Open Propositions

Consider the statement

$$x + 3 = 8.$$

Here, we aren’t asking what value of $x$ solves the equation (we could, but that isn’t our focus). Instead, we’re asking whether, given a specific value of $x$, the statement is a true proposition or a false proposition.

We start by modeling the statement using a letter, like we’ve done up to this point, but we also use parentheses to denote the variable $x$:

\[ \begin{array}{rl} p(x)\text{: } &x + 3 = 8. \end{array} \]

We can substitute values in for $x$ to get different propositions:

\[ \begin{array}{rl} p(5)\text{: } &5 + 3 = 8 \\ p(3)\text{: } &3 + 3 = 8 \\ p(-4.73)\text{: } &-4.73 + 3 = 8 \end{array} \]

We can evaluate some of these propositions:

\[ \begin{array}{ll} p(5) &= 1 \\ p(3) &= 0 \\ p(-4.73) &= 0 \end{array} \]

Consider the statement

\[ \begin{array}{rl} p(x)\text{: } &x \text{ was the 30th president of the United States.} \end{array} \]

We can evaluate this statement with many values substituted in for $x$:

\[ \begin{array}{ll} p(\text{James K. Polk}) &= 0 \\ p(\text{Cecil Rhodes}) &= 0 \\ p(\text{Grover Cleveland}) &= 0 \\ p(\text{Calvin Coolidge}) &= 1 \\ p(\text{Bill Clinton}) &= 0 \end{array} \]

The U.S. Postal Service will only ship a package in a box that meets certain requirements: the sum of the box’s length and girth must not exceed 108 inches, where the girth is the perimeter of the box’s rectangular cross section.

Letting $\ell$, $w$, and $h$ represent a box’s length, width, and height respectively, the perimeter of that cross section is $2w + 2h$. So, to comply with USPS shipping requirements, we need

$$\ell + 2w + 2h \leq 108 \text{ inches.}$$

We can model this situation using a propositional function with three variables:

\[ \begin{array}{rl} s(\ell, w, h)\text{: } &\ell + 2w + 2h \leq 108 \text{ inches.} \end{array} \]

Can we ship a box with dimensions $\ell = 32$ inches, $w = 16$ inches, and $h = 18$ inches?

\[ \begin{array}{lll} \boldsymbol{s(32, 16, 18)} & = & (32) + 2(16) + 2(18) \leq 108 \\ & = & 32 + 32 + 36 \leq 108 \\ & = & 100 \leq 108 \\ & = & 1 \end{array} \]

Because $100 \leq 108$, we have $s(32, 16, 18) = 1$, so we can ship a box with these dimensions.

What about a box with dimensions $\ell = 20$ inches, $w = 18$ inches, and $h = 30$ inches?

\[ \begin{array}{lll} \boldsymbol{s(20, 18, 30)} & = & (20) + 2(18) + 2(30) \leq 108 \\ & = & 20 + 36 + 60 \leq 108 \\ & = & 116 \leq 108 \\ & = & 0 \end{array} \]

Since $116 \not\leq 108$, we can’t ship a box with length $20$ inches, width $18$ inches, and height $30$ inches with USPS.

For the propositional function

\[ \begin{array}{rl} p(x)\text{: } &x \text{ was the 30th president of the United States,} \end{array} \]

we could argue $p(3)$ is either false or undefined. We can eliminate this ambiguity by specifying what kinds of values we’re allowed to substitute into $p(x)$. If a value $a$ is allowed as an input, then $p(a)$ is either true or false. If a value $b$ is not allowed, then $p(b)$ is undefined.

Let’s restrict the values allowed as inputs to $p(x)$ to proper names. This means values like “James K. Polk,” “Cecil Rhodes,” “Grover Cleveland,” “Calvin Coolidge,” and “Bill Clinton” can be substituted for $x$ and will yield true or false.

However, when substituting a value for $x$ that isn’t a proper name, we’ll say $p(x)$ is undefined. Since $3$ isn’t a proper name, $p(3)$ is undefined.

We could further restrict the inputs to be names of U.S. presidents only, rather than just any proper name. In that case, $p(\text{Cecil Rhodes})$ would be undefined, since “Cecil Rhodes” isn’t the name of any U.S. president.

Reconsider the propositional function

\[ \begin{array}{rl} p(x)\text{: } &x \text{ was the 30th president of the United States.} \end{array} \]

Suppose we restrict the allowable values to names of U.S. presidents, except “Calvin Coolidge.” Then $p(\text{Calvin Coolidge})$ would be undefined. Substituting any other U.S. president’s name yields $0$, since only Calvin Coolidge was the 30th U.S. president. Under this restriction, $p(x)$ never yields $1$ — only $0$, or undefined.

Restricting the allowable names to Spanish monarchs would also only ever yield $0$ or undefined, since no Spanish monarch was ever the 30th president of the United States. We could similarly restrict the allowable values to even integers — again, $p(x)$ would always be either false, or undefined.

Reconsider the propositional function

\[ \begin{array}{rl} p(x)\text{: } &x + 3 = 8. \end{array} \]

If we restrict our inputs to integers only, we can make $p(x)$ true by substituting $5$ in for $x$. Any other integer yields $0$. Non-integers yield undefined values, since they aren’t allowed as inputs — so $p(3.14159265)$ is undefined.

However, if we restrict allowed inputs to any real number, $p(3.14159265)$ is no longer undefined — it’s equal to $0$.

Consider the propositional function

\[ \begin{array}{rl} r(x)\text{: } &x \text{ has a right angle,} \end{array} \]

with universe of discourse $\mathcal{U}$ the collection of all planar polygons.

Since the number $2$ isn’t a planar polygon, $r(2)$ is undefined.

Suppose $s_1$ represents a square with side length $1$. $s_1$ is a planar polygon, meaning $r(s_1)$ is either $0$ or $1$. Since every square has a right angle, $r(s_1) = 1$.

Suppose $s_2$ represents an equilateral triangle with side length $1$. $s_2$ is also a planar polygon. Since no equilateral triangle has a right angle, $r(s_2) = 0$.

Consider the propositional function

\[ \begin{array}{rl} e(x, y)\text{: } &x + y \text{ is an even integer,} \end{array} \]

with universe of discourse $\mathcal{U}$ for both $x$ and $y$ the integers. This means both $x$ and $y$ must be integers.

We’d have $e(2, 4) = 1$, $e(3, 7) = 1$, $e(1, 2) = 0$, and $e(4, 13) = 0$.

$e(2, 2.3)$, $e(2.18, 3.14)$, and $e(4.411, 10)$ would all be undefined, since $2.3$, $2.18$, $3.14$, and $4.411$ aren’t integers.

$e(2, \text{Monday})$ would also be undefined, since “Monday” isn’t an integer. Similarly, $e(\text{Red}, \text{Cactus})$ would also be undefined.

For $e(x, y)$ to yield $0$ or $1$, both $x$ and $y$ need to be integers.

Consider the propositional function

\[ \begin{array}{rl} q(x, y)\text{: } &x \div y \text{ is larger than 5.2,} \end{array} \]

with universe for $x$, denoted $\mathcal{U}_x$, all real numbers, and universe for $y$, denoted $\mathcal{U}_y$, all real numbers except $0$.

Here, $q(2.2, 0.0001) = 1$, $q(10, 2) = 0$, $q(0.52, 0.01) = 1$, and $q(0.52, 0.1) = 0$.

$q(1, 0)$ would be undefined, since $0 \notin \mathcal{U}_y$.

$q(\text{Monday}, 1)$ would be undefined, since $\text{Monday}$ is not a real number.

$q(\text{Friday}, 0)$ would be undefined, since $\text{Friday}$ is not a real number, and $0$ is not a non-zero real number.

Consider the propositional function

\[ \begin{array}{rl} q(x, y, z)\text{: } &(x + y) \div z = 1, \end{array} \]

with universe for $x$ and $y$, denoted $M$, all integers, and universe for $z$, denoted $N$, all real numbers except $0$.

We have $q(1, 0, 1) = 1$, $q(-2, 7, 5) = 1$, $q(-1, 1, 1) = 0$, and $q(10, -23, 2) = 0$.

$q(0.1, 1, 1)$ is undefined, since the supplied value for $x$ ($0.1$) isn’t in $M$.

Similarly, $q(23, -3.14, 10)$ is undefined, since the supplied value for $y$ ($-3.14$) isn’t in $y$’s universe, $M$.

Finally, $q(20, -10, 0)$ is undefined, since the supplied value for $z$ ($0$) isn’t in $z$’s universe, $N$.