Sets Of Numbers

(a) $\mathbb{Z},\mathbb{Q},ℝ$   (b) $\mathbb{Q},ℝ$   (c) $\mathbb{Z},\mathbb{Q},ℝ$   (d) $ℝ$ only   (e) $\mathbb{Q},ℝ$

Keep the four definitions in front of you as you work:

• $\mathbb{N}=\{1,2,3,\dots \}$, the counting numbers;
• $\mathbb{Z}$, all integers, positive, negative and zero;
• $\mathbb{Q}$, all numbers that can be written as one integer divided by another;
• $ℝ$, all rational numbers together with all irrational numbers.

It also helps to remember the chain recorded in this section,

which means that as soon as a number lands in one of these sets, it automatically lies in every set further to the right.

(a) $-7$. It is not a counting number, since counting numbers start at $1$ and go up, so $-7\notin \mathbb{N}$. It is a whole number with a minus sign, so it is a negative integer: $-7\in \mathbb{Z}$. By the chain, it then also lies in $\mathbb{Q}$ and $ℝ$. (Directly, $-7={\displaystyle \frac{-7}{1}}$, a ratio of two integers.)

(b) ${\displaystyle \frac{5}{8}}$. This is not an integer, since $8$ does not divide $5$ evenly, so it is in neither $\mathbb{N}$ nor $\mathbb{Z}$. It is written as one integer over another, which is exactly the definition of a rational number, so ${\displaystyle \frac{5}{8}}\in \mathbb{Q}$, and therefore also ${\displaystyle \frac{5}{8}}\in ℝ$.

(c) $0$. A common sticking point. Zero is not a natural number as this section defines them, because the list $\{1,2,3,\dots \}$ begins at $1$. But the section explicitly includes zero among the integers, so $0\in \mathbb{Z}$, and hence $0\in \mathbb{Q}$ and $0\in ℝ$. (Directly, $0={\displaystyle \frac{0}{1}}$.)

(d) $\sqrt{2}$. The section states that $\sqrt{2}$ cannot be written as $p/q$ with $p$ and $q$ integers, and gives the proof. So it is irrational: it is in none of $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$. But irrational numbers are real numbers, so $\sqrt{2}\in ℝ$ and nothing more.

(e) $3.75$. Its decimal expansion stops, and the section says a terminating decimal is rational. Writing it out,

a ratio of two integers. So $3.75\in \mathbb{Q}$ and $3.75\in ℝ$. It is not an integer, because it sits strictly between $3$ and $4$.

(a) Commutative property of addition   (b) Distributive property   (c) Associative property of multiplication   (d) Multiplicative inverse property   (e) Additive identity property   (f) Commutative property of multiplication

The trick is to look at what actually changed from one side of the equals sign to the other.

(a) $7+x=x+7$. Nothing has been grouped or expanded; the two things being added have simply swapped places. Swapping the order of a sum is the commutative property of addition, $a+b=b+a$, with $a=7$ and $b=x$.

(b) $3(x+5)=3x+15$. On the left, a single multiplication acts on a bracket containing a sum; on the right, the multiplication has been carried out on each term separately. That is the distributive property, $a(b+c)=ab+ac$, with $a=3$, $b=x$, $c=5$. Note $3\cdot 5=15$, which is where the $15$ comes from.

(c) $(2\cdot 4)\cdot y=2\cdot (4\cdot y)$. The same three things are multiplied in the same order on both sides; only the parentheses have moved. Re-grouping a product is the associative property of multiplication, $a(bc)=(ab)c$.

(d) $9\cdot {\displaystyle \frac{1}{9}}=1$. A number is multiplied by its reciprocal and the result is $1$. That is the multiplicative inverse property, $a\cdot {\displaystyle \frac{1}{a}}=1$ for $a\ne 0$, with $a=9$.

(e) $x+0=x$. Zero has been added and has left the number unchanged. That is the additive identity property, $a+0=a$.

(f) $x\cdot 4=4\cdot x$. Two factors have swapped places, so this is the commutative property of multiplication, $ab=ba$.

The pair to keep straight is commutative versus associative. Commutative means the items changed order; associative means the items kept their order but were grouped differently. Compare (a) and (f), where things moved past each other, with (c), where only the brackets moved.

