Chapter 2: Relations and Functions

Two ordered pairs are equal if and only if their corresponding components are equal. Therefore, comparing the components of the given equal ordered pairs,

$$x + 1 = 3$$ and $$y - 2 = 1$$.

Solving the first equation,

$$x = 3 - 1 = 2$$.

Solving the second equation,

$$y = 1 + 2 = 3$$.

Hence $$x = 2$$ and $$y = 3$$.

By definition, $$P \times Q$$ is the set of all ordered pairs $$(p, q)$$ with $$p \in P$$ and $$q \in Q$$. Forming all such pairs,

$$P \times Q = \{(a, r), (b, r), (c, r)\}$$.

Similarly, $$Q \times P$$ is the set of all ordered pairs $$(q, p)$$ with $$q \in Q$$ and $$p \in P$$,

$$Q \times P = \{(r, a), (r, b), (r, c)\}$$.

Although both sets have the same number of elements (three), the ordered pairs themselves are different. For instance, $$(a, r) \in P \times Q$$ but $$(a, r) \notin Q \times P$$ because the first coordinate must come from $$Q = \{r\}$$.

Hence $$P \times Q \ne Q \times P$$. In general, the Cartesian product is not commutative.

By definition, $$P \times P \times P$$ is the set of all ordered triples $$(a, b, c)$$ with $$a, b, c \in P$$.

Since $$P$$ has $$2$$ elements, the number of such ordered triples is $$2 \times 2 \times 2 = 8$$.

Listing all triples with each coordinate chosen from $$\{1, 2\}$$,

$$P \times P \times P = \{(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)\}$$.

The Cartesian product $$\mathbf{R} \times \mathbf{R}$$ is the set of all ordered pairs $$(x, y)$$ where $$x, y \in \mathbf{R}$$. Each such ordered pair corresponds to a unique point in the coordinate plane, so

$$\mathbf{R} \times \mathbf{R} = \{(x, y) : x, y \in \mathbf{R}\}$$ represents the set of all points of the coordinate (Cartesian) plane.

The Cartesian product $$\mathbf{R} \times \mathbf{R} \times \mathbf{R}$$ is the set of all ordered triples $$(x, y, z)$$ with $$x, y, z \in \mathbf{R}$$. Each ordered triple corresponds to a unique point in three-dimensional space, so

$$\mathbf{R} \times \mathbf{R} \times \mathbf{R} = \{(x, y, z) : x, y, z \in \mathbf{R}\}$$ represents the set of all points of three-dimensional space.

$$A$$ is the set of all first coordinates appearing in the ordered pairs of $$A \times B$$, and $$B$$ is the set of all second coordinates.

Collecting the first coordinates: $$p, p, m, m$$. Removing duplicates,

$$A = \{p, m\}$$.

Collecting the second coordinates: $$q, r, q, r$$. Removing duplicates,

$$B = \{q, r\}$$.

As a check, $$A \times B = \{p, m\} \times \{q, r\} = \{(p, q), (p, r), (m, q), (m, r)\}$$, which matches the given set.

Two ordered pairs are equal precisely when their corresponding components are equal. Equating components,

$$\dfrac{x}{3} + 1 = \dfrac{5}{3}$$ and $$y - \dfrac{2}{3} = \dfrac{1}{3}$$.

Solving the first equation,

$$\dfrac{x}{3} = \dfrac{5}{3} - 1 = \dfrac{5 - 3}{3} = \dfrac{2}{3}$$.

Multiplying both sides by $$3$$,

$$x = 2$$.

Solving the second equation,

$$y = \dfrac{1}{3} + \dfrac{2}{3} = \dfrac{3}{3} = 1$$.

Therefore $$x = 2$$ and $$y = 1$$.

If a set $$A$$ has $$p$$ elements and a set $$B$$ has $$q$$ elements, then $$A \times B$$ has $$p \times q$$ elements (each element of $$A$$ pairs with each element of $$B$$).

