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The domain of a function $f:X\to Y$ is the specified set $X$ of allowed inputs. A function must assign a value to every member of that set. Its codomain $Y$ separately specifies where the outputs are allowed to lie.

The formula alone need not determine the domain. The rule $x\mapsto x^2$ on the real numbers and its restriction to the nonnegative real numbers have different allowed inputs. The first is not injective, while the second is. Similarly, the real-valued rule $x\mapsto 1/x$ excludes zero from its domain of a function.

Changing a formula by cancellation can alter where it is defined. The principle preserve the domain when simplifying a function records why an extension across a cancelled denominator is a separate operation.

A function $f:X\to Y$ is injective if $f(x)=f(x^{\prime })$ implies $x=x^{\prime }$. Thus each element of its codomain has at most one preimage. The function $n\mapsto 2n$ from $\mathbb{Z}$ to $\mathbb{Z}$ is injective, whereas $x\mapsto x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is not, since $1$ and $-1$ have the same image.

The choice of domain of a function matters: restricting $x\mapsto x^2$ to the nonnegative real numbers makes it injective. A function is bijective exactly when it is both injective and surjective; injectivity alone does not require reaching every element of the codomain.

A metric space $(X,d)$ is interpolating if, for every $x,y\in X$ and every pair of nonnegative real numbers $r,s$ with $d(x,y)=r+s$, there is $z\in X$ with $d(x,z)=r$ and $d(z,y)=s$. Thus each distance can be split at any prescribed fraction. The nonnegativity requirement is essential.

A convex set in a normed vector space is interpolating: when $r+s>0$, take $z=x+\frac{r}{r+s}(y-x)$; when $r+s=0$, take $z=x=y$. A two-point metric space with the discrete metric is not interpolating, because a distance of one has no midpoint at distance $1/2$ from both endpoints. Interpolation makes the metric exponential candidate satisfy the triangle inequality, so every bounded interpolating metric space is an exponentiable object of the category of bounded metric spaces and non-expansive maps.

A positive real number is a real number $x$ with $x>0$. Zero is excluded; allowing zero gives a nonnegative real number. Both $1/2$ and $\sqrt{2}$ are positive.

The sum, product, and reciprocal of positive real numbers are positive. Consequently the positive real numbers form a group under multiplication, and the natural logarithm converts that multiplication into addition: $\log (xy)=\log x+\log y$. Positivity is also the domain restriction that makes the real logarithm well-defined.

For a nonnegative real number $x$, the squarefree counting function counts positive square-free integers:For example, $Q(10)=7$, counting $1,2,3,5,6,7,10$. The Möbius function gives the exact formulaTo see why, consider ${\sum }_{d^2\mid n}\mu (d)$. Only square-free integers $d$ contribute. Choosing their prime factors from the primes whose squares divide $n$ gives the product ${\prod }_{p^2\mid n}(1-1)$. This is one when there are no such prime factors and zero otherwise, so it is the indicator function of squarefreeness. Summing over the positive integers $n\le x$ and exchanging two finite sums leaves $\lfloor x/d^2\rfloor$ multiples of each $d^2$. The floor function makes the formula exact for nonintegral $x$ too.

For a nonnegative integer $n$, its square root $\sqrt{n}$ is the unique nonnegative real number whose square is $n$. It is an integer exactly when $n$ is a perfect square; otherwise it is an irrational number. Indeed, if $\sqrt{n}=a/b$ with coprime positive integers $a,b$, then $a^2=n{b}^2$ forces every prime divisor of $b$ to divide $a$, so $b=1$.

For example, $\sqrt{49}=7$, whereas $\sqrt{12}=2\sqrt{3}$ is irrational. The integer square root is a different operation: it rounds $\sqrt{n}$ down, so the integer square root of $12$ is $3$.

A real- or complex-valued function is unbounded when no finite constant bounds the moduli of all its values. Equivalently, for every $M>0$ there is an allowed input $x$ with $|f(x)|>M$.

Unboundedness does not say that all sufficiently large inputs produce large values. For example, $f(x)=x\sin x$ on the nonnegative real numbers is zero at $x=n\pi$, yet $f(\pi /2+2n\pi )=\pi /2+2n\pi$ grows without bound. Thus this function is unbounded but has no limit $+\infty$ as $x\to \infty$.