(a) ${\displaystyle \frac{-6}{1}}$    (b) ${\displaystyle \frac{7}{10}}$    (c) ${\displaystyle \frac{225}{100}}={\displaystyle \frac{9}{4}}$

The definition to satisfy is the one given in this section: a number is rational when it can be written as ${\displaystyle \frac{p}{q}}$ with $p$ and $q$ integers. So for each number we must actually exhibit such a fraction.

(a) $-6$. The section points out that every integer is rational, because it can be given a denominator of $1$:

Here $p=-6$ and $q=1$, both integers, so $-6$ is rational. Check: dividing $-6$ by $1$ gives back $-6$.

(b) $0.7$. This decimal terminates after one digit, and that digit sits in the tenths place. So it is seven tenths:

Both $7$ and $10$ are integers, so $0.7$ is rational. Check: dividing $7$ by $10$ gives $0.7$.

(c) $2.25$. This decimal terminates after two digits, so the last digit is in the hundredths place, and we put the number over $100$:

That already answers the question, since $225$ and $100$ are integers. If we want it in lowest terms, divide top and bottom by $25$:

Check: $9$ divided by $4$ is $2.25$.

The general rule behind (b) and (c): a decimal that stops after $n$ digits can be written over $1$ followed by $n$ zeros. One digit gives a denominator of $10$; two digits give $100$; three would give $1000$. This is why the section says every terminating decimal is a rational number.

$x=4$ or $x=-9$

The expression on the left is a product of two things: the number $x-4$ and the number $x+9$. We are told this product equals $0$.

The Zero-Product Property stated in this section says: if $ab=0$, then $a=0$ or $b=0$ (or both). Here $a$ is $x-4$ and $b$ is $x+9$, so at least one of these must be zero. That gives two possibilities, and we examine each.

Possibility 1: $x-4=0$. Adding $4$ to both sides,

Possibility 2: $x+9=0$. Subtracting $9$ from both sides,

So the possible values are $x=4$ and $x=-9$.

Check both. Put $x=4$ into the original product:

The last step uses the rule from this section that the product of any number and $0$ is $0$. Now put $x=-9$:

Both work.

A warning about how this property is used. It only works when the product equals zero. If instead we were told $(x-4)(x+9)=12$, we could not conclude that $x-4=12$ or $x+9=12$, because there are many pairs of numbers whose product is $12$. Zero is special: it is the only number that forces one of the factors to be it.

(a) Defined, equals $0$   (b) No meaning   (c) No meaning   (d) No meaning

The rule from this section is short and absolute: division by zero is not a valid operation. So the only thing to look at in each part is the denominator.

(a) ${\displaystyle \frac{0}{5}}$. The denominator is $5$, which is not zero, so the expression is fine. Its value is $0$: as the section notes, if $a\ne 0$ then ${\displaystyle \frac{0}{a}}=0$, because $a\cdot 0=0$.

(b) ${\displaystyle \frac{5}{0}}$. The denominator is zero, so this has no meaning. Here is the reason the section gives. If ${\displaystyle \frac{5}{0}}$ were equal to some number $b$, then multiplying back would give $5=0\cdot b$. But $0\cdot b=0$ for every $b$, so we would need $5=0$, which is false. No value of $b$ can work.

(c) ${\displaystyle \frac{0}{0}}$. Again the denominator is zero, so again there is no meaning. The reason is different and worth understanding. If ${\displaystyle \frac{0}{0}}$ were equal to some number $b$, we would need $0=0\cdot b$. This time the equation is true for every $b$ — for $b=1$, for $b=7$, for $b=-53$, for all of them. So instead of no candidate, we have too many, and there is no way to single out one answer. Either way, no value can be assigned.

(d) ${\displaystyle \frac{x}{x-3}}$ when $x=3$. Substitute $x=3$ everywhere:

The denominator has come out to zero, so this has no meaning. This is the case to be most careful about, because the denominator $x-3$ does not look like zero — it only becomes zero for one particular value of $x$. Whenever a letter appears in a denominator, you must ask which values of that letter would make the denominator vanish, and rule them out.

(a) $3700$    (b) $5(x+y)$

The distributive property says

Since this is an equality, it may be read from left to right (expanding a bracket) or from right to left (collecting a common factor into a bracket). Part (a) uses it right to left, and part (b) uses it right to left as well — the direction that is usually the harder one to spot.