Here $$n(A) = 3$$ and $$B = \{3, 4, 5\}$$, so $$n(B) = 3$$.

$$n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$$.

$$G \times H$$ consists of all ordered pairs $$(g, h)$$ with $$g \in G$$ and $$h \in H$$. Pairing each element of $$G$$ with each element of $$H$$,

$$G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$$.

Similarly, $$H \times G$$ consists of all ordered pairs $$(h, g)$$ with $$h \in H$$ and $$g \in G$$,

$$H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$$.

Both products contain $$n(G) \times n(H) = 2 \times 3 = 6$$ ordered pairs.

$$A \times A \times A$$ is the set of all ordered triples $$(x, y, z)$$ with $$x, y, z \in A$$.

Since $$n(A) = 2$$, the number of such ordered triples is $$2 \times 2 \times 2 = 8$$. Each coordinate is chosen from $$\{-1, 1\}$$, giving

$$A \times A \times A = \{(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1),$$ $$(1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)\}$$.

$$A$$ is the set of all first coordinates that appear in the ordered pairs of $$A \times B$$, and $$B$$ is the set of all second coordinates.

First coordinates: $$a, a, b, b$$. Removing repetitions, $$A = \{a, b\}$$.

Second coordinates: $$x, y, x, y$$. Removing repetitions, $$B = \{x, y\}$$.

Check: $$A \times B = \{a, b\} \times \{x, y\} = \{(a, x), (a, y), (b, x), (b, y)\}$$, which matches.

Pairing each element of $$A$$ with each element of $$B$$,

$$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$$.

This set has $$n(A \times B) = 4$$ elements. A set with $$n$$ elements has exactly $$2^n$$ subsets, so

Number of subsets $$= 2^4 = 16$$.

Listing all $$16$$ subsets (the empty set, four singletons, six pairs, four triples, and the whole set):

$$\phi,$$
$$\{(1, 3)\}, \{(1, 4)\}, \{(2, 3)\}, \{(2, 4)\},$$
$$\{(1, 3), (1, 4)\}, \{(1, 3), (2, 3)\}, \{(1, 3), (2, 4)\}, \{(1, 4), (2, 3)\}, \{(1, 4), (2, 4)\}, \{(2, 3), (2, 4)\},$$
$$\{(1, 3), (1, 4), (2, 3)\}, \{(1, 3), (1, 4), (2, 4)\}, \{(1, 3), (2, 3), (2, 4)\}, \{(1, 4), (2, 3), (2, 4)\},$$
$$\{(1, 3), (1, 4), (2, 3), (2, 4)\}.$$

If $$(x, 1), (y, 2), (z, 1) \in A \times B$$, then the first coordinates $$x, y, z$$ belong to $$A$$ and the second coordinates $$1, 2$$ belong to $$B$$.

Since $$x, y, z$$ are distinct, $$A$$ contains at least the three elements $$x, y, z$$. But $$n(A) = 3$$, so $$A$$ contains exactly these three elements:

$$A = \{x, y, z\}$$.

The second coordinates seen are $$1$$ and $$2$$, so $$B$$ contains $$1$$ and $$2$$. Since $$n(B) = 2$$, $$B$$ contains exactly these two elements:

$$B = \{1, 2\}$$.

If $$n(A) = p$$, then $$n(A \times A) = p^2$$. Since $$A \times A$$ has $$9$$ elements,

$$p^2 = 9 \implies p = 3$$ (as $$p \ge 0$$).

So $$A$$ has $$3$$ elements. The pairs $$(-1, 0)$$ and $$(0, 1)$$ lie in $$A \times A$$, so the first coordinates $$-1, 0$$ and the second coordinates $$0, 1$$ are all elements of $$A$$. This gives the three distinct elements $$-1, 0, 1$$ in $$A$$.

Since $$n(A) = 3$$,

$$A = \{-1, 0, 1\}$$.

Now $$A \times A = \{-1, 0, 1\} \times \{-1, 0, 1\}$$ has $$9$$ ordered pairs:

$$A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}$$.

Excluding the two already listed, the remaining $$7$$ elements are

$$(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)$$.

