Which numbers do measurements and calculations need?

A square panel with sides of 1 m needs a diagonal brace. The geometric model gives a length of √2 m, while a calculator displays 1.41421356… m. Can we simply write “equals 1.414 m”? No: the shorter value is already a different length. Understanding the difference requires both number sets and a convention about accuracy.

From counting to parts of a unit

To count panels, natural numbers are enough: 1, 2, 3, … . In this course we denote them by ℕ and start at 1; some other sources include 0 in ℕ. That is a convention about notation.

To record an empty stock or a debt in whole units, we add 0 and negative numbers. These give the integers, ℤ: …, −2, −1, 0, 1, 2, … . But integers cannot express half a metre.

A rational number can be written as a fraction $p/q$ , where $p$ and $q$ are integers and $q\ne0$ . For instance, $0.5=1/2$ , $-2=-2/1$ and $0=0/1$ . Thus integers are rational too. Division by zero does not create a new member of this set: it is undefined.

A rational number has a terminating or eventually repeating decimal representation: $1/8=0.125$ , whereas $1/3=0.333\ldots$ . An infinite decimal expansion alone does not make a number irrational. Definitions and classification in OpenStax, §1.1.

A length no fraction can express

The symbol √2 denotes the nonnegative number whose square is 2. In a square with 1 m sides, that is the numerical value of the diagonal in metres: the sum of the squares of the two sides equals the square of the diagonal. Here we assume an ideal square; real cutting errors are a separate issue.

√2 cannot be expressed as a ratio of two integers. Such real numbers are called irrational; their decimal expansions never terminate or become periodic. Other examples are π and −√2. In contrast, √9 = 3 is rational: a radical sign alone tells us nothing decisive.

Rational and irrational numbers together form the real numbers, ℝ. Each real number corresponds to one point on the number line. Farther right means larger: −2 < −0.5 < 0 < 1 < √2 < 2. Zero is neither positive nor negative.

ℝ · Real

ℚ · Rational1/3, −0.5

ℤ · Integers0, −2

ℕ · Natural1, 2, 3, …

ℝ ∖ ℚIrrational√2, π, −√2

Each inner box is part of the outer box. All real numbers share one number line; only a few points are shown. Box areas are schematic; they do not represent the number of elements.

A set groups numbers together

Here a set is a collection of objects, and an element is one of those objects. We read $2\in\mathbb N$ as “2 belongs to ℕ”. If every element of one set belongs to another, the first is a subset of the second. Therefore $\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R$ : calling 2 rational does not stop it from belonging to ℕ.

The union collects elements from either set; the intersection keeps only shared elements. For $A=\{1,2\}$ and $B=\{2,3\}$ , the union contains 1, 2, 3, while the intersection contains only 2. Curly braces enclose a set’s listed elements. A set with no elements is the empty set, ∅. This language will help us describe all solutions of a problem, including the absence of any solution.

What rounding actually loses

Check two nearby lengths by squaring them:

1.414^2=1.999396<2<2.002225=1.415^2.

Both numbers are positive, so $1.414<\sqrt2<1.415$ . Writing √2 ≈ 1.414 indicates an approximation; √2 = 1.414 would be false. The difference between the exact value and this approximation is about 0.000214 m, or 0.214 mm.

Rounding to the nearest thousandth leaves a distance of at most half a step from the chosen number: 0.0005 m, or 0.5 mm. This bounds the rounding error of the exact number; it guarantees nothing about a tape measure or the finished panel. Choosing the nearest representable number and the distinction from chopping digits are explained in NIST DLMF, §3.1(i), Rounding.

Branch · ProofYou can skip this proof and continue reading.

Why √2 cannot be a fraction

Suppose $\sqrt2=p/q$ , where the fraction is already in lowest terms and $p,q$ are positive integers. Then $p^2=2q^2$ , so $p^2$ is even. An odd number has an odd square, so $p$ must also be even: $p=2k$ . Substitution gives $4k^2=2q^2$ , hence $q^2=2k^2$ . Thus $q$ is even too. Both numbers are divisible by 2, contradicting the fraction being in lowest terms. Therefore no such fraction exists.

You can choose a suitable approximation when ordering the brace, while retaining √2 until the final step of a mathematical calculation. Consider −4, 0, 7/8, √9, √2: the first four are rational, the last is irrational, and all five lie on the same real number line.