Part (a). Look at $37\cdot 99+37\cdot 1$. Both terms are a product with $37$ in them, so this matches the right-hand side $ab+ac$ with

Reading the property from right to left,

Now the bracket is easy:

Check by the long route: $37\cdot 99=3663$ and $37\cdot 1=37$, and $3663+37=3700$. The answers agree, but the first route needed no multiplication by $99$ at all.

Part (b). Look at $5x+5y$. Again both terms contain a common factor, this time $5$. Matching to $ab+ac$ with

and reading right to left,

Check by expanding again: $5(x+y)=5\cdot x+5\cdot y=5x+5y$, which is what we started with.

The section notes that these rules hold "for real numbers, variables, and algebraic expressions", which is why the same property that handled plain numbers in part (a) also handles the letters in part (b).

No. ${\displaystyle \frac{2}{7}}$ is rational.

Go back to the definition. This section says a number is rational when it can be written as a fraction of two integers. It does not say anything about the decimal having to stop.

The number ${\displaystyle \frac{2}{7}}$ is already written as a fraction of two integers: $p=2$ and $q=7$. So by the definition it is rational, and that settles the matter — no examination of the decimal is needed at all.

So what does the decimal tell us? Look at the digits carefully:

The block $285714$ comes round again and again. This is exactly the second case the section describes: a decimal that "does not terminate but has repeating blocks of digits", and such numbers are rational. The section's own example is $0.3333\cdots ={\displaystyle \frac{1}{3}}$, where the repeating block is just the single digit $3$.

An irrational number is different in a specific way. The section says its decimal "goes on forever and never repeats". Both conditions matter. For $\pi =3.141592\dots$ the digits go on forever with no block ever settling into a cycle, and that is why $\pi$ is irrational.

So the summary is:

• decimal stops — rational, for example $3.89={\displaystyle \frac{389}{100}}$;
• decimal runs on but repeats a block — rational, for example ${\displaystyle \frac{2}{7}}$;
• decimal runs on and never repeats — irrational, for example $\pi$ and $\sqrt{2}$.

The mistake to avoid is treating "the decimal never ends" as though it meant irrational. It is the failure to repeat, not the failure to stop, that makes a number irrational.

$-2,-{\displaystyle \frac{3}{2}},0,\sqrt{2},\pi$

This section explains that every real number matches exactly one point on the line, that positive numbers lie to the right of $0$ and negative numbers lie to the left, and that the further right a point is, the larger its number.

So the task is to compare the five numbers by size. The safest way is to convert each into decimal form, because decimals are easy to compare digit by digit.

Now sort them.

The two negative numbers go on the left of $0$. Between them, $-2$ is further from $0$ than $-1.5$ is, so $-2$ sits further to the left. This is the step where students most often go wrong: with negative numbers, the one that looks bigger is the one further left. Picture two debts — owing $2$ dollars is a worse position than owing $1.50$.

Then comes $0$ at the origin.

The two positive numbers go to the right of $0$, with the larger further right. Since $1.414$ is less than $3.142$, the number $\sqrt{2}$ comes first and $\pi$ comes last.

Left to right, the order is

On a sketch, mark $0$ and $1$ first to fix the scale, then place $\sqrt{2}$ a little under half way between $1$ and $2$, $\pi$ just past $3$, $-{\displaystyle \frac{3}{2}}$ half way between $-1$ and $-2$, and $-2$ at the whole-number mark.

Note that two of these five numbers, $\sqrt{2}$ and $\pi$, are irrational. They still have their own single point on the line — the section says every real number does, rational or not.

(a) True   (b) False   (c) True   (d) False   (e) True

Recall from the previous section that $B\subset A$ means every element of $B$ is also an element of $A$. To confirm such a claim we check that nothing in $B$ escapes $A$; to refute it, a single escaping element is enough.

(a) $\mathbb{N}\subset \mathbb{Z}$: true. The natural numbers are $1,2,3,\dots$, and the integers are $\dots ,-2,-1,0,1,2,\dots$. Every natural number appears in the integer list, since the integers include all the positive whole numbers. So nothing in $\mathbb{N}$ escapes $\mathbb{Z}$.

(b) $\mathbb{Z}\subset \mathbb{N}$: false. For this we only need one counterexample. Take $-1$. It is an integer, so $-1\in \mathbb{Z}$. But the natural numbers begin at $1$ and go upward, so $-1\notin \mathbb{N}$. We have found an element of $\mathbb{Z}$ that is not in $\mathbb{N}$, so the claim fails. (The number $0$ would serve as a counterexample just as well.)

Parts (a) and (b) together make an important point: $B\subset A$ and $A\subset B$ are different statements. One can hold while the other fails. Indeed the section records the chain in one direction only,

and it should not be read backwards.

(c) $\mathbb{Q}\subset ℝ$: true. The section defines the real numbers as all rational numbers together with all irrational numbers. So every rational number is by construction a real number.

(d) $\sqrt{2}\in \mathbb{Q}$: false. This section proves that $\sqrt{2}$ cannot be written as $p/q$ with $p$ and $q$ integers, so it fails the definition of a rational number. Note this is a statement about membership ($\in$), not about subsets ($\subset$) — read the symbol carefully.

(e) $\mathbb{N}\subset ℝ$: true. We could argue directly, but it is quicker to chain the results we already have. By (a), everything in $\mathbb{N}$ is in $\mathbb{Z}$. Every integer is rational, since $n={\displaystyle \frac{n}{1}}$. And by (c), every rational is real. Following an element of $\mathbb{N}$ along this route, it ends up in $ℝ$. So $\mathbb{N}\subset ℝ$.

The rule the student is quoting is ${\displaystyle \frac{a}{a}}=1$, and it is a perfectly good rule — but only for $a\ne 0$. It comes from the multiplicative inverse property in the table of this section, which is stated as

with the condition $a\ne 0$ written right into it. The student has applied the rule at the one value where it is not allowed to be applied. Whenever a rule carries a condition, the first thing to do is check the condition.

Why the condition is there is worth spelling out. Suppose, for the sake of argument, that ${\displaystyle \frac{0}{0}}$ did equal some number $b$. Division and multiplication undo each other, so this would mean

Now use the fact proved in this section, that $0\cdot a=0$ for every $a$. It tells us $0\cdot b=0$ no matter what $b$ is. So the equation $0=0\cdot b$ is satisfied by every number $b$:

• $b=1$ works, since $0=0\cdot 1$;
• $b=7$ works, since $0=0\cdot 7$;
• $b=-53$ works too.

The student picked out $1$ from this crowd, but nothing whatsoever singles $1$ out. The value $7$ has just as much claim. Since a symbol like ${\displaystyle \frac{0}{0}}$ must stand for one definite number if it is to stand for anything, and here no one number can be chosen over the others, the symbol is given no meaning at all.

This is why the section says flatly that division by zero is not a valid operation and that $a/0$ has no meaning for any $a$ — the case $a=0$ included.

It is worth noticing how this differs from ${\displaystyle \frac{5}{0}}$. There the trouble is the opposite: $5=0\cdot b$ would require $5=0$, so no value of $b$ works. One case has no candidates and the other has too many, and neither can be assigned a value.

Sometimes true.

The answer is sometimes. To establish that, we need one example where the sum is irrational and one where it is rational.

An example where the sum is irrational. Take $\sqrt{2}$ and $\sqrt{2}$, both irrational by this section. Their sum is

using the multiplicative identity and then the distributive property from the table.

Is $2\sqrt{2}$ irrational? Suppose it were rational. Then we could write

for some integers $p$ and $q$ with $q\ne 0$. Dividing both sides by $2$,

But $p$ is an integer and $2q$ is an integer, and $2q\ne 0$. So this would make $\sqrt{2}$ a ratio of two integers, that is, rational — contradicting what this section proved. Hence $2\sqrt{2}$ is irrational, and here the sum of two irrationals is irrational.

An example where the sum is rational. Take $\sqrt{2}$ and $-\sqrt{2}$.

First check that $-\sqrt{2}$ really is irrational, since otherwise the example does not count. Suppose $-\sqrt{2}$ were rational, say $-\sqrt{2}={\displaystyle \frac{p}{q}}$. Multiplying both sides by $-1$ gives $\sqrt{2}={\displaystyle \frac{-p}{q}}$, and $-p$ is an integer, so $\sqrt{2}$ would be rational — again a contradiction. So $-\sqrt{2}$ is irrational.

Now add them. By the additive inverse property,

and $0={\displaystyle \frac{0}{1}}$ is rational. So here the sum of two irrationals is rational.

Conclusion. Both outcomes are possible, so the statement is sometimes true.

The general lesson is that the irrational numbers, unlike the sets $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, are not tidy under addition: adding two of them can land you back inside $\mathbb{Q}$. Never assume a property of individual numbers automatically survives an operation — test it.