Examine the arrows. Each $$x \in P$$ is paired with the two square roots of $$x$$ in $$Q$$: $$9$$ pairs with $$\pm 3$$, $$4$$ with $$\pm 2$$, and $$25$$ with $$\pm 5$$. In every case, $$x = y^2$$, i.e. $$y$$ is a square root of $$x$$.

(i) Set-builder form:

$$R = \{(x, y) : x \in P,\ y \in Q,\ y\text{ is a square root of }x\}$$, equivalently $$R = \{(x, y) : x \in P,\ y \in Q,\ x = y^2\}$$.

(ii) Roster form: Listing each ordered pair shown by an arrow,

$$R = \{(9, 3), (9, -3), (4, 2), (4, -2), (25, 5), (25, -5)\}$$.

Domain: set of first coordinates $$= \{9, 4, 25\}$$.

Range: set of second coordinates $$= \{3, -3, 2, -2, 5, -5\}$$.

A relation from $$A$$ to $$B$$ is, by definition, any subset of $$A \times B$$.

First find $$A \times B$$:

$$A \times B = \{1, 2\} \times \{3, 4\} = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$$.

So $$n(A \times B) = 2 \times 2 = 4$$.

A set with $$n$$ elements has $$2^n$$ subsets, so the number of relations from $$A$$ to $$B$$ is

$$2^{n(A \times B)} = 2^4 = 16$$.

The condition $$3x - y = 0$$ gives $$y = 3x$$. We require both $$x$$ and $$y = 3x$$ to lie in $$A = \{1, 2, \ldots, 14\}$$, i.e. $$1 \le 3x \le 14$$, so $$x \in \{1, 2, 3, 4\}$$ (since $$3 \times 5 = 15 > 14$$).

Listing the ordered pairs,

So $$R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$$.

Domain $$= \{1, 2, 3, 4\}$$ (set of first coordinates).

Codomain $$= A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$$.

Range $$= \{3, 6, 9, 12\}$$ (set of second coordinates).

The natural numbers less than $$4$$ are $$x = 1, 2, 3$$. Compute $$y = x + 5$$ for each:

$$x = 1 \Rightarrow y = 6$$; $$x = 2 \Rightarrow y = 7$$; $$x = 3 \Rightarrow y = 8$$.

Roster form: $$R = \{(1, 6), (2, 7), (3, 8)\}$$.

Domain $$= \{1, 2, 3\}$$ (first coordinates).

Range $$= \{6, 7, 8\}$$ (second coordinates).

The difference $$x - y$$ is odd precisely when one of $$x, y$$ is even and the other is odd. Examine every pair $$(x, y) \in A \times B$$:

Collecting the pairs whose difference is odd,

$$R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$$.

From the arrows, each $$x \in P$$ is connected to $$y = x - 2$$ in $$Q$$:

$$5 \to 3$$ (since $$5 - 2 = 3$$); $$6 \to 4$$ (since $$6 - 2 = 4$$); $$7 \to 5$$ (since $$7 - 2 = 5$$).

(i) Set-builder form:

$$R = \{(x, y) : y = x - 2,\ x \in P,\ y \in Q\}$$.

(ii) Roster form:

$$R = \{(5, 3), (6, 4), (7, 5)\}$$.

Domain $$= \{5, 6, 7\}$$; Range $$= \{3, 4, 5\}$$.

For each $$x \in \{0, 1, 2, 3, 4, 5\}$$ compute $$x + 5$$.

$$x = 0, 1, 2, 3, 4, 5 \Rightarrow x + 5 = 5, 6, 7, 8, 9, 10$$.

So $$R = \{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$$.

Domain $$= \{0, 1, 2, 3, 4, 5\}$$ (first coordinates).

Range $$= \{5, 6, 7, 8, 9, 10\}$$ (second coordinates).

The prime numbers less than $$10$$ are $$2, 3, 5, 7$$. Cube each:

$$2^3 = 8,\ 3^3 = 27,\ 5^3 = 125,\ 7^3 = 343$$.

Hence $$R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}$$.

A relation from $$A$$ to $$B$$ is any subset of $$A \times B$$. The number of subsets of a set with $$n$$ elements is $$2^n$$.

$$n(A \times B) = n(A) \times n(B) = 3 \times 2 = 6$$.

