Skip to content

Why each family of numbers was invented

Numbers · level 7, lesson 1 of 6

A question to start

Here are four equations. Try to solve each one.

\[x + 5 = 2 \qquad\quad 3x = 2 \qquad\quad x^2 = 2 \qquad\quad x^2 = -1\]

The first has the answer \(x = -3\), the second \(x = \tfrac23\), the third \(x = \sqrt2\) (or \(-\sqrt2\)). The fourth seems to have no answer at all.

But go back in time. To someone who only knew the counting numbers 1, 2, 3, ..., the first equation had no answer either: you cannot take 5 away from 2. To someone who only knew whole numbers, the second had no answer. To the ancient Greeks, who thought every length was a ratio of whole numbers, the third was a shock. Each "impossible" equation was made possible by inventing a bigger family of numbers. This lesson tells that story as one pattern, and shows why the fourth equation is the next step on the same ladder.

The idea

Being closed under an operation

A family of numbers is closed under an operation if doing that operation to any two members of the family always gives another member. "Closed" means you cannot escape the family by doing that operation.

  • The natural numbers \(\mathbb{N} = \{1, 2, 3, \ldots\}\) (the counting numbers) are closed under addition: add two counting numbers and you get a counting number. They are also closed under multiplication.
  • They are not closed under subtraction: \(2 - 5\) is not a counting number.
  • They are not closed under division: \(2 \div 3\) is not a counting number.

The letters are standard: \(\mathbb{N}\) for natural, \(\mathbb{Z}\) for the integers (from the German Zahlen, "numbers"), \(\mathbb{Q}\) for the rationals (from quotient), \(\mathbb{R}\) for the reals and \(\mathbb{C}\) for the complex numbers.

The whole history of number systems can be told with one sentence: each time a family was not closed under an operation people needed, they built a bigger family that was. And each time, they insisted that the old rules of arithmetic must still work in the new family.

Step 1: subtraction needs the negative numbers

To make subtraction always possible, we add zero and the negatives and get the integers

\[\mathbb{Z} = \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}\]

One neat way to think of a new integer is as a difference of two counting numbers that we cannot yet work out. The pair "2 take away 5" names the number \(-3\). So do "1 take away 4" and "10 take away 13". Different pairs can name the same integer, just as different fractions can name the same number.

Negative numbers were used long ago. The Chinese text The Nine Chapters on the Mathematical Art (about 2000 years old) used red and black counting rods for positive and negative amounts, and the Indian mathematician Brahmagupta wrote rules for them, as fortunes and debts, in the year 628. Yet in Europe they were distrusted for centuries; some writers in the 1500s called them "absurd numbers".

The old rules decide the new facts. Why is \((-1)\times(-1) = 1\)? Nobody voted for it. It is forced, if we want the distributive law \(a(b + c) = ab + ac\) to keep working:

\[0 = (-1)\times 0 = (-1)\times\big(1 + (-1)\big) = (-1)\times 1 + (-1)\times(-1) = -1 + (-1)\times(-1)\]

The only number that adds to \(-1\) to give 0 is 1, so \((-1)\times(-1) = 1\). This idea, "keep the old rules and see what they force", is how every later family is built too.

The integers are closed under addition, subtraction and multiplication. They are still not closed under division: \(2 \div 3\) is not an integer.

Step 2: division needs the fractions

The rational numbers \(\mathbb{Q}\) are all the numbers \(\dfrac{p}{q}\) where \(p\) and \(q\) are integers and \(q \ne 0\). Again a new number is a pair of old ones, and again different pairs can name the same number: \(\dfrac{2}{3} = \dfrac{4}{6} = \dfrac{-10}{-15}\). The test for "same number" is cross-multiplying:

\[\frac{p}{q} = \frac{r}{s} \quad\text{exactly when}\quad ps = qr\]

The rationals are closed under all four operations (apart from dividing by zero, which no family allows). For example, the sum of two rationals is rational, because

\[\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}\]

and the top and bottom are integers, with \(bd \ne 0\) because neither \(b\) nor \(d\) is 0. A family closed under all four operations, with the usual rules, has a name in university mathematics: a field. \(\mathbb{Q}\) is the smallest field that contains the counting numbers.

Step 3: roots need the irrational numbers

The rationals are closed under \(+\), \(-\), \(\times\) and \(\div\), but not under taking square roots. In level 6 you proved that \(\sqrt2\) is not any fraction. So the equation \(x^2 = 2\), which asks for the side of a square of area 2, has no rational answer.

There is a second, deeper way to see the gap. The decimals

\[1, \quad 1.4, \quad 1.41, \quad 1.414, \quad 1.4142, \quad \ldots\]

are all rational, and their squares get closer and closer to 2. They are heading somewhere, but the place they are heading is not a rational number. The rational number line is full of holes like this. The real numbers \(\mathbb{R}\) fill every hole: every endless decimal is a real number. This "no holes" property is called completeness, and it is what calculus is built on.

