An external force is a force exerted on a member inside the chosen system boundary by a body outside it. Changing that boundary can change which forces are external. Their sum changes the system's total momentum when its internal forces cancel.

An internal force is a force between members inside the chosen system boundary. Changing that boundary can change which forces are internal. Pairwise Newtonian mechanics interactions cancel through third-law interaction pairs in the system's total momentum balance.

A free-body diagram isolates a body and shows the forces exerted on it. A third-law interaction pair acts on different bodies and belongs in separate body diagrams.

A zygote is the biological cell formed by fusion of gametes during fertilization. In a typical haploid-diploid life cycle, two haploid contributions form one diploid cell nucleus, restoring the two sets separated by meiosis.

The zygote begins a new developmental sequence. In land plants, it develops into the sporophyte; the preceding gametophyte generation instead produces the fusing gametes.

In the common angiosperm arrangement, one sperm fuses with the ovum to form a diploid zygote; another fuses with the central biological cell's two polar cell nuclei, initiating endosperm with three chromosome sets. The zygote develops into a plant embryo, and the ovule develops into a seed.^[ref]

For an original counting example, stipulate a plant with $2n=14$ chromosomes and the common two-polar-nucleus arrangement. Each haploid sperm and ovum contributes $n=7$. The zygote therefore has $7+7=14$ chromosomes, while the initiating endosperm cell nucleus has $7+7+7=21$. The maternal seed coat remains derived from $2n=14$ plant tissue; the whole seed has no single uniform chromosome count. This is bookkeeping for a stipulated example, not identification of a particular species. It follows by counting the chromosome sets entering each fusion, rather than assigning one count to everything enclosed by a seed coat.

For the Fermat primality test, a passing base satisfies $a^{n-1}\equiv 1(\mathrm{mod}n)$. Passing proves only this modular congruence, not that $n$ is a prime number. An odd composite number that passes a coprime base is a Fermat pseudoprime to that base. A Carmichael number passes every coprime base, so even testing all the units would not certify that $n$ is a prime number. ^[ref]

This small table uses only the bases $2,3,5,11$, each reduced modulo $n$ first. A numeric entry is $a^{n-1}\mathrm{mod}n$: one passes, while any other numeric entry gives a Fermat witness. A "factor" entry records a proper greatest common divisor and stops before modular exponentiation. The prime/composite classification is independently established by trial division or the displayed prime factorization, not by the passing entries. These four bases are illustrative, with no asserted universal coverage or error probability.

For $341=11\cdot 31$, $\gcd (2,341)=1$ and $2^{10}=1024=3\cdot 341+1$, so $2^{340}=(2^{10}{)}^{34}\equiv 1(\mathrm{mod}341)$. Yet $\gcd (3,341)=1$ and $3^{340}\equiv 56(\mathrm{mod}341)$, so three is a Fermat witness. One coprime base passes and another fails: being a Fermat pseudoprime depends on the base. Choosing eleven instead reveals the proper factor through $\gcd (11,341)=11$.

For $561=3\cdot 11\cdot 17$, every coprime base passes: each of $2,10,16$ divides $560$, so Fermat's little theorem gives $a^{560}\equiv 1$ modulo each prime factor; the Chinese remainder theorem combines the three modular congruences. Equivalently, the Carmichael function gives $\lambda (561)=80\mid 560$. Bases two and five have greatest common divisor one and pass, but $\gcd (3,561)=3$, $\gcd (11,561)=11$ and $\gcd (17,561)=17$ expose factors. The qualification "coprime" is therefore crucial even for a Carmichael number.

The Fermat primality test is useful because a Fermat witness can prove compositeness using repeated squaring without finding a factor. Its failure to certify a passing candidate motivates stronger primality testing methods such as the Miller-Rabin primality test.

A Fermat witness for an integer $n>1$ is a nonzero residue class represented by $1\le a<n$ for which $a^{n-1}\not\equiv 1(\mathrm{mod}n)$. It proves that $n$ is a composite number by Fermat's little theorem. Here the definition includes nonunits; some conventions restrict Fermat witnesses to bases coprime to $n$. The Fermat primality test can first compute the greatest common divisor: if $1<\gcd (a,n)<n$, it already has a proper factor. A zero residue class is excluded. ^[ref]

For a prime number $p$ and $p\nmid a$, multiplication by $a$ gives a permutation of the $p-1$ nonzero residue classes. For $a=3$ modulo seven, the representatives $1,2,3,4,5,6$ map to $3,6,2,5,1,4$: the same collection in a different order. Repeating this permutation from one visits all six representatives before returning.

For $a=2$ modulo seven, the permutation cycles are instead $(1,2,4)$ and $(3,6,5)$. The powers of two visit only the permutation cycle containing one, so their multiplicative order is three. Both permutation cycles return after three steps; applying the permutation six times still fixes every nonzero residue class. Fermat's little theorem guarantees a return at six, not that every base visits every nonzero residue class.

More generally, every permutation cycle of this map has length equal to the multiplicative order of $a$: starting at a nonzero residue class $x$, a return means $a^{k}x\equiv x(\mathrm{mod}p)$, and its modular multiplicative inverse allows cancellation of $x$. Thus the $p-1$ nonzero residue classes split into equal-length permutation cycles, which also explains the divisibility of the multiplicative order into $p-1$.

