Lesson 1.1 · Unit 1 · N-RN.3
The real number system, and the properties every later step will cite
Algebra is arithmetic done with letters, and it only works because the letters obey the same rules numbers do. Those rules have names. Learning the names now is not bookkeeping: every justification you write for the rest of the course, and both of the argument writing tasks at the end of it, will point back to one of them.
- The sets nest inside one another. Natural numbers \( 1, 2, 3, \ldots \) sit inside whole numbers (add 0), which sit inside integers (add the negatives), which sit inside rationals, which together with the irrationals make the real numbers.
- A rational number is one that can be written as a ratio of two integers with a nonzero denominator. Every terminating decimal and every repeating decimal is rational, and so is every integer, since \( 7 = \frac{7}{1} \).
- An irrational number cannot be written that way. Its decimal expansion never terminates and never repeats. \( \pi \) and \( \sqrt{2} \) are the standard examples.
- To classify a number, simplify it first. \( \sqrt{49} \) is not irrational; it is 7. The form a number is written in does not decide which set it belongs to.
- The commutative properties say order does not matter for addition or multiplication: \( a + b = b + a \) and \( ab = ba \). Neither holds for subtraction or division.
- The associative properties say grouping does not matter for addition or multiplication: \( (a + b) + c = a + (b + c) \) and \( (ab)c = a(bc) \).
- The distributive property connects the two operations: \( a(b + c) = ab + ac \). It is the only property linking addition and multiplication, and it is the most used property in the whole course.
- Identities and inverses. Adding 0 changes nothing and multiplying by 1 changes nothing. Every number has an additive inverse summing to 0, and every number except 0 has a multiplicative inverse multiplying to 1.
Where students lose marks: assuming subtraction is commutative. \( 7 - 3 \) and \( 3 - 7 \) are not equal, and neither are \( 12 \div 4 \) and \( 4 \div 12 \). Only addition and multiplication commute, and forgetting that is what produces sign errors when terms are moved around.
The problem. Classify each of these into every set it belongs to: \( \sqrt{49} \), \( -\tfrac{3}{4} \), \( \sqrt{10} \), \( 0.\overline{36} \), \( \sqrt[3]{-8} \). Then name the property that justifies rewriting \( 4 + (6 + x) \) as \( (4 + 6) + x \).
Step one: simplify before classifying. This is the step students skip. \( \sqrt{49} = 7 \) and \( \sqrt[3]{-8} = -2 \), because \( (-2)^3 = -8 \). Both are now recognizable integers. \( \sqrt{10} \) does not simplify, since 10 is not a perfect square.
Step two: classify \( \sqrt{49} = 7 \). It is a counting number, so it is natural. Everything natural is whole, every whole number is an integer, and every integer is rational since \( 7 = \frac{7}{1} \). Every rational is real. So: natural, whole, integer, rational, real.
Step three: classify \( -\tfrac{3}{4} \). It is already a ratio of two integers with a nonzero denominator, so it is rational, and therefore real. It is not an integer, so it is not whole or natural either.
Step four: classify \( \sqrt{10} \). Ten lies between the perfect squares 9 and 16, so \( \sqrt{10} \) is between 3 and 4 and is not an integer. The square root of a whole number is either a whole number or irrational, with nothing in between, so \( \sqrt{10} \) is irrational, and real.
Step five: classify \( 0.\overline{36} \). The bar means the block 36 repeats forever. Every repeating decimal is rational, and this one equals \( \frac{36}{99} = \frac{4}{11} \). Check it: \( 4 \div 11 = 0.3636\ldots \) So: rational and real.
Step six: classify \( \sqrt[3]{-8} = -2 \). It is a negative integer, so it is an integer, rational and real. It is not whole and not natural, because both of those sets exclude negatives.
Step seven: name the property. Going from \( 4 + (6 + x) \) to \( (4 + 6) + x \) changes only the grouping. The order of the three terms is unchanged, so this is not commutativity. It is the associative property of addition. It is worth doing, because the left side cannot be simplified further while the right side becomes \( 10 + x \) immediately.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- List the number sets, from smallest to largest, that contain the natural numbers.
Show the full solution
Natural, whole, integer, rational, real
- Classify \( \sqrt{81} \) into every set it belongs to.
Show the full solution
\( \sqrt{81} = 9 \), a counting number. Natural, whole, integer, rational, real
- Classify \( -5 \) into every set it belongs to.
Show the full solution
Negative, so not whole or natural. It is an integer, and \( -5 = \frac{-5}{1} \). Integer, rational, real
- Name the property shown by \( 3(x + 5) = 3x + 15 \).
Show the full solution
The distributive property
- Name the property shown by \( 7 \cdot 4 = 4 \cdot 7 \).
Show the full solution
The commutative property of multiplication
- Is \( \sqrt{16} + \sqrt{2} \) rational or irrational? Justify your answer.
Show the full solution
Simplify first: \( \sqrt{16} = 4 \), which is rational, while \( \sqrt{2} \) is irrational. A rational plus an irrational is always irrational, because if the sum were rational you could subtract the rational part and be left with a rational value for \( \sqrt{2} \), which is impossible. Irrational
- Give a counterexample showing that subtraction is not associative.
Show the full solution
Test \( (10 - 4) - 3 \) against \( 10 - (4 - 3) \). The left gives \( 6 - 3 = 3 \). The right gives \( 10 - 1 = 9 \). Since \( 3 \neq 9 \), regrouping changed the result, so subtraction is not associative. One counterexample is enough to disprove a universal claim. \( (10 - 4) - 3 = 3 \) but \( 10 - (4 - 3) = 9 \)
- A student says every square root is irrational. Give two counterexamples and state the
rule correctly.
Show the full solution
\( \sqrt{25} = 5 \) and \( \sqrt{100} = 10 \), both rational. The correct rule is that the square root of a whole number is rational exactly when that number is a perfect square, and irrational otherwise. There is nothing in between, which is why simplifying before classifying matters. \( \sqrt{25} \) and \( \sqrt{100} \); a square root is rational when the radicand is a perfect square
- Name the property justifying each step:
\( 2(x + 3) + 4x = 2x + 6 + 4x = 2x + 4x + 6 = (2 + 4)x + 6 = 6x + 6 \).
Show the full solution
The first step distributes the 2 across the sum, which is the distributive property. The second moves the 6 past \( 4x \), changing the order of terms, which is the commutative property of addition. The third groups the two \( x \) terms and factors out the \( x \), which is the distributive property again, read right to left. The last is arithmetic. Distributive, commutative (addition), distributive, then arithmetic
- Explain why 0 has no multiplicative inverse, and why every other real number does.
Show the full solution
The multiplicative inverse of \( a \) is the number \( \frac{1}{a} \) satisfying \( a \cdot \frac{1}{a} = 1 \). For any nonzero \( a \) that number exists and is a real number. For \( a = 0 \), the requirement would be a number \( b \) with \( 0 \cdot b = 1 \). But zero times anything is zero, so no such \( b \) exists, and the requirement is not merely hard to satisfy but impossible. This is exactly why division by zero is undefined: dividing by \( a \) means multiplying by its multiplicative inverse, and zero has none. No number multiplied by 0 gives 1, since zero times anything is zero, which is why division by zero is undefined