Note something new here. The first two steps were driven by equations (\(x + 5 = 2\) and \(3x = 2\)). The real numbers add far more than roots of equations. The number \(\pi\) is not a solution of any polynomial equation with whole number coefficients (this was proved by Ferdinand von Lindemann in 1882), yet it is a perfectly good real number. The reals are about filling gaps, not only about solving equations.

Step 4: \(x^2 = -1\) needs something new

Now the real numbers. They are closed under \(+\), \(-\), \(\times\), \(\div\), and every positive real has a square root. But no real number has a negative square.

Why not? Take any real number \(x\).

  • If \(x \ge 0\), then \(x \times x\) is a product of two numbers that are not negative, so \(x^2 \ge 0\).
  • If \(x < 0\), then \(-x > 0\), and \(x^2 = (-x)\times(-x)\) (two minus signs cancel, by step 1), which is positive.

Either way \(x^2 \ge 0\). So \(x^2 = -1\) has no real solution, and neither does \(x^2 = -4\) or \(x^2 + 2x + 5 = 0\) (more on that one in lesson 5).

The same move as before solves it: invent a number \(i\) with \(i^2 = -1\), keep all the old rules, and see what they force. That gives the complex numbers \(\mathbb{C}\), which are the subject of the rest of this level and all of level 8.

The families sit inside each other

ℕ1, 2, 3 ℤ0, −4 ℚ2/3, −0.5 ℝ√2, π ℂi, 3 − 2i
\[\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}\]

The symbol \(\subset\) means "is contained in". Every counting number is an integer, every integer is a fraction (\(5 = \tfrac51\)), every fraction is a real number, and (as you will see) every real number is a complex number.

What we gain and what we lose

Each step gains an operation. But each step also loses something that the smaller family had.

Step We gain We lose
\(\mathbb{N} \to \mathbb{Z}\) subtraction always works a smallest number (there is no smallest integer)
\(\mathbb{Z} \to \mathbb{Q}\) division always works (not by 0) a "next" number (there is no next fraction after \(\tfrac12\))
\(\mathbb{Q} \to \mathbb{R}\) no holes; limits exist being listable one by one (see Infinity)
\(\mathbb{R} \to \mathbb{C}\) every polynomial equation has a root order: no "bigger" or "smaller" that fits the rules

The last line is a surprise worth checking. Could we decide that \(i > 0\)? Then multiplying both sides of \(i > 0\) by the positive number \(i\) would give \(i^2 > 0\), that is \(-1 > 0\). False. Could we decide \(i < 0\)? Then \(-i > 0\), and \((-i)(-i) > 0\) gives \(i^2 > 0\) again, the same contradiction. So there is no way to order the complex numbers that keeps the usual rules for inequalities. Level 8's last lesson shows that the next step after \(\mathbb{C}\) loses even more.

Worked examples

1. Which family do you need? For each equation, find the smallest family from \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\), \(\mathbb{C}\) that has a solution.

  • \(x + 7 = 3\). The solution is \(x = -4\), an integer: \(\mathbb{Z}\).
  • \(5x = -2\). The solution is \(x = -\tfrac25\): \(\mathbb{Q}\).
  • \(x^3 = -8\). The solution \(x = -2\) is an integer: \(\mathbb{Z}\). (Odd powers of negatives are negative, so cube roots of negative numbers are real.)
  • \(x^2 = 5\). The solutions are \(\pm\sqrt5\), irrational: \(\mathbb{R}\).
  • \(2x^2 = 8\). Then \(x^2 = 4\) and \(x = \pm 2\). The answer \(x = 2\) is a counting number, so \(\mathbb{N}\) already contains a solution.
  • \(x^2 + 4 = 0\). Then \(x^2 = -4\), which no real number solves: \(\mathbb{C}\) (the solutions are \(\pm 2i\)).

2. Closed or not? Is the set of odd integers closed under addition? Under multiplication?

  1. Addition: \(3 + 5 = 8\), which is even. One counterexample is enough: not closed.
  2. Multiplication: write two odd numbers as \(2m + 1\) and \(2n + 1\). Their product is \(4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1\), which is odd. This works for every pair, so closed.

Notice the difference: to show "not closed" you need one example; to show "closed" you need an argument that covers every case.

3. Are the irrational numbers closed under multiplication? No: \(\sqrt2\) is irrational, but \(\sqrt2 \times \sqrt2 = 2\) is rational. Under addition? Also no: \(\sqrt2 + (-\sqrt2) = 0\). The irrationals are not a "family" in the sense of this lesson at all; they are what is left over when you take \(\mathbb{Q}\) out of \(\mathbb{R}\).