Always true.

The answer is always. Since the claim covers every pair of rational numbers, examples cannot settle it; we must argue from the definition.

Setting up. Let $r$ and $s$ be any two rational numbers. By the definition in this section, each can be written as a ratio of integers, so there are integers $p,q,m,n$ with

and with $q\ne 0$ and $n\ne 0$, since a zero denominator would have no meaning.

Multiplying. Multiplying fractions means multiplying the tops together and the bottoms together:

Checking this is a legitimate rational number. Two things must be confirmed.

First, the top and bottom must be integers. The number $pm$ is a product of two integers, and multiplying two integers always gives an integer. The same applies to $qn$. So both $pm$ and $qn$ are integers.

Second, the denominator must not be zero. Here we can use the Zero-Product Property from this section. Suppose $qn=0$. The property says that then $q=0$ or $n=0$. But we arranged at the start that neither $q$ nor $n$ is zero. So $qn=0$ is impossible, and therefore $qn\ne 0$.

Conclusion. We have written $rs$ as a ratio of two integers with non-zero denominator, so $rs$ is rational. Since $r$ and $s$ were arbitrary rational numbers, this holds in every case, and the statement is always true.

A numerical illustration of the argument: with $r={\displaystyle \frac{3}{4}}$ and $s={\displaystyle \frac{-2}{5}}$,

which is again a ratio of two integers, as promised.

Contrast this with the previous exercise. Rational numbers stay rational when multiplied, but irrational numbers need not stay irrational when added. Whether a property survives an operation has to be checked case by case.

Either $a=b$ or $a=-b$; and $a=5$ or $a=-5$.

First part. The left-hand side is a product of the two numbers $a-b$ and $a+b$, and we are told it equals $0$. The Zero-Product Property says at least one factor must be zero, so there are two cases.

Case 1: $a-b=0$. Adding $b$ to both sides gives

Case 2: $a+b=0$. Subtracting $b$ from both sides gives

So the conclusion is: $a=b$ or $a=-b$. In words, $a$ and $b$ are either the same number or opposite numbers.

Second part. We want all real $a$ with

Since $25=5^2$, we can rewrite this as

and then subtract $5^2$ from both sides to get everything on one side:

Now recall the identity from the first section of this chapter,

Reading it from right to left with $b=5$,

so our equation becomes

This is exactly the situation of the first part, with $b=5$. Therefore $a=5$ or $a=-5$.

Check both. If $a=5$ then $a^2=25$. If $a=-5$ then $a^2=(-5)(-5)=25$, since a negative times a negative is positive. Both work.

The point to carry away is that the second solution is easy to lose. Many students see $a^2=25$ and write only $a=5$. Moving everything to one side and using the Zero-Product Property produces both cases automatically, so nothing gets forgotten.

First be clear about what has to be shown. The symbol $-a$ means the additive inverse of $a$, that is, the number which added to $a$ gives $0$. There is only one such number for each $a$. So to prove $(-1)\cdot a=-a$, it is enough to show that the number $(-1)\cdot a$ has this property: that adding it to $a$ gives $0$.

So let us compute $a+(-1)\cdot a$, justifying every step.

Step 1. Rewrite the first $a$ using the multiplicative identity property, $a\cdot 1=a$, read from right to left:

This looks like it is making things worse, but it is the key move: it puts both terms into the shape "something times $a$", which is what lets us use the distributive property next.

Step 2. Apply the distributive property. The table states it as $a(b+c)=ab+ac$; combined with the commutative property of multiplication this is the same as $ba+ca=(b+c)a$. Using it with $b=1$ and $c=-1$:

Step 3. Simplify inside the bracket using the additive inverse property, which says $1+(-1)=0$:

Step 4. Use the fact proved in this section that the product of any number and $0$ is $0$:

Putting the steps together. Following the chain from beginning to end,

Conclusion. The number $(-1)\cdot a$ added to $a$ gives $0$, which is exactly the defining property of the additive inverse of $a$. Hence

A check with numbers, to see the result is sensible: with $a=7$ we get $(-1)\cdot 7=-7$, and indeed $7+(-7)=0$. With $a=-3$ we get $(-1)\cdot (-3)=-(-3)=3$, which is the familiar rule that a negative times a negative is positive — and notice that it has now been derived rather than simply remembered.