So the number of relations from $$A$$ to $$B$$ is

$$2^{6} = 64$$.

The difference of any two integers is itself an integer. So for every $$a, b \in \mathbf{Z}$$ we have $$a - b \in \mathbf{Z}$$, which means every ordered pair $$(a, b) \in \mathbf{Z} \times \mathbf{Z}$$ belongs to $$R$$.

Therefore $$R = \mathbf{Z} \times \mathbf{Z}$$.

Every integer appears as a first coordinate (take any $$a \in \mathbf{Z}$$ and $$b = 0$$; then $$(a, 0) \in R$$). Similarly every integer appears as a second coordinate.

Domain of $$R = \mathbf{Z}$$.

Range of $$R = \mathbf{Z}$$.

For every natural number $$x$$, the value $$y = 2x$$ is also a natural number, so every $$x \in \mathbf{N}$$ contributes the pair $$(x, 2x)$$ to $$R$$.

Domain $$= \mathbf{N}$$ (every $$x \in \mathbf{N}$$ appears as a first coordinate).

Codomain $$= \mathbf{N}$$ ($$R \subseteq \mathbf{N} \times \mathbf{N}$$).

Range $$=$$ set of values of $$2x$$ as $$x$$ ranges over $$\mathbf{N} = \{2, 4, 6, 8, \ldots\}$$, i.e. the set of even natural numbers.

Is $$R$$ a function? A relation is a function from its domain if each domain element has exactly one image. Here each $$x \in \mathbf{N}$$ gives exactly one $$y$$, namely $$y = 2x$$. Hence $$R$$ is a function from $$\mathbf{N}$$ to $$\mathbf{N}$$.

Substitute each value of $$x$$ into $$f(x) = 2x + 1$$.

$$f(1) = 2(1) + 1 = 3$$.
$$f(2) = 2(2) + 1 = 5$$.
$$f(3) = 2(3) + 1 = 7$$.
$$f(4) = 2(4) + 1 = 9$$.
$$f(5) = 2(5) + 1 = 11$$.
$$f(6) = 2(6) + 1 = 13$$.
$$f(7) = 2(7) + 1 = 15$$.

Completed table:

Compute $$y = x^2$$ for each tabulated $$x$$.

Domain: $$x^2$$ is defined for every real $$x$$, so the domain is $$\mathbf{R}$$.

Range: Since $$x^2 \ge 0$$ for all real $$x$$, and every non-negative real $$y$$ equals $$x^2$$ for $$x = \sqrt{y}$$, the range is $$[0, \infty) = \{y \in \mathbf{R} : y \ge 0\}$$.

Graph: Plot the points $$(-4, 16), (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), (4, 16)$$ and join them with a smooth curve. The resulting curve is a parabola opening upward with its vertex at the origin and symmetric about the $$y$$-axis.

Tabulate a few values of $$y = x^3$$.

Plot these points and join them with a smooth curve.

Description of the graph: The curve passes through the origin. For $$x > 0$$ it rises steeply (more steeply than the line $$y = x$$ for $$|x| > 1$$), and for $$x < 0$$ it descends symmetrically (the graph is symmetric about the origin because $$(-x)^3 = -x^3$$, i.e. $$f$$ is an odd function). The curve has a point of inflection at the origin (it is tangent to the $$x$$-axis there) and extends without bound in both directions.

Domain $$= \mathbf{R}$$; range $$= \mathbf{R}$$.

Compute $$y = \dfrac{1}{x}$$ for each tabulated $$x$$.

$$\dfrac{1}{-2} = -0.5$$; $$\dfrac{1}{-1.5} = -\dfrac{2}{3} \approx -0.667$$; $$\dfrac{1}{-1} = -1$$; $$\dfrac{1}{-0.5} = -2$$; $$\dfrac{1}{0.25} = 4$$; $$\dfrac{1}{0.5} = 2$$; $$\dfrac{1}{1} = 1$$; $$\dfrac{1}{1.5} = \dfrac{2}{3} \approx 0.667$$; $$\dfrac{1}{2} = 0.5$$.

Domain: $$\dfrac{1}{x}$$ is defined for every real $$x$$ except $$x = 0$$, so the domain is $$\mathbf{R} - \{0\}$$.

Range: If $$y = \dfrac{1}{x}$$ then $$x = \dfrac{1}{y}$$, which is real and non-zero iff $$y \ne 0$$. Every non-zero real $$y$$ is attained. Hence the range is $$\mathbf{R} - \{0\}$$.

By the algebra of real functions, $$(f + g)(x) = f(x) + g(x)$$, etc.

Sum:

$$(f + g)(x) = f(x) + g(x) = x^2 + (2x + 1) = x^2 + 2x + 1$$.

Difference:

$$(f - g)(x) = f(x) - g(x) = x^2 - (2x + 1) = x^2 - 2x - 1$$.

Product:

$$(fg)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2$$.

Quotient:

$$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)} = \dfrac{x^2}{2x + 1}$$, valid whenever $$2x + 1 \ne 0$$, i.e. $$x \ne -\dfrac{1}{2}$$.

Both $$f$$ and $$g$$ are defined on $$[0, \infty)$$.

Sum:

$$(f + g)(x) = \sqrt{x} + x$$, for $$x \ge 0$$.

Difference:

$$(f - g)(x) = \sqrt{x} - x$$, for $$x \ge 0$$.

Product:

$$(fg)(x) = \sqrt{x} \cdot x = x\sqrt{x} = x^{3/2}$$, for $$x \ge 0$$.

Quotient:

$$\left(\dfrac{f}{g}\right)(x) = \dfrac{\sqrt{x}}{x} = \dfrac{1}{\sqrt{x}}$$, valid for $$x > 0$$ (we must exclude $$x = 0$$).

Tabulate a few values of $$y = x + 10$$.

Graph: Plot the points $$(-4, 6), (-3, 7), \ldots, (4, 14)$$. They lie on a straight line. The graph of $$f(x) = x + 10$$ is a straight line with slope $$1$$ and $$y$$-intercept $$10$$ — it passes through $$(0, 10)$$ and $$(-10, 0)$$, rising at $$45^{\circ}$$ to the $$x$$-axis.

Domain $$= \mathbf{R}$$; range $$= \mathbf{R}$$.

Since $$f$$ is linear, write $$f(x) = ax + b$$ for some constants $$a, b$$.

Use two of the given ordered pairs to set up equations. From $$(1, 1)$$: $$f(1) = a + b = 1$$. From $$(0, -1)$$: $$f(0) = b = -1$$.

Substitute $$b = -1$$ into $$a + b = 1$$:

$$a - 1 = 1 \implies a = 2$$.

Hence $$f(x) = 2x - 1$$.

Verify with the remaining pairs: $$f(2) = 2(2) - 1 = 3$$ ✓; $$f(-1) = 2(-1) - 1 = -3$$ ✓.

The function is defined wherever the denominator is non-zero. Factorise the denominator:

$$x^2 - 5x + 4 = (x - 1)(x - 4)$$.

So the denominator vanishes when $$x = 1$$ or $$x = 4$$.

The numerator $$x^2 + 3x + 5$$ is a polynomial defined for every real $$x$$, so it imposes no additional restriction.

Hence the domain of $$f$$ is

$$\mathbf{R} - \{1, 4\}$$, i.e. all real numbers except $$1$$ and $$4$$.

The function is defined piecewise, so examine each piece separately and plot a few sample points.

For $$x \lt 0$$: $$f(x) = 1 - x$$. This is the part of the line $$y = 1 - x$$ (slope $$-1$$, $$y$$-intercept $$1$$) lying strictly to the left of the $$y$$-axis. Since $$x \lt 0$$ on this piece, $$-x \gt 0$$, so $$f(x) = 1 - x \gt 1$$ here. Sample points: $$f(-1) = 2$$ and $$f(-2) = 3$$. As $$x \to 0^-$$, $$f(x) \to 1$$.

For $$x = 0$$: $$f(0) = 1$$. This contributes the single point $$(0, 1)$$.

For $$x \gt 0$$: $$f(x) = x + 1$$. This is the part of the line $$y = x + 1$$ (slope $$1$$, $$y$$-intercept $$1$$) lying strictly to the right of the $$y$$-axis. Since $$x \gt 0$$ on this piece, $$f(x) = x + 1 \gt 1$$ here. Sample points: $$f(1) = 2$$ and $$f(2) = 3$$. As $$x \to 0^+$$, $$f(x) \to 1$$.

The left piece approaches $$(0, 1)$$ as $$x \to 0^-$$, the middle case gives exactly $$f(0) = 1$$, and the right piece approaches $$(0, 1)$$ as $$x \to 0^+$$. All three agree at $$(0, 1)$$, so the graph is a single continuous curve with no break or jump.

Graph: The graph is a V-shape opening upward with its vertex at $$(0, 1)$$. The left ray (drawn for $$x \lt 0$$) has slope $$-1$$ and rises to the left; the right ray (drawn for $$x \gt 0$$) has slope $$1$$ and rises to the right; the vertex point $$(0, 1)$$ is supplied by the value $$f(0) = 1$$. This is exactly the graph of $$y = |x| + 1$$.

For a relation to be a function, every input in its domain must have a single, well-defined image. The only point where ambiguity could arise in a piecewise definition is where the two pieces overlap.

For $$f$$: The pieces overlap only at $$x = 3$$. Check the two definitions there:

$$x^2 \big|_{x = 3} = 9$$ and $$3x \big|_{x = 3} = 9$$.

Both pieces give the same value $$9$$, so $$f(3) = 9$$ is unambiguous. Elsewhere, only one piece applies. Hence every $$x$$ in the domain $$[0, 10]$$ has a unique image, and $$f$$ is a function.

For $$g$$: The pieces overlap at $$x = 2$$. Check:

$$x^2 \big|_{x = 2} = 4$$ and $$3x \big|_{x = 2} = 6$$.

The two pieces give different values $$4$$ and $$6$$ at $$x = 2$$, so the element $$2$$ has two different images under $$g$$. This violates the definition of a function.

Hence $$g$$ is not a function.

Compute the numerator and denominator separately.

$$f(1.1) = (1.1)^2 = 1.21$$ and $$f(1) = 1^2 = 1$$.

So $$f(1.1) - f(1) = 1.21 - 1 = 0.21$$.

And $$1.1 - 1 = 0.1$$.

Therefore

$$\dfrac{f(1.1) - f(1)}{1.1 - 1} = \dfrac{0.21}{0.1} = 2.1$$.

The function is defined wherever the denominator is non-zero. Factorise:

$$x^2 - 8x + 12 = (x - 2)(x - 6)$$.

The denominator vanishes at $$x = 2$$ and $$x = 6$$.

The numerator $$x^2 + 2x + 1 = (x + 1)^2$$ is a polynomial defined for every real $$x$$, so no additional restriction comes from it.

Hence the domain is $$\mathbf{R} - \{2, 6\}$$.

Domain: For $$f(x)$$ to be real, the radicand must be non-negative:

$$x - 1 \ge 0 \iff x \ge 1$$.

So the domain is $$[1, \infty)$$.

Range: By convention, $$\sqrt{\,\cdot\,} \ge 0$$. As $$x$$ ranges over $$[1, \infty)$$, $$x - 1$$ ranges over $$[0, \infty)$$, and so $$\sqrt{x - 1}$$ ranges over $$[0, \infty)$$. Every non-negative real $$y$$ is attained at $$x = y^2 + 1$$.

Hence the range is $$[0, \infty)$$.

Domain: The absolute value is defined for every real number, so the domain is $$\mathbf{R}$$.

Range: By definition, $$|x - 1| \ge 0$$ for every real $$x$$. The value $$0$$ is attained at $$x = 1$$; every $$y > 0$$ is attained at $$x = 1 + y$$ (or $$x = 1 - y$$). So every non-negative real is achieved.

Hence the range is $$[0, \infty)$$.

Let $$y = \dfrac{x^2}{1 + x^2}$$. We find every value of $$y$$ for which this equation has a real solution $$x$$.

