The Fermat-Euler theorem states $a^{\varphi (n)}\equiv 1(\mathrm{mod}n)$ when $(a,n)=1$. For prime $p$, $\varphi (p)=p-1$, giving Fermat's little theorem for $p\nmid a$; the form $a^p\equiv a(\mathrm{mod}p)$ also covers $p\mid a$.

If $a\equiv b(\mathrm{mod}p)$, induction and the binomial theorem showIndeed, write $a^{p^n}=b^{p^n}+c{p}^{n+1}$ and raise to the $p$th power; every nonleading binomial term gains enough powers of $p$.

Fix $a>1$ and choose any odd prime $p\nmid a^2-1$. PutFermat's theorem gives $N_p\equiv 1(\mathrm{mod}p)$, and $N_p$ is odd, so $2p\mid N_p-1$. Also $a^{2p}\equiv 1(\mathrm{mod}{N}_p)$, hence $a^{N_p-1}\equiv 1(\mathrm{mod}{N}_p)$. For odd $p$,is composite. This generalized repunit pseudoprime construction gives infinitely many distinct base-$a$ pseudoprimes because the values are unbounded.

With $\gamma (v)=(1-v^2/c^2{)}^{-1/2}$,Adding and subtracting $c{t}^{\prime }$ givesThe product of the two multipliers is one, so $x_{+}^{\prime }{x}_{-}^{\prime }=x_{+}{x}_{-}$, which is $x^{\prime 2}-c^2{t}^{\prime 2}=x^2-c^2{t}^2$.

Successive transformations multiply the $x_{+}$ factors. Equatingand solving gives the relativistic velocity-addition law

During a short time $dt$, the rocket loses mass $-dm>0$. Conservation of upward momentum, including gravity's impulse and exhaust velocity $v-U$, gives to first orderwhich is the stated rocket equation.

Here $dm/dt=-\alpha$ and $U=U_0{m}_0/m$, soLift-off from rest requires positive initial acceleration,With $v(0)=0$, integration givesThe dimensions are $[m_0]=[m]=M$, $[\alpha ]=M/T$, $[U_0]=[U]=[v]=L/T$, $[g]=L/T^2$, and $[t]=T$. Both displayed terms in $v$ therefore have dimension $L/T$.

The Chinese remainder theorem says that for pairwise coprime positive integers $m_i$, the mapis a bijection. For two moduli, choose $u,v$ with $um+vn=1$; thenhas residues $a$ modulo $m$ and $b$ modulo $n$. Uniqueness follows because the difference of two solutions is divisible by both coprime moduli, hence by their product. Induction proves the general case.

The two given congruences are compatible modulo $\gcd (6,8)=2$, and checking modulo $\mathrm{lcm}(6,8)=24$ gives

Write $d=2^r{3}^{s}m$ with $(m,6)=1$. Use the Chinese remainder theorem to chooseThen $4a^2+9b^2\equiv 1$ modulo each of the pairwise coprime factors $2^r$, $3^s$, and $m$, hence modulo $d$.

Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore