Adding two whole numbers always gives a whole number, but does not. Which operations can be carried out inside a set of numbers without the answer ever leaving it?

Algebra 1 · N.RN.B.3

Closure

Adding two whole numbers always gives a whole number, but 4 - 9 does not. Which operations can be carried out inside a set of numbers without the answer ever leaving it?

You need

Sorting a number into its setsMath 8 Subtracting integersMath 7

Where you are 3 in Properties of real numbers

Number9

Irrational and real numbers5

Properties of real numbers3

Closure

Commutative, associative and distributive properties Identity and inverse elements

Why this exists

Start with the whole numbers 0, 1, 2, 3, \ldots and do arithmetic with them. Add two of them and you get another whole number. Multiply two of them and you get another whole number. Subtract, and sometimes it works, 9 - 4 = 5, and sometimes it does not: 4 - 9 has no answer among the whole numbers at all. Figure 1 shows why. Starting at 4 and moving 9 to the left runs off the end of the dots.

Figure 1The same subtraction, 4 - 9, on two sets. Among the whole numbers (top) the answer lands where no member sits. Among the integers (bottom) it lands on a member.

Negative numbers fill that gap. Put each whole number’s opposite beside it and you have the Integers, and among the integers every subtraction has an answer: 4 - 9 = -5. Division then breaks in the same way, since 9 \div 4 is no integer, and fractions fill that gap, giving the rational numbers. Square roots break it once more, because no fraction squares to 2 (Irrational numbers), and the irrational numbers fill that one.

Each new set was made so that an operation which used to throw its answer outside no longer does. That property is called closure.

Definition

A set of numbers is closed under an operation when carrying out that operation on any two members of the set, including a member with itself, always gives a member of the set.

For addition and a set S, that reads

\begin{gathered} \text{if } a \in S \text{ and } b \in S, \\ \text{then } a + b \in S \end{gathered}

for every choice of a and b, and the same with -, \times or \div in place of +.

The words any and always make the two answers unlike each other.

To show a set is closed, you need an argument that covers every pair at once, usually with letters for the members. Examples, however many, are not enough, unless the set is small enough for every pair to be listed.
To show a set is not closed, one pair whose answer lands outside is enough. A single case that breaks a claim about every case is a counterexample.

The sets this lesson uses are the ones in Sorting a number into its sets: the natural numbers 1, 2, 3, \ldots; the whole numbers 0, 1, 2, \ldots; the integers \ldots, -2, -1, 0, 1, 2, \ldots; the rational numbers \mathbb{Q}; the irrational numbers; and the real numbers.

Intuition

Think of the set as a fenced field: the set is closed when every pair of numbers from inside lands its answer inside, and one answer thrown over the fence is enough to say it is not.

On the top line of Figure 1 the members are the dots. Adding a whole number moves right from a dot by a whole number of steps, and to the right there are dots all the way, so addition stays inside. Subtracting moves left, and to the left the dots stop at 0. On the bottom line the dots carry on in both directions, so neither move can leave.

Results

Closure of the familiar sets. Division always means division by a nonzero member, since no set has an answer for division by 0.

Set	Closed under	Not closed under
natural numbers	+, \times	-, \div
whole numbers	+, \times	-, \div
integers	+, -, \times	\div
rational numbers	all four	none
irrational numbers	none	all four
real numbers	all four	none

Each “not closed” needs only one counterexample:

natural numbers: 2 - 2 = 0 and \displaystyle 2 \div 5 = \frac{2}{5};
whole numbers: 4 - 9 = -5 and \displaystyle 9 \div 4 = \frac{9}{4};
integers: \displaystyle 9 \div 4 = \frac{9}{4};
irrational numbers: \displaystyle \sqrt{2} + (-\sqrt{2}) = 0, \displaystyle \sqrt{2} - \sqrt{2} = 0, \displaystyle \sqrt{2} \times \sqrt{2} = 2 and \displaystyle \sqrt{2} \div \sqrt{2} = 1, where \displaystyle -\sqrt{2} is irrational as a nonzero rational times \displaystyle \sqrt{2}.

A rational number with an irrational one, from Irrational numbers. Let r be rational and x irrational. Then r + x, r - x and x - r are irrational, and r \times x and x \div r are irrational as long as r \neq 0.

Closure does not pass up or down between the sets. The integers are closed under subtraction although the whole numbers inside them are not, and the irrational numbers are closed under none of the four operations although the real numbers around them are closed under all four.

Why it is true

Whole numbers and natural numbers, adding and multiplying. Adding counts two collections put together and multiplying counts equal groups, and a count is a whole number. If neither number is 0 the count is at least 1, which covers the natural numbers.

Integers under subtraction. Subtracting is adding the opposite, a - b = a + (-b) (Subtracting integers). If b is an integer, so is -b, and the sum of two integers is an integer. So the integers are closed under subtraction because they are closed under addition and contain the opposite of every member. That second part is exactly what the whole numbers lack, and it is why the negative numbers were needed.

Rational numbers, a reminder. What makes a number rational proves this with the fraction rules, such as \displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} and \displaystyle \frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}. Each new top is built from integers by multiplying, adding and subtracting, so it is an integer; each new bottom is a product of nonzero integers, so it is a nonzero integer. The closure of \mathbb{Q} rests on the closure of the integers.

A rational number with an irrational one, a reminder. If r + x were a rational number s, then x = s - r would be rational by closure of \mathbb{Q}, which it is not. Products and quotients go the same way with r \neq 0. This is a Proof by contradiction with closure as its engine.

Real numbers. That adding, subtracting, multiplying and dividing by a nonzero number never leave the The real number line is a basic property of the real numbers, taken here as a starting fact rather than proved.

Worked examples

Example 1

When subtraction leaves the whole numbers

Decide whether the whole numbers are closed under (a) addition, (b) subtraction and (c) division by a nonzero whole number. Whenever an answer escapes, say which set from the table does contain it.

Solution

(a) Closed. A pair such as 6 + 8 = 14 only illustrates this; the counting argument in “Why it is true” covers every pair.

(b) Not closed. 9 - 4 = 5 lands inside, but 4 - 9 = -5 does not, and that single pair is the counterexample; pairs that work do not rescue the claim. The answer -5 is an integer (Figure 1).

(c) Not closed. 8 \div 4 = 2 lands inside, but \displaystyle \frac{9}{4} = 2.25 lies between 2 and 3 and is no whole number. It is a rational number.

Example 2

Checking every pair of a small set

Is the set \{-1, 0, 1\} closed under (a) multiplication, (b) addition, (c) subtraction?

Solution

(a) The set has three members, so there are only nine ordered pairs, and all of them can be checked. The table lists every product, with the first number down the side and the second along the top.

\times	-1	1
-1	1	-1
0	0	0
1	-1	1

Every entry is -1, 0 or 1, so the set is closed under multiplication. For a set this small, listing every pair is an argument that covers every pair.

(b) Not closed: 1 + 1 = 2, and 2 \notin \{-1, 0, 1\}. Using the same member twice is allowed by the definition.

Example 3

Decimals that stop

A terminating decimal is a decimal that stops, such as 2.5, 0.04 or -3.125. Is the set of terminating decimals closed under addition, subtraction and multiplication? Under division by a nonzero member?

Solution

A terminating decimal with k digits after the point is an integer divided by 10^k: for example \displaystyle 0.04 = \frac{4}{100}. Take two of them, \displaystyle \frac{n}{10^k} and \displaystyle \frac{m}{10^j} with n and m integers, and name them so that k \geq j.

Adding and subtracting. Rewrite the second over 10^k by multiplying top and bottom by 10^{k - j}:

\begin{aligned} &\frac{n}{10^k} \pm \frac{m}{10^j} \\[4pt] &= \frac{n \pm m \cdot 10^{k - j}}{10^k} \end{aligned}

The top is an integer, since the integers are closed under multiplication, addition and subtraction. An integer over a power of 10 is a terminating decimal, so the set is closed under both.

Multiplying. The tops multiply and the bottoms multiply:

\frac{n}{10^k} \times \frac{m}{10^j} = \frac{nm}{10^{k + j}}

again an integer over a power of 10. Closed. For instance \displaystyle 2.5 \times 0.04 = \frac{25}{10} \times \frac{4}{100} = \frac{100}{1000} = 0.1.

Dividing. Not closed. 0.7 and 0.3 both stop, but \displaystyle 0.7 \div 0.3 = \frac{7}{3}, and long division of 7 by 3 leaves remainder 1 at every step, so \displaystyle \frac{7}{3} = 2.333\ldots never stops. The answer is still rational, as closure of \mathbb{Q} promised. Like the integers, the terminating decimals fail only at division.

Example 4

Two irrational numbers

Let \displaystyle p = 3 + \sqrt{5} and \displaystyle q = 3 - \sqrt{5}. Find p + q, p - q and p \times q, and decide whether each is rational or irrational.

Solution

Both p and q are irrational: \displaystyle \sqrt{5} is irrational because 5 is not a perfect square (Irrational numbers), and a rational number plus or minus an irrational one is irrational.

For the sum, the \displaystyle \sqrt{5} parts cancel:

p + q = (3 + \sqrt{5}) + (3 - \sqrt{5}) = 6

which is rational. For the difference, the 3s cancel instead:

p - q = (3 + \sqrt{5}) - (3 - \sqrt{5}) = 2\sqrt{5}

which is irrational, a nonzero rational times \displaystyle \sqrt{5}. For the product, multiply each part of the first bracket by each part of the second (the distributive property) and use \displaystyle \sqrt{5} \times \sqrt{5} = 5:

\begin{aligned} p \times q &= 9 - 3\sqrt{5} + 3\sqrt{5} - 5 \\ &= 4 \end{aligned}

which is rational. So p + q and p \times q escape the irrational numbers while p - q stays inside. “Not closed” means that some answer escapes, not that every answer does.

Example 5

Using closure to prove a number irrational

Show that \displaystyle 7 - 3\sqrt{2} is irrational.

Solution

One way, step by step. 3 is a nonzero rational number and \displaystyle \sqrt{2} is irrational (Irrational numbers), so \displaystyle 3\sqrt{2} is irrational. Then 7 is rational, so \displaystyle 7 - 3\sqrt{2} is a rational number minus an irrational one, which is irrational.

Another way, straight from closure. Suppose \displaystyle 7 - 3\sqrt{2} were a rational number s. Solve for \displaystyle \sqrt{2}:

\begin{aligned} 7 - 3\sqrt{2} &= s \\ 3\sqrt{2} &= 7 - s \\ \sqrt{2} &= \frac{7 - s}{3} \end{aligned}

7 - s is rational, because the rational numbers are closed under subtraction, and dividing it by the nonzero rational number 3 keeps it rational, because they are closed under division. So \displaystyle \sqrt{2} would be rational, which it is not. The supposition was false, and \displaystyle 7 - 3\sqrt{2} is irrational. Every step of solving for \displaystyle \sqrt{2} stayed inside \mathbb{Q}; that is what closure buys.

Example 6

A new set that is closed

Let S be the set of all numbers \displaystyle a + b\sqrt{2} with a and b rational. It contains every rational number (take b = 0), and \displaystyle \sqrt{2}, and numbers such as \displaystyle 3 - \frac{1}{2}\sqrt{2}. Show that S is closed under addition, subtraction and multiplication, and find \displaystyle (1 + 2\sqrt{2})(3 - \sqrt{2}) in the form \displaystyle a + b\sqrt{2}.

Solution

Take two members, \displaystyle a + b\sqrt{2} and \displaystyle c + d\sqrt{2}, with a, b, c and d rational.

Adding and subtracting. Regroup the rational parts together and the \displaystyle \sqrt{2} parts together, using the commutative and associative properties of addition, then the distributive property:

\begin{aligned} &(a + b\sqrt{2}) \pm (c + d\sqrt{2}) \\ &= (a \pm c) + (b \pm d)\sqrt{2} \end{aligned}

a \pm c and b \pm d are rational by closure of \mathbb{Q}, so the answer is in S.

Multiplying. The distributive property gives four products, and \displaystyle \sqrt{2} \times \sqrt{2} = 2:

\begin{aligned} &(a + b\sqrt{2})(c + d\sqrt{2}) \\ &= ac + ad\sqrt{2} + bc\sqrt{2} + 2bd \\ &= (ac + 2bd) + (ad + bc)\sqrt{2} \end{aligned}

ac + 2bd and ad + bc are made from rational numbers by multiplying and adding, so they are rational by closure of \mathbb{Q}, and the product is in S.

The product asked for. Here a = 1, b = 2, c = 3 and d = -1:

\begin{aligned} ac + 2bd &= 3 + 2(2)(-1) = -1 \\ ad + bc &= -1 + 6 = 5 \end{aligned}

so \displaystyle (1 + 2\sqrt{2})(3 - \sqrt{2}) = -1 + 5\sqrt{2}.

Check by multiplying out directly: \displaystyle 3 - \sqrt{2} + 6\sqrt{2} - 2 \times 2 = -1 + 5\sqrt{2}.

The irrational numbers on their own are not closed, but putting \displaystyle \sqrt{2} together with all the rational numbers builds a set that is. Division is taken up in Exercise 20.

Boundaries

Closure is about a set and an operation together. “The integers are closed” says nothing until the operation is named; they are closed under subtraction and not under division.

Division by 0 is left out from the start. No set has an answer for 5 \div 0, so “closed under division” always means division by a nonzero member.

Not closed does not mean every answer escapes. \displaystyle \sqrt{2} + \sqrt{3} and \displaystyle \sqrt{2} + \sqrt{2} = 2\sqrt{2} are irrational. The irrational numbers are not closed under addition because some sums are rational, so each case has to be worked out.

The rule for products needs r \neq 0. \displaystyle 0 \times \sqrt{2} = 0 is rational. For sums and differences no exception is needed.

Closure is about where the answer lands, nothing more. It says that a + b is a real number; whether a + b equals b + a is a different property, the commutative one.

Where this goes next

Closure is the first of the Properties of real numbers: it says a real-number calculation lands on a real number. The commutative, associative and distributive properties and identities and inverses then say how those answers relate to each other.

Polynomials behave like the integers: adding and subtracting polynomials gives a polynomial, and so does multiplying them, but dividing usually does not. Facts such as “an odd number times an odd number is odd” are closure facts proved with letters, the work of Proving statements about even and odd numbers.

The real numbers are not closed under taking square roots: no real number squares to -4. Adding a number whose square is -1 fills that gap, the same move that added the negative numbers, and gives the Complex numbers. The set of Example 6 under division leads to rationalizing a denominator.

Exercises

Closed or not

Decide whether each set is closed under the operation given. If it is, give a reason that covers every pair; if it is not, give a counterexample.

The integers, under division by a nonzero integer.
The natural numbers 1, 2, 3, \ldots, under subtraction.
The even integers, under addition.
The odd integers, under addition.
The odd integers, under multiplication.
The negative integers, under multiplication.
The multiples of 5 (the integers 5n, with n an integer), under subtraction.
The set \{0, 1\}, under multiplication and under addition.

Inside the rational numbers

Use the fraction rules from What makes a number rational to write each answer as an integer over a nonzero integer, and say why the top and the bottom are integers and the bottom is not 0. Give the answer in lowest terms.

\displaystyle \frac{3}{8} + \frac{5}{12}
\displaystyle \frac{5}{14} \div \left(-\frac{10}{7}\right)
0.\overline{6} - 1.75, after writing each decimal as a fraction (Turning a repeating decimal into a fraction)

Rational or irrational

Each number is made from irrational numbers, or from a rational number and an irrational one. Simplify where you can, then decide whether it is rational or irrational.

\displaystyle (4 + \sqrt{7}) + (4 - \sqrt{7})
\displaystyle (\sqrt{7} + 4)(\sqrt{7} - 4)
\displaystyle 3\sqrt{2} \div \sqrt{2}
\displaystyle \frac{5}{8} - \pi
\displaystyle (\sqrt{5} - 1)^2

Reasoning

Tarquin multiplies ten different pairs of numbers from a set, and every product lands in the set. Saffron finds one pair whose product does not. Whose evidence settles whether the set is closed under multiplication? Explain.
Take the rational numbers strictly between 0 and 1, the numbers q with 0 < q < 1. Is this set closed under multiplication? Under addition? Explain.
Take the irrational numbers together with 0. Adding 0 to the set takes care of \displaystyle \sqrt{2} + (-\sqrt{2}) = 0. Is this new set closed under addition? Explain.
Show that \displaystyle (4 + \sqrt{2})(4 - \sqrt{2}) = 14, and use it to write \displaystyle \frac{1}{4 + \sqrt{2}} in the form \displaystyle a + b\sqrt{2} with a and b rational. Explain what this suggests about the set S of Example 6 under division.

Answers

Not closed. \displaystyle (-6) \div 4 = -\frac{3}{2} = -1.5, which is not an integer.
Not closed. 4 - 4 = 0, and 0 is not a natural number.
Closed. Two even integers are 2m and 2n with m and n integers, and 2m + 2n = 2(m + n) by the distributive property. m + n is an integer, so the sum is even.
Not closed. 7 + 9 = 16, which is even. In fact every sum escapes: (2m + 1) + (2n + 1) = 2(m + n + 1).
Closed. Multiplying out, (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1, which is 2(2mn + m + n) + 1. Since 2mn + m + n is an integer, the product is odd.
Not closed. (-2) \times (-3) = 6, which is positive. In fact every product escapes.
Closed. 5m - 5n = 5(m - n) by the distributive property, and m - n is an integer because the integers are closed under subtraction.
Closed under multiplication: the four products are 0 \times 0 = 0, 0 \times 1 = 0, 1 \times 0 = 0 and 1 \times 1 = 1, all in the set. Not closed under addition: 1 + 1 = 2.
\displaystyle \frac{19}{24}. \displaystyle \frac{3 \times 12 + 8 \times 5}{8 \times 12} = \frac{36 + 40}{96} = \frac{76}{96}, which reduces to \displaystyle \frac{19}{24}. The top is built from integers; the bottom is a product of nonzero integers.
\displaystyle -\frac{1}{4}. By \displaystyle \frac{ad}{bc}, \displaystyle \frac{5 \times 7}{14 \times (-10)} = \frac{35}{-140}, which reduces to \displaystyle -\frac{1}{4}. Both are products of integers, and the bottom is not 0 because neither factor is.
\displaystyle -\frac{13}{12}. \displaystyle 0.\overline{6} = \frac{2}{3} and \displaystyle 1.75 = \frac{7}{4}, so the difference is \displaystyle \frac{2 \times 4 - 3 \times 7}{3 \times 4} = \frac{8 - 21}{12} = -\frac{13}{12}.
Rational: the \displaystyle \sqrt{7} parts cancel and the sum is 8.
Rational: \displaystyle 7 - 4\sqrt{7} + 4\sqrt{7} - 16 = -9.
Rational: \displaystyle 3\sqrt{2} \div \sqrt{2} = 3. Both numbers are irrational, so this is another counterexample for division.
Irrational. \displaystyle \frac{5}{8} is rational and \pi is irrational, and a rational number minus an irrational one is irrational.
Irrational. \displaystyle (\sqrt{5} - 1)^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5}, a rational number minus a nonzero rational times \displaystyle \sqrt{5}.
Saffron’s. Closed means every pair gives a product in the set, so one failing pair makes the statement false. Tarquin’s ten pairs leave other pairs unchecked.
Closed under multiplication, not under addition. If q and t are in the set, qt is rational by closure of \mathbb{Q} and positive as a product of positive numbers. Multiplying both sides of t < 1 by the positive number q keeps the inequality, so qt < q < 1. But \displaystyle \frac{1}{2} + \frac{2}{3} = \frac{7}{6}, which is greater than 1.
No. \displaystyle 1 + \sqrt{2} and \displaystyle -\sqrt{2} are both irrational, so both are in the set, but \displaystyle (1 + \sqrt{2}) + (-\sqrt{2}) = 1, which is rational and not 0, so it is outside the set.
\displaystyle (4 + \sqrt{2})(4 - \sqrt{2}) = 16 - 4\sqrt{2} + 4\sqrt{2} - 2 = 14. Dividing both sides by \displaystyle 14(4 + \sqrt{2}) gives \displaystyle \frac{1}{4 + \sqrt{2}} = \frac{4 - \sqrt{2}}{14}, which is \displaystyle \frac{2}{7} - \frac{1}{14}\sqrt{2}, in S. It suggests S is closed under division by a nonzero member, since dividing is multiplying by the reciprocal. In general \displaystyle (a + b\sqrt{2})(a - b\sqrt{2}) = a^2 - 2b^2 is rational, and it is not 0 unless a = b = 0, since otherwise \displaystyle \sqrt{2} would be a fraction.