Clear the denominator and collect the $$x^2$$ terms:

$$y(1 + x^2) = x^2 \implies y + y x^2 = x^2 \implies y = x^2 - y x^2 = x^2(1 - y)$$.

If $$y \ne 1$$, then

$$x^2 = \dfrac{y}{1 - y}$$.

A real number $$x$$ exists exactly when $$x^2 \ge 0$$, that is when

$$\dfrac{y}{1 - y} \ge 0$$.

This fraction is non-negative precisely when the numerator $$y$$ and the denominator $$1 - y$$ have the same sign, or when the numerator is $$0$$. Checking each case:

• If $$y \lt 0$$: the numerator satisfies $$y \lt 0$$ while the denominator satisfies $$1 - y \gt 0$$, so the ratio is negative. Not allowed.
• If $$y = 0$$: the ratio equals $$0$$, so $$x^2 = 0$$ and $$x = 0$$. Allowed.
• If $$0 \lt y \lt 1$$: the numerator satisfies $$y \gt 0$$ and the denominator satisfies $$1 - y \gt 0$$, so the ratio is positive. Allowed.
• If $$y = 1$$: substituting $$y = 1$$ in the original equation gives $$1 + x^2 = x^2$$, i.e. $$1 = 0$$, which is impossible. Not allowed.
• If $$y \gt 1$$: the numerator satisfies $$y \gt 0$$ while the denominator satisfies $$1 - y \lt 0$$, so the ratio is negative. Not allowed.

So a real $$x$$ exists exactly for the values $$0 \le y \lt 1$$.

Hence the range of $$f$$ is $$[0, 1)$$.

Apply the definitions of sum, difference, and quotient of real functions.

Sum:

$$(f + g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2$$.

Difference:

$$(f - g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4$$.

Quotient:

$$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)} = \dfrac{x + 1}{2x - 3}$$, defined wherever $$g(x) \ne 0$$, i.e. $$2x - 3 \ne 0$$, i.e. $$x \ne \dfrac{3}{2}$$.

Use the ordered pairs to set up equations in $$a$$ and $$b$$.

From $$(0, -1)$$: $$f(0) = a(0) + b = b = -1$$. So $$b = -1$$.

From $$(1, 1)$$: $$f(1) = a + b = 1$$. With $$b = -1$$,

$$a - 1 = 1 \implies a = 2$$.

So $$a = 2$$ and $$b = -1$$, giving $$f(x) = 2x - 1$$.

Verify using the remaining pairs:

$$f(2) = 2(2) - 1 = 3$$ ✓; $$f(-1) = 2(-1) - 1 = -3$$ ✓.

A subset of $$\mathbf{Z} \times \mathbf{Z}$$ is a function from $$\mathbf{Z}$$ to $$\mathbf{Z}$$ only if each first coordinate is paired with a unique second coordinate. We test whether the same product $$ab$$ can give different sums $$a + b$$.

Take the first coordinate $$6$$:

• $$a = 2,\ b = 3$$: $$ab = 6$$ and $$a + b = 5$$. So $$(6, 5) \in f$$.
• $$a = 1,\ b = 6$$: $$ab = 6$$ and $$a + b = 7$$. So $$(6, 7) \in f$$.

The same first coordinate $$6$$ is paired with two different second coordinates $$5$$ and $$7$$. This violates the definition of a function.

Hence $$f$$ is not a function from $$\mathbf{Z}$$ to $$\mathbf{Z}$$.

Find the highest prime factor of each $$n \in A$$.

• $$9 = 3 \times 3$$. Highest prime factor $$= 3$$.
• $$10 = 2 \times 5$$. Highest prime factor $$= 5$$.
• $$11$$ is prime. Highest prime factor $$= 11$$.
• $$12 = 2^2 \times 3$$. Highest prime factor $$= 3$$.
• $$13$$ is prime. Highest prime factor $$= 13$$.

So $$f(9) = 3,\ f(10) = 5,\ f(11) = 11,\ f(12) = 3,\ f(13) = 13$$.

The range is the set of image values (without repetition):

Range of $$f = \{3, 5, 11, 13\}$$.