Real Numbers

How to use this guide

Attempt before reading each worked solution. Then hide the guide and retrieve the method on a later day.

1. Day 0: Do the cold start. Read §§2.1–2.2 and reconstruct the number-system map and the definitions of supremum and completeness.
2. Day 1: Read §§2.3–2.4. Re-solve their worked examples from a blank page, then try mixed practice 1–8.
3. Day 3: Study §2.5 and the derivation list. Solve practice 9–16 without notes; check each answer and revisit its cited section.
4. About day 7: Shuffle practice 17–25. Explain aloud why each method applies, then try one exam problem timed.
5. About day 14: Rebuild all derivations and retry every marked question. Complete the self-check only with evidence from unaided work.
6. Before the exam: Do both exam problems under time limits, then review only the skills your evidence shows are weak.

How the chapter fits together

Number systems lead to the complete ordered field ℝ. Its structure supports inequalities, counting formulas and induction.

Choose a tool

2.1 Number Systems

The sets grow by solving equations that smaller number systems cannot always solve. Knowing which set contains a number determines which tools are valid.

In plain words: each number system fixes an equation the smaller one cannot always solve. In ℕ you can count but not always subtract; ℤ adds the negatives, so x + 3 = 1 has a solution; ℚ adds fractions, so 2x = 1 does; ℝ fills the remaining gaps, so x² = 2 does. Each set sits inside the next.

Integers add negatives to ℕ. Rationals add quotients with non-zero integer denominator. The real numbers add limits such as √2 that cannot be expressed as a rational quotient. For example, √2 is irrational: if √2=p/q in lowest terms, then p²=2q² makes p even; substituting p=2m makes q even too, contradicting lowest terms.

Binary uses place values that are powers of two. A bit string [a_m…a₀]₂ represents ∑_k=0^ma_k2^k, where each a_k is 0 or 1. For a fixed-point example, [10.11]₂=2+1/2+1/4=2.75. The field ℤ₂ uses arithmetic modulo 2; ordinary binary integer addition and multiplication instead represent integer arithmetic and do not make the set of finite bit strings a field.

Trap: zero is not in ℕ under these notes’ convention. Write ℕ₀ when zero is included. Also, an integer written as a decimal or fraction can still belong to a smaller set, such as 6/3=2∈ℤ.

Near miss: √4 is written with a root sign, but its value is the natural number 2; √2 has no fraction form at all. Decide which set a number belongs to by its value, not by how it is written.

Try it: binary place-value experiments.

2.2 Real Numbers

The real-number rules explain why algebraic transformations are legitimate and why limits do not leave gaps in the number line.

In plain words: the real numbers are the number line with no holes. The field axioms say you can add, subtract, multiply and divide (except by 0) with the usual rules; the order axioms say how < behaves; and completeness says every non-empty set that is bounded above has a smallest upper bound, its supremum. That last rule is what makes √2 a real number.

The field axioms encode commutativity, associativity, neutral elements, inverses and distributivity. The order axioms make < a total, transitive order compatible with addition and multiplication by positive numbers. Together these support the familiar algebra of inequalities. ℤ is not a field because, for example, 1/3 is not an integer. ℚ is an ordered field, but it is not complete.

For M={1, 1/2, 1/3,…}, sup(M)=1 and inf(M)=0. The maximum is 1 because 1∈M; there is no minimum because 0∉M. The completeness axiom says every non-empty real subset bounded above has a supremum in ℝ. It is the property that fills the “gaps” missing in ℚ.

Trap: a supremum is not automatically a maximum. First establish the least or greatest bound, then check whether it belongs to the set.

Near miss: the set M = {x : x < 1} has upper bounds 1, 2, 10 and so on, and its supremum is 1, but 1 is not in M, so M has no maximum. A bounded set always has a supremum; it may not have a maximum.

Try it: supremum and completeness experiments.

2.3 Absolute Values and Inequalities

Absolute value turns signs into distance, while the order axioms let you solve distance constraints as intervals.

In plain words: absolute value is distance: |x| is how far x is from 0, and |x − a| is how far x is from a. So an inequality such as |x − 3| < 2 simply says “x is less than 2 away from 3”, which is the interval (1, 5).

For c≥0, |x−a|≤c is equivalent to a−c≤x≤a+c. It describes the closed interval of points no more than c from a. For c>0, |x−a|<c gives an open interval. An inequality with a “greater than” sign usually describes the outside of an interval. Splitting at the point where the expression changes sign is a reliable general method.

Sign check: squaring both sides can introduce solutions unless both sides are known to be non-negative. Also, √x² is |x|, not always x.

