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You stop breaking down a factor once it is prime because prime numbers only have two factors, 1 and the number itself, meaning they cannot be broken down further into smaller positive integers.

Conditions: The current branch value is prime.; The goal is to express the original number as a product of primes only.

In the two-child problem, the conditioning event 'at least one girl' reduces the original sample space of four equally likely ordered outcomes {GG, GB, BG, BB} to three outcomes {GG, GB, BG} by eliminating BB. This restriction changes the effective sample size from 4 to 3, directly altering the probability calculation's denominator.

Conditions: Original sample space: {GG, GB, BG, BB}.; Conditioning event: At least one girl.

The example defines a subspace $W$ spanned by the first two standard basis vectors of $\mathbb{R}^4$, and its orthogonal complement $W^\perp$ spanned by the last two standard basis vectors. Together, they form a direct sum that equals the entire space $\mathbb{R}^4$.

Conditions: The ambient space is $\mathbb{R}^4$.; The bases are taken as given on the board.

In the 'surface points in time' visualization, the vertical axis represents the distribution in time, not the actual three-dimensional physical location or depth. Each point retains its x, y horizontal location, while the plotted vertical extent encodes elapsed time, as illustrated by lengths corresponding to 250 days and around 500 days.

Conditions: The visualization is the left panel labeled “surface points in time”.; The data represents groundwater measurements over time.

The detailed balance condition requires that for any two states $x$ and $y$, the probability flow from $x$ to $y$ exactly equals the flow from $y$ to $x$. This bidirectional equilibrium guarantees that the target distribution is stationary for the Markov chain.

Conditions: Must hold for all pairs of states $x$ and $y$ in the state space.; $P(x)$ and $P(y)$ are probabilities under the target distribution.; $T(y|x)$ and $T(x|y)$ are transition probabilities.

The key difference between a jump discontinuity and a removable discontinuity lies in the existence of the two-sided limit. In a jump discontinuity, the left and right limits exist but are unequal, so the two-sided limit does not exist. In a removable discontinuity, the left and right limits are equal, so the two-sided limit exists, but it does not match the function value (or the function is undefined).

Conditions: Comparing two types of discontinuities at an interior point.; One-sided limits are finite in both cases.

A natural transformation $\eta$ between two functors $F$ and $G$ (both mapping from category $C$ to category $D$) is a rule that assigns to every object $x$ in $C$ a morphism $\eta_x$ in $D$, specifically from $F(x)$ to $G(x)$. This morphism $\eta_x$ is called the component of the natural transformation at object $x$.

Conditions: $F$ and $G$ are functors from category $C$ to category $D$.; For each object $x$ in $C$, $\eta_x$ is a morphism in $D$ from $F(x)$ to $G(x)$.; For every morphism $h: x \to y$ in $C$, the commutative square condition $\eta_y \circ F(h) = G(h) \circ \eta_x$ must hold.

The indefinite integral of $x^2$ yields a family of functions $F(x) = \frac{x^3}{3} + C$. By defining the accumulation function $A(x)$ such that $A(0) = \int_0^0 t^2 dt = 0$, we impose an initial condition.

Conditions: Integrand is $f(t) = t^2$; Lower limit of integration is 0; Accumulation starts from zero

To express a vector $y$ as the sum of two perpendicular components, compute its projection $p = \frac{y \cdot v}{v \cdot v} v$ onto the line spanned by a nonzero vector $v$. The residual $r = y - p$ is then calculated, yielding the decomposition $y = p + r$, where $p$ is parallel to $v$ and $r$ is perpendicular to $v$.

Conditions: $v$ is a nonzero vector defining the line.; $y$ is the vector to be decomposed.

You can check a prime factorization by multiplying the prime factors back together. If the product equals the original number, the factorization is correct.

Conditions: The factors being multiplied must be the claimed prime factors.; The product must equal the original number exactly.

The derivative $f'(x)$ is formally defined as the finite limit of the difference quotient $\frac{f(x+h)-f(x)}{h}$ as $h$ approaches zero. This definition requires evaluating the limit rather than substituting $h=0$ directly, and it only applies when the resulting limit exists and is finite.

Conditions: The finite limit must exist.; $h$ approaches zero.

The unit $m^4$ arises from the defining integral $I = \int r^2 dA$, where the differential area $dA$ contributes $m^2$ and the squared distance $r^2$ contributes another $m^2$. Multiplying these terms results in $m^2 \cdot m^2 = m^4$, assuming lengths are measured in meters.

Conditions: Lengths are measured in meters in the specific example.; The integral represents an area moment of inertia.