Up until now in this text, we have been exclusively special kinds of mappings called functions. In this section, we broaden our horizons to study more general mappings called relations. The reader is encouraged to revisit Definition functiondefn in Section FunctionsandtheirRepresentations before proceeding with the definition of relation below.

Unlike Definition functiondefn, Definition relationdefn puts no conditions on the process which maps elements of \(A\) to elements of \(B\). This means that while all functions are relations, not all relations need be functions. For example, consider the mappings \(f\) and \(g\) below from Section FunctionsandtheirRepresentations.

Figure 1: Mapping diagram for \(f\)

Figure 2: Mapping diagram for \(g\)

Both \(f\) and \(g\) are relations. More specifically, \(f\) is a function from \(N\) to \(T\) while \(g\) is merely relation from \(T\) to \(N\). As with functions, we may describe general relations in a variety of different ways: verbally, as mapping diagrams, or a set of ordered pairs. For example, just as we may describe the function \(f\) above as

\[ f = \{ (\text {White Paw}, \text {cat}), (\text {Cooper}, \text {cat}), (\text {Bingo}, \text {lizard}), (\text {Kennie}, \text {turtle}) \}, \]

we may represent \(g\) as

\[ g = \{ (\text {cat}, \text {White Paw}), (\text {cat}, \text {Cooper}), ( \text {lizard}, \text {Bingo}), (\text {turtle}, \text {Kennie}) \}. \]

Note here the grammar ‘\(g\) is a relation from \(T\) to \(N\)’ is evidenced by the elements of \(T\) being listed first in the ordered pairs (i.e., the abscissae) and the elements of \(N\) being listed second (i.e., the ordinates.)

Unlike functions, we do not use function notation when describing the input/output relationship for general relations. For example, we may write ‘\(f(\text {White Paw}) = \text {cat}\)’ since \(f\) maps the input ‘White Paw’ to only one output, ‘cat.’ However, \(g(\text {cat})\) is ambiguous since it could mean ‘White Paw’ or ‘Cooper.’

As with functions, our focus in this course will rest with relations of real numbers. Consider the relation \(R\) described as follows: \(R = \{ (-1,3), (0,-3), (4,-2), (4,1) \}\). Below is a mapping diagram of \(R\).

Figure 3: Mapping diagram for \(R\)

However, since \(R\) relates real numbers, we can also create the graph of \(R\) in the same way we graphed functions - by interpreting the ordered pairs which comprise \(R\) as points in the plane. Since we have no context, we use the default labels ‘\(x\)’ for the horizontal axis and ‘\(y\)’ for the vertical axis.

Figure 4: The graph of \(R\)

Our next example focuses on using relations to describe sets of points in the plane and vice-versa.

As with functions, we can describe relations algebraically using equations. For example, the equation \(v^2+w^3 = 1\) relates two variables \(v\) and \(w\) each of which represent real numbers. More formally, we can express this sentiment by defining the relation \(R = \{ (v,w) \, | \, v^2+w^3 = 1\}\). An ordered pair \((v,w) \in R\) means \(v\) and \(w\) are related by the equation \(v^2+w^3 = 1\); that is, the pair \((v,w)\) satisfy the equation.

For example, to show \((3,-2) \in R\), we check that when we substitute \(v=3\) and \(w=-2\), the equation \(v^2+w^3 = 1\) is true. Sure enough, \((3)^2+(-2)^3 = 9 - 8 = 1\). Hence, \(R\) maps \(3\) to \(-2\). Note, however, that \((-2,3) \notin R\) since \((-2)^2+(3)^3 = -8+27 \neq 1\) which means \(R\) does not map \(-2\) to \(3\).

When asked to ‘graph the equation’ \(v^2+w^3 = 1\), we really have two options. We could graph the relation \(R\) above. In this case, we would be graphing \(v^2+w^3 = 1\) on the \(vw\)-plane. Alternatively, we could define \(S = \{ (w,v) \, | \, v^2+w^3 = 1 \}\) and graph \(S\). This is equivalent to graphing \(v^2+w^3 = 1\) on the \(wv\)-plane. We do both in our next example.

