Classify real numbers as rational or irrational, and reason about the rationality of sums and products of rational and irrational numbers (A.NR, Numerical Reasoning).

What this topic is asking

The Numerical Reasoning (A.NR) strand opens the course by sorting the real numbers into rational and irrational, and then asking you to reason about what happens when you combine them. On the Georgia Milestones EOC this shows up as quick classification items and as short justification (constructed-response or drag-and-drop) items where you explain why a sum or product lands in one set or the other. It is foundational: the same rational-versus-irrational distinction returns when you simplify radicals and when you decide whether a quadratic has nice or messy solutions.

Rational numbers

A rational number is any number that can be written as $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. This set is larger than it first looks. It includes:

• All integers, since $7 = \frac{7}{1}$.
• All terminating decimals, since $0.25 = \frac{1}{4}$.
• All repeating decimals, since $0.\overline{3} = \frac{1}{3}$ and $0.\overline{45} = \frac{45}{99} = \frac{5}{11}$.

The decimal test is the fastest classifier on the EOC: if the decimal stops or settles into a repeating block, the number is rational.

Irrational numbers

An irrational number cannot be written as a ratio of integers. Its decimal expansion is non-terminating and non-repeating. The two families you will meet most are:

• Roots of non-perfect powers: $\sqrt{2}$, $\sqrt{3}$, $\sqrt{20}$, $\sqrt[3]{5}$. A square root is rational only when the radicand is a perfect square ($\sqrt{16} = 4$, $\sqrt{49} = 7$), and a cube root is rational only when the radicand is a perfect cube ($\sqrt[3]{27} = 3$).
• Special constants like $\pi$ and $e$.

Reasoning about sums and products

The EOC rewards knowing the closure rules, not just memorizing examples.

• Rational $+$ rational $=$ rational. $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. The integers are closed under addition and multiplication, so their ratios are too.
• Rational $\times$ rational $=$ rational. $\frac{2}{3} \cdot \frac{3}{5} = \frac{2}{5}$.
• Rational $+$ irrational $=$ irrational. $5 + \sqrt{3}$ is irrational. If the sum were rational, subtracting the rational part would force the irrational part to be rational, which is impossible.
• Nonzero rational $\times$ irrational $=$ irrational. $3\sqrt{2}$ is irrational. The "nonzero" matters: $0 \cdot \sqrt{2} = 0$ is rational.

Two irrational numbers can combine to give either set, so there is no fixed rule there: $\sqrt{2} + (-\sqrt{2}) = 0$ (rational) and $\sqrt{2} \cdot \sqrt{2} = 2$ (rational), but $\sqrt{2} + \sqrt{3}$ is irrational. This is exactly the kind of distinction a justification item probes.

How the Milestones examines this topic

• Multiple choice. "Which number is irrational?" with a perfect-square root and a repeating decimal among the distractors.
• Drag and drop. Sort a list of numbers into rational and irrational bins, or match each combination to its result type.
• Constructed response. State and justify whether a given sum or product is rational or irrational; full credit needs the general reason plus the specific number.

Why the closure reasoning works

The justification items reward an argument by contradiction, which is worth internalizing. To see why rational $+$ irrational is always irrational, suppose the sum $r + i$ were some rational number $s$. Rearranging gives $i = s - r$. The right side is a difference of two rational numbers, and rationals are closed under subtraction, so $s - r$ is rational, which would make $i$ rational. That contradicts the assumption that $i$ is irrational, so the sum could not have been rational in the first place. The same one-line argument, with multiplication and a nonzero factor, explains why a nonzero rational times an irrational stays irrational, and recognizing the structure of the argument is faster than testing examples.

Connecting to the rest of the course

This classification is not trivia. When you simplify a radical, you are pulling the rational part out of an irrational number, writing $\sqrt{20} = 2\sqrt{5}$ as a rational coefficient times an irrational root. When you solve a quadratic, the discriminant tells you whether the solutions are rational (a perfect-square discriminant) or irrational (a non-perfect-square discriminant). So the rational-versus-irrational lens you build here is the same lens you use to predict whether later answers will be clean integers or exact radicals.

Try this

Q1. Classify $\sqrt{45}$, $\frac{7}{8}$, and $0.\overline{6}$ as rational or irrational. [1 point]

• Cue. $\sqrt{45} = 3\sqrt{5}$ irrational; $\frac{7}{8}$ rational; $0.\overline{6} = \frac{2}{3}$ rational.

Q2. Is $\sqrt{5} \cdot \sqrt{5}$ rational or irrational? Explain. [1 point]

• Cue. $\sqrt{5} \cdot \sqrt{5} = 5$, which is rational, so two irrationals can multiply to a rational.