Back to Exercise: Explain operations on rational and irrational numbers

Exercises: Explain Why Sums and Products of Rational and Irrational Numbers Follow Specific Rules

Work through each section in order. For explanation problems, write in complete sentences and include the reasoning (proof template, counterexample, or construction). Classification alone is not sufficient — explain why.

Grade 9·21 problems·~32 min·Common Core Math - HS Number and Quantity·standard·hsn-rn-b-3

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Warm-Up: Classifying Real Numbers

Classify each number and explain your reasoning. Simplify any radicals before classifying.

Which of the following is an irrational number?

$\sqrt{25}$

$0.121212\ldots$

$\sqrt{7}$

$-\frac{3}{4}$

Which of the following is rational?

$\sqrt{11}$

$\sqrt{36}$

$\sqrt{2}$

$\pi$

State the definition of a rational number. Then explain what makes a number irrational. Give one example of each.

Fluency Practice

For each problem, classify the expression and provide the required argument or proof.

Classify $5 + \sqrt{9}$ . (Simplify the radical first.)

Irrational, because it is a sum involving a square root.

Rational, because $\sqrt{9} = 3$ and $5 + 3 = 8$ , which is rational.

Irrational, because adding 5 to a square root always gives an irrational number.

Rational, but only because the rational part (5) is a whole number.

Use the algebraic construction to explain why the sum of two rational numbers is always rational. Your argument must work for ALL pairs of rational numbers, not just a specific example. Let $r = \frac{p}{q}$ and $s = \frac{m}{n}$ be any two rational numbers.

Prove by contradiction that $4 + \sqrt{3}$ is irrational. Use the proof-by-contradiction template: (1) state what you want to prove, (2) assume the OPPOSITE, (3) derive a contradiction, (4) state your conclusion.

Using the same proof-by-contradiction template, prove that $\frac{1}{2} + \sqrt{5}$ is irrational.

Prove by contradiction that $3 \cdot \sqrt{2}$ is irrational. Then explain specifically why the proof requires the factor 3 to be nonzero — what goes wrong if we try to apply the same argument to $0 \cdot \sqrt{2}$ ?

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