4. What the old rules force for \(i\). Suppose \(i^2 = -1\) and the usual rules still hold. What must \((2i)^2\) and \((1 + i)^2\) be?

  1. \((2i)^2 = 2^2 \times i^2 = 4 \times (-1) = -4\). So \(2i\) is a square root of \(-4\).
  2. \((1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i\). So \(1 + i\) is a square root of \(2i\), a number that is not even on the real line.

We never had to decide these answers; the rules decided them. That is the method of the whole level.

Try it

  1. Is \(\mathbb{N}\) closed under the operation "take the larger of the two numbers"? Under "subtract the smaller from the larger"?
  2. Find the smallest family (\(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\) or \(\mathbb{C}\)) containing a solution of: (a) \(4x = 10\); (b) \(x + 10 = 4\); (c) \(x^2 = 49\); (d) \(x^2 = 0.49\); (e) \(x^2 = 0.4\); (f) \(x^2 + 9 = 0\).
  3. Is the set of even integers closed under addition? Under division (when the bottom is not 0)?
  4. Show that the product of two rational numbers is rational.
  5. Use the distributive law to show that \((-2)\times(-3) = 6\), in the same way as the proof of \((-1)\times(-1) = 1\).
  6. The square numbers \(\{1, 4, 9, 16, \ldots\}\): closed under multiplication? Under addition?
  7. Show that if \(x\) is rational and \(y\) is irrational, then \(x + y\) is irrational. (Hint: suppose \(x + y\) were rational and look at \((x + y) - x\).)
  8. Explain why no ordering of \(\mathbb{C}\) can have \(2i > 0\), using the same idea as the lesson's argument for \(i\).
Answers
  1. "Take the larger": yes, the larger of two counting numbers is one of them, so it is a counting number. "Subtract the smaller from the larger": no, because \(3 - 3 = 0\) and 0 is not in \(\mathbb{N} = \{1, 2, 3, \ldots\}\). (If you count 0 as natural, it is closed.)
  2. (a) \(x = \tfrac{10}{4} = \tfrac52\): \(\mathbb{Q}\). (b) \(x = -6\): \(\mathbb{Z}\). (c) \(x = 7\): \(\mathbb{N}\). (d) \(x = 0.7 = \tfrac{7}{10}\): \(\mathbb{Q}\). (e) \(x = \sqrt{0.4} = \sqrt{\tfrac{2}{5}}\), which is irrational (\(\sqrt{0.4} = \tfrac{\sqrt{10}}{5}\) and \(\sqrt{10}\) is irrational): \(\mathbb{R}\). (f) \(x = \pm 3i\): \(\mathbb{C}\).
  3. Addition: \(2m + 2n = 2(m + n)\), even, so closed. Division: \(2 \div 4 = \tfrac12\), not an integer, so not closed.
  4. \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}\). The top \(ac\) and the bottom \(bd\) are integers, and \(bd \ne 0\) because \(b \ne 0\) and \(d \ne 0\). So the product is rational.
  5. \(0 = (-2)\times 0 = (-2)\times\big(3 + (-3)\big) = (-2)\times 3 + (-2)\times(-3) = -6 + (-2)\times(-3)\). The only number that adds to \(-6\) to give 0 is 6, so \((-2)\times(-3) = 6\).
  6. Multiplication: \(a^2 \times b^2 = (ab)^2\), a square, so closed. Addition: \(1 + 4 = 5\) is not a square, so not closed.
  7. Suppose \(x + y = r\) with \(r\) rational. Then \(y = r - x\). The difference of two rationals is rational (the rationals are closed under subtraction), so \(y\) would be rational. But \(y\) is irrational, a contradiction. So \(x + y\) is irrational.
  8. Suppose \(2i > 0\). Multiplying both sides by the positive number \(2i\) keeps the inequality, giving \((2i)^2 > 0\), that is \(-4 > 0\). False. If instead \(2i < 0\), then \(-2i > 0\) and \((-2i)^2 > 0\) gives \(-4 > 0\) again. So neither works.

Common mistake

Thinking "imaginary" means "fake". The names "negative", "irrational" and "imaginary" all come from times when people distrusted the new numbers. Each new family is built from the old one by a precise rule and obeys the same laws of arithmetic. Complex numbers are as solid as fractions; engineers use them every day for electric circuits and signals.

Proving "closed" with examples. \(1 \times 3 = 3\), \(3 \times 5 = 15\) and \(5 \times 7 = 35\) are all odd, but three examples prove nothing about all odd numbers. Closure needs an argument for every pair, such as the \(2m + 1\) argument in worked example 2. One counterexample, though, is enough to show a family is not closed.

Where it leads