Explain why sums and products of rationals are rational, and when a sum or product involving an irrational number is guaranteed to be irrational

Learning goal: Explain why the sum or product of two rational numbers is rational, why the sum of a rational number and an irrational number is irrational, and why the product of a nonzero rational number and an irrational number is irrational, using the definitions and proof by contradiction.

The definitions that carry the whole argument. A rational number is a number that can be written as p/q, where p and q are integers and q ≠ 0. An irrational number is a real number that cannot be written that way. Every real number is one or the other. The proofs in this lesson use nothing beyond these definitions and the fact that integers are closed under addition, subtraction and multiplication: the sum, difference or product of two integers is always an integer. That closure fact is what students often use without noticing, and it deserves to be said out loud.

Why the sum of two rational numbers is rational. Let the two numbers be a/b and c/d with b ≠ 0 and d ≠ 0. Then a/b + c/d = (ad + bc)/(bd). The numerator ad + bc is an integer, because products and sums of integers are integers. The denominator bd is an integer and it is not zero, because a product of two nonzero integers is nonzero. So the sum is an integer over a nonzero integer, which is exactly the definition of rational. A concrete instance: 3/4 + 5/6 = (3·6 + 4·5)/(4·6) = (18 + 20)/24 = 38/24 = 19/12. The formula does not need lowest terms; it only needs to exhibit some integer over some nonzero integer.

Why the product of two rational numbers is rational. (a/b)(c/d) = (ac)/(bd). Again ac is an integer and bd is a nonzero integer, so the product is rational. Instance: (3/4)(5/6) = 15/24 = 5/8. Note the pattern of both proofs: start from the definition, perform the operation, and show that the result still has the required shape. The set of rational numbers is said to be closed under addition and multiplication.

Why a rational plus an irrational is irrational — proof by contradiction. Let r be rational and x be irrational, and suppose, to reach a contradiction, that the sum s = r + x is rational. Then x = s − r. Both s and r are rational, and the difference of two rationals is rational by the same argument as for sums (a/b − c/d = (ad − bc)/(bd)). So x would be rational. But x was assumed irrational, and a number cannot be both. The supposition was false, so r + x is irrational. Concrete instance: 2 + √3 is irrational, because if it were some rational s, then √3 = s − 2 would be rational, and √3 is not. This proof style, assume the opposite and derive an impossibility, is the standard tool for "cannot be written as" statements, because there is no direct way to check every possible fraction.

Why a nonzero rational times an irrational is irrational. Let r be a nonzero rational and x irrational, and suppose the product p = rx is rational. Since r ≠ 0, we may divide: x = p/r. The quotient of two rationals with a nonzero divisor is rational: (a/b) ÷ (c/d) = (ad)/(bc), and bc ≠ 0 when both b and c are nonzero. So x would be rational, a contradiction. Therefore rx is irrational. Instance: 5√2 is irrational, and so is (1/2)π. The word nonzero is not decoration. If r = 0, then 0 · √2 = 0, which is rational; the division step x = p/r is impossible when r = 0, and the theorem is simply false there.

Where the argument does not reach. The three statements say nothing about two irrationals combined with each other, and students routinely over-extend them. The sum of two irrationals can be rational: √2 + (−√2) = 0. The product of two irrationals can be rational: √2 · √8 = √16 = 4. And (2 + √3) + (2 − √3) = 4 is rational even though each addend is irrational. None of these contradict the lesson, because the theorems have a rational number as one of the two inputs. The correct general statement about two irrationals is that the result may be rational or irrational, and you must examine the specific numbers.

A worked model inside the lesson. Decide whether 0.272727… + 1/4 is rational or irrational and justify it. The repeating decimal is rational: with x = 0.2727…, 100x = 27.2727…, so 99x = 27 and x = 27/99 = 3/11. Then 3/11 + 1/4 = (3·4 + 11·1)/(11·4) = (12 + 11)/44 = 23/44, an integer over a nonzero integer. The sum is rational because both addends are rational, and the closure argument guarantees it before any arithmetic is done; the arithmetic just names the fraction.

What you may take as known. These proofs need a supply of irrational numbers to start from. In this course you may take √2, √3, √5 (and any square root of a whole number that is not a perfect square) and π as known irrationals. The classic proof that √2 is irrational is given in the worked examples; it uses the same contradiction pattern and the fact that a square of an integer is even only when the integer is even.

Why this matters beyond the proofs. When you solve x² − 4x + 1 = 0 and get x = 2 ± √3, the theorems tell you immediately that both roots are irrational: a rational (2) plus a nonzero rational multiple (±1) of an irrational (√3). When you compute the circumference of a circle with a rational diameter, you know the answer is irrational because it is a nonzero rational times π. The theorems let you classify numbers without a calculator and without hoping a decimal display will "show" a pattern, which it never can.

Worked closure proof for a difference: Show that the difference of two rational numbers is rational. Write them as a/b and c/d with b, d ≠ 0. Then a/b − c/d = (ad − bc)/(bd). The numerator ad − bc is an integer because products and differences of integers are integers; the denominator bd is a nonzero integer because a product of nonzero integers is nonzero. Hence the difference is rational. Instance: 5/6 − 3/4 = (20 − 18)/24 = 2/24 = 1/12. This difference result is the step the "rational plus irrational" proof relies on.

Worked contradiction for a sum: Prove that 3 − √7 is irrational. Suppose it is rational and call it s. Then √7 = 3 − s. The right side is the difference of two rationals, so it is rational. But √7 is irrational (7 is not a perfect square). This contradiction means the supposition was wrong, so 3 − √7 is irrational. Numerically 3 − √7 ≈ 0.354, and no decimal display could have decided this; only the argument does.

Worked contradiction for a product, with the nonzero condition: Prove that (2/3)π is irrational, and explain why the same argument fails for 0 · π. Suppose (2/3)π = p for some rational p. Since 2/3 ≠ 0, divide: π = p ÷ (2/3) = (3/2)p, a rational number. That contradicts the known irrationality of π, so (2/3)π is irrational. For 0 · π the division by the rational factor is impossible, and indeed 0 · π = 0 is rational; the theorem requires the rational factor to be nonzero.

Worked proof that √2 is irrational: Suppose √2 = p/q in lowest terms, with p and q integers and q ≠ 0. Squaring, 2 = p²/q², so p² = 2q². Then p² is even, and a square is even only when its root is even, so p = 2k for some integer k. Substituting, 4k² = 2q², so q² = 2k², making q² even and therefore q even. Now p and q are both even, contradicting lowest terms. So no such fraction exists and √2 is irrational.

Classification record: Make a table with columns number, rational or irrational, and the reason. Classify 7/3 + 2/5, (7/3)(2/5), 4 + √11, 4√11, 0 · √11, √3 · √12, (1 + √5) + (1 − √5), and 0.4545… + 1/2. For every rational result write the fraction of integers that proves it; for every irrational result name the theorem used and the known irrational it starts from; for the two-irrational cases state explicitly that the theorems in this lesson do not apply and show the computation that decides them.

Independent performance: Write a complete proof, in full sentences, that if r is rational and x is irrational then r + x is irrational, and a second complete proof that if r is a nonzero rational and x is irrational then rx is irrational. Each proof must state what is assumed, where the definition of rational is used, where closure of the integers is used, and where the contradiction appears. Then construct your own example of two irrational numbers whose sum is rational and two irrational numbers whose product is rational, and explain in one paragraph why those examples do not contradict either theorem. Success criteria: both proofs are by contradiction and name the exact step that would fail if r = 0 in the product case; every example is checked by computation; the paragraph correctly identifies that the theorems require one rational input.