The product argument gives a short proof outline. Since multiplication by $a$ is a permutation of the nonzero residue classes,Every factor is a unit, so their product has a modular multiplicative inverse and can be cancelled. This leaves $a^{p-1}\equiv 1(\mathrm{mod}p)$. The whole argument works for $p=2$, where the product is just one.^[ref]

For bits $x_1,\dots ,x_n$, repeated exclusive or computes the parity of the number of ones:An induction proves this: the first bit gives its own parity, and appending a zero preserves it while appending a one changes it. The binary string $10110$ has three ones, so its parity bit is one. Append that parity bit to obtain $101101$, whose six bits have even parity. Flipping the second bit produces $111101$ and changes the combined exclusive or to one, detecting an error. Flipping the third bit as well produces $110101$, with four ones and combined exclusive or zero: an even number of flips can escape this check.

For three inputs, repeated exclusive or means an odd number of true inputs, rather than exactly one: $1\oplus 1\oplus 1=1$. The “exactly one” description applies to the two-input exclusive or. This distinction matters when translating a requirement that exactly one option be selected into a program condition.

Two equally spaced families of perpendicular lines partitioning the Euclidean plane into congruent squares. A general affine map can turn this grid into parallelograms, or collapse directions if its linear part of an affine map is not injective.

A zeitgeber is an external timing cue capable of entraining a circadian rhythm.

A racemate contains the same amount of substance of each member of an enantiomer pair. Their equal and opposite contributions cancel in ordinary solution polarimetry under identical conditions. Zero net optical rotation therefore does not establish achirality of the individual molecules.^[ref]

An electron microscope forms a spatially enlarged representation of a specimen using electrons, electron lenses, and a detector. Its appeal is access to structure too fine for an ordinary optical microscope, but a short de Broglie wavelength alone does not guarantee fine electron microscope resolution. Increasing magnification can enlarge an existing blur without revealing any additional structure.

In transmission electron microscopy, an electron gun illuminates a thin specimen; magnetic electron lenses form and enlarge its transmitted image on a fluorescent screen or detector. A vacuum reduces collisions with gas. Lens electric currents control focusing and magnification.^[ref]

In his 1888 reflection experiments, Heinrich Hertz excited a straight conductor with a central spark gap driven by an induction coil. The receiver was a wire circle of $35\mathrm{cm}$ radius with an adjustable small spark gap; tiny sparks were observed in darkness. There was no connecting transmission wire. ^[ref]

He removed hanging metal fixtures, installed a $4\mathrm{m}$ by $2\mathrm{m}$ zinc reflecting sheet, and moved and rotated the receiver. Spark strength alternated with position and gap orientation, supporting specular reflection and wave interference. He checked observations from farther away because the observer's body perturbed the response. Changing oscillator dimensions changed the spatial pattern. ^[ref]

These tests supported propagating electromagnetic waves, rather than mere detection of nearby induction. They did not establish unlimited reception distance. His wire-versus-air wave interference experiment addressed finite velocity; he later judged its quantitative velocity comparison doubtful because of room effects. ^[ref]

The physical inference combines transmission with a spatially structured response to a reflected disturbance. In an ideal monochromatic incident-and-reflected model, standing wave nodes are separated by $\lambda /2$; combining a reliable wavelength with independently known frequency gives phase velocity $v=\nu \lambda$. A spark is a threshold event, however: its presence or length is not a calibrated wave intensity or a direct arrival-time measurement. The actual loop also responds to its orientation and finite size, so the simple standing-wave model is an explanation of the test's logic, rather than a reconstruction of Hertz's measured points. Poynting theorem explains how source energy can reach and drive the receiver's discharge.

An outgoing radiation pulse carries a finite amount of electromagnetic energy through successively larger enclosing spheres. In vacuum, far from a localized source and without absorption or additional reflections, let $P_{\mathrm{emit}}(t)$ be the emitted power. The retarded time at radius $r$ is $s=t-r/c$. Define ${\overline{S}}_r$ as the angular average of the radial Poynting vector over the sphere, not a cycle average:Here $\overline{u}$ is the angularly averaged energy density in the leading radiation zone approximation. Multiplying by shell area removes geometric dilution: the energy per unit radial thickness translates outward without losing its shape. These angular averages do not assume that an actual electric dipole emits equally in all directions. Nor do they specify an exact spherically symmetric transverse solution of the Maxwell equations.

For an explicitly stipulated model, emitAfter emission ends, $t>\tau$, the pulse lies between $r=c(t-\tau )$ and $r=ct$, provided this shell is in the radiation zone. Its radial thickness remains $c\tau$. Substituting $s=t-r/c$ in the shell integral $\int 4\pi r^2\overline{u}dr$ gives $U=P_0\tau /2$ at every such time. The same substitution in the time integral of $4\pi r^2{\overline{S}}_r$ at any enclosing sphere gives that same total crossing energy. This is conservation of energy in two different views.

Along any fixed direction with nonzero outgoing power, compare the same part of the pulse at two radii, hence at different arrival times. The leading electric field amplitude falls as $1/r$, while its squared wave intensity falls as $1/r^2$. At $2r$ the field is one half and the energy flux one quarter; the signal arrives $r/c$ later. A small fixed area receiver therefore intercepts less energy at greater distance even though every enclosing sphere receives the same total pulse energy. The inverse-square law expresses spreading over area, rather than destruction of energy. Close to the source, the neglected near and far field terms prevent extrapolating these formulas down to $r=0$.