Which number types survive arithmetic?

Try

A mixture uses 1/2 litre of one ingredient and 2/3 litre of another. The total, 7/6 litre, is rational because it is a ratio of integers with a nonzero denominator. This example suggests that rational numbers are closed under addition: adding any two of them stays in the same collection. A proof must explain “any,” not merely confirm this recipe.

An irrational number is a real number that cannot be expressed as an integer divided by a nonzero integer. A decimal that has merely been rounded does not establish irrationality. The exact number and its displayed approximation are different objects.

Predict

Predict whether adding a rational number to an irrational one can produce a rational result. Try to reason backward: if the result were rational, could subtracting the rational input recover an irrational number?

For multiplication, keep zero visible. A proposed rule saying “rational times irrational is irrational” fails when the rational factor is zero. Decide whether this is the only rational-factor exception, and explain why a division step in your proof would care about it.

Experiment

Use the known irrationality of √5 to complete this evidence table. The status column requires a reason, not an estimate of the decimal.

Operation	Exact result	Status and reason
1/2+2/3	7/6	Rational integer fraction
√5+(−√5)	0	Rational despite two irrational inputs
√5·√5	5	Rational despite two irrational inputs
2+√5	2+√5	Irrational by the subtraction argument below
0·√5	0	The zero exception

Construct another irrational-plus-irrational example whose result is irrational. You can use 2√5: its irrationality follows from the nonzero-rational-factor argument. This differs from claiming that every combination of irrational inputs behaves alike.

Understand

Write two arbitrary rational numbers as a/b and c/d, with a,b,c,d integers and b,d nonzero. Their sum is (ad+bc)/(bd), and their product is ac/(bd). The new numerators and denominator are integers, and bd is nonzero. This proves both closure statements. Replacing c by −c also proves closure under subtraction. A nonzero rational a/b has rational reciprocal b/a because then a is nonzero.

Now let r be rational and u irrational. Suppose r+u were rational, say q. Rearranging gives u=q−r, which is rational by the subtraction result. That contradicts the assumption on u, so r+u is irrational. This is a contradiction proof: assume the opposite outcome, use established facts, and identify the impossible conclusion.

For multiplication assume additionally r≠0. If ru=q were rational, then u=q/r would be rational because division by a nonzero rational is multiplication by its rational reciprocal. Again there is a contradiction. If r=0, the product is rational and the division step is forbidden. The exception is part of the theorem, not a technicality to omit.

Notice what has not been proved: no one rule classifies all sums or products of two irrational numbers. The table already supplies rational outcomes; expressions such as √5+√5 and √5(1+√5)=5+√5 supply irrational outcomes.

Apply

Write a short proof for one arbitrary pair before calculating your fresh numerical cases. Identify each closure fact you use. “Its decimal goes on forever” is insufficient: rational numbers such as 1/3 also have infinite decimal expansions, while a calculator only displays finitely many digits.

When supplying a counterexample involving two irrational inputs, justify both inputs. For example, knowing √5 is irrational implies 1+√5 is irrational by the mixed-sum theorem. That makes the product example above legitimate; simply labeling a complicated expression irrational would not.

Continue

Optional depth: √2 is irrational because a reduced fraction p/q with square 2 would satisfy p²=2q². An integer square is even only when its base is even, so p is even. Writing p=2k then forces q²=2k² and q even, contradicting the fraction being reduced. This proves one foundational example without relying on its decimal expansion.

The four general closure arguments and their exceptions address N-RN.B.3. Next, extend beyond real numbers with a new number i whose square is −1, while retaining distributivity and careful arithmetic.