Note that regardless of which geometric depiction we choose for \(v^2+w^3 = 1\), the graph appears to be symmetric about the \(w\)-axis. To prove this is the case, consider a generic point \((v,w)\) on the graph of \(v^2+w^3 = 1\) in the \(vw\)-plane.

To show the point symmetric about the \(w\)-axis, \((-v,w)\), is also on the graph of \(v^2+w^3 = 1\), we need to show that the coordinates of the point \((-v,w)\) satisfy the equation \(v^2+w^3 = 1\). That is, we need to show \((-v)^2+w^3 = 1\). Since \((-v)^2+w^3 = v^2 + w^3\), and we know by assumption \(v^2 + w^3 = 1\), we get \((-v)^2+w^3 = v^2+w^3 = 1\), proving \((-v,w)\) is also on the graph of the equation.

The key reason our proof above is successful is that algebraically, the equation \(v^2+w^3 = 1\) is unchanged if \(v\) is replaced with \(-v\). Geometrically, this means the graph is the same if it undergoes a reflection across the \(w\)-axis. We generalize this reasoning in the following result. Note that, as usual, we default to the more common \(x\) and \(y\)-axis labels.

Parts of Theorem symmetrytestequations should look familiar from our work with even and odd functions. Indeed if a function \(f\) is even, \(f(-x) = f(x)\). Hence, the equation \(y=f(-x)\) reduces to the equation \(y=f(x)\), so the graph of \(f\) is symmetric about the \(y\)-axis.

Likewise if \(f\) is odd, then \(f(-x) = -f(x)\). In this case, the equation \(-y = f(-x)\) reduces to \(-y = -f(x)\), or \(y = f(x)\), proving the graph is symmetric about the origin.

When it comes to symmetry about the \(x\)-axis, most of the time this indicates a violation of the Vertical Line Test, which is why we haven’t discussed that particular kind of symmetry until now.

We put Theorem symmetrytestequations to good use in the following example.

Looking at the graphs of the equations \(x^2-y^2 = 4\) and \((x-1)^2+4y^2 = 16\) in Example graphssymmex, it is evident neither of these equations represents \(y\) as a function of \(x\) nor \(x\) as a function of \(y\). (Do you see why?)

With the concept of ‘function’ being touted in the opening remarks of Section FunctionsandtheirRepresentations as being one of the ‘universal tools’ with which scientists and engineers solve a wide variety of problems, you may well wonder if we can’t somehow apply what we know about functions to these sorts of relations. It turns out that while, taken all at once, these equations do not describe functions, taken in parts, they do.

For example, consider the equation \(x^2 - y^2 = 4\) whose graph is reproduced below.

Solving for \(y\), we obtained \(y = \pm \sqrt {x^2-4}\). Defining \(f_{1}(x) = \sqrt {x^2-4}\), we get a functional description for the ‘upper half’ of this graph.

Likewise, setting \(f_{2}(x) = -\sqrt {x^2-4}\), we get a functional description for the ‘lower half’ of the curve.

If, for instance, we wanted to analyze this curve near \((3, -\sqrt {5})\), we could use the function \(f_{2}\) and all the associated function tools to do just that.

In this way we say the equation \(x^2 - y^2 = 4\) implicitly describes \(y\) as a function of \(x\) meaning that given any point \((x_{0},y_{0})\) on \(x^2 - y^2 = 4\), we can find a function \(f\) defined (on an interval) containing \(x_{0}\) so that \(f(x_{0}) = y_{0}\) and whose graph lies on the curve \(x^2 - y^2 = 4\).

Note that in this case, we are fortunate to have two explicit formulas for functions that cover the entire curve, namely \(f_{1}(x) = \sqrt {x^2-4}\) and \(f_{2}(x) = -\sqrt {x^2-4}\). We explore this concept further in the next example.

Not all equations implicitly define \(y\) as a function of \(x\). For a quick example, take \(x = 117\) or any other vertical line. Even if an equation implicitly describes \(y\) as a function of \(x\) near one point, there’s no guarantee we can find an explicit algebraic representation for that function.

While the theory of implicit functions is well beyond the scope of this text, we will nevertheless see this concept come into play in Section InverseFunctions. For our purposes, it suffices to know that just because a relation is not a function doesn’t mean we cannot find a way to apply what we know about functions to analyze the relation locally through a functional lens.