Near miss: solving |x| = −2 with “x = 2 or x = −2” looks routine, but a distance cannot be negative, so there is no solution. Check the sign of the right-hand side first.

Try it: absolute-value distance experiments.

2.4 Sums, Products, Binomial Coefficient

Summation and product notation compress repeated operations. Factorials count arrangements; binomial coefficients count selections and become the weights in expansions.

In plain words: ∑ is shorthand for “add up” and ∏ for “multiply together”. The index under the sign says where to start, the number on top where to stop, and the expression after the sign is what to take for each value of the index. n! counts the orders in which n different things can be lined up; the binomial coefficient counts the ways to choose k of them when order does not matter.

Useful coefficient properties are C(n,0)=C(n,n)=1 and C(n,k)=C(n,n−k). Pascal’s identity says adjacent entries combine to form the next row. The binomial theorem generalises the familiar square and cube patterns.

Memory hook: in the term indexed by k, the exponent on a is n−k and the exponent on b is k. Check that each term’s exponents add to n.

Near miss: “the sum of k² for k = 1 to 3” looks as if it might be (1 + 2 + 3)² = 36, but it means 1² + 2² + 3² = 14. The expression after ∑ is worked out for each k first, then the results are added.

Try it: Pascal triangle and binomial experiments.

2.5 Mathematical Induction

Induction proves infinitely many linked claims using a finite argument: establish a starting case, then show each true case passes the claim to the next one.

In plain words: induction works like a row of dominoes. Show that the first one falls (the base case), and show that whenever one falls it knocks over the next (the step). Then every domino falls, so the claim holds for every n from the starting value on.

Write the induction step as a conditional: assume A(n) for an arbitrary allowed n, and use it to prove A(n+1). Do not assume the result you are trying to prove. The base case starts the chain; the implication advances it. For two-step induction, verify two consecutive base cases and advance from n to n+2, which covers both parities. In strong induction (the notes’ “extended induction hypothesis”) you assume A(n₀), A(n₀+1), …, A(n) all hold and use any of them to prove A(n+1); this helps when the next case depends on an earlier case rather than only on the one before.

In the extended step, (1+x)ⁿ⁺²=(1+x)²(1+x)ⁿ. The square factor is non-negative. Applying the hypothesis and expanding leaves the extra term x²(2n+1+nx), which is non-negative for x≥−2 and n≥1.

Trap: checking several values is not an induction proof. In the step, n is arbitrary—not one particular example—and the induction hypothesis must be stated before it is used.

Near miss: n² + n + 41 is prime for n = 0, 1, 2, …, 39, yet at n = 40 it gives 1681 = 41². Forty checked cases are still not a proof; only the step from n to n + 1 covers every case.

Try it: induction proof experiments.

Formula map and checks

Derive it cold

Rebuild these on blank paper before opening the answers in the relevant sections.

1. Prove √2 is irrational by contradiction. §2.1
2. Derive √x²=|x| by splitting x≥0 and x<0. §2.3
3. Derive the factorial expression for C(n,k) by counting ordered selections and dividing by k!. §2.4
4. Prove the sum 1+…+n=n(n+1)/2 by induction. §2.5
5. Use the induction hypothesis and Pascal’s identity to prove the binomial theorem for n+1 from n. §2.4–2.5

Explain it

Answer aloud in complete sentences. Then compare with the model answer. (Retrieval and explanatory questioning [MIS, AMN, NEW].)

1. Why does being expressible as p/q not guarantee a unique written fraction?
2. Why are field and order axioms not enough to define ℝ?
3. When is a supremum a maximum?
4. Why is √x² not always x?
5. When should you use a binomial coefficient instead of a permutation count?
6. Why does induction need both a base case and a step?

Mixed practice

Try all 25 problems before opening the answer key. They are interleaved, so identify the method before calculating. (Desirable difficulty and varied retrieval [MIS, AMN].)

1. Classify −5/2 as natural, integer, rational, irrational or real; list every applicable category.
2. Solve |x−4|<3 and write the answer as an interval.
3. Prove by induction that ∑_k=1ⁿk²=n(n+1)(2n+1)/6 for n∈ℕ.
4. Expand (a+b)³.
5. For M={1−1/n:n∈ℕ}, find sup(M), max(M), inf(M), and min(M).
6. Evaluate ∑_k=1⁴(2k−1) and ∏_k=1⁴k.
7. In one sentence, state the contradiction in the proof that √2 is irrational.
8. Solve |3x−2|>7.
9. Why is ℤ not a field under ordinary addition and multiplication? Give one specific failed field property.
10. How many 2-element subsets can be chosen from an 8-element set?
11. Convert [101101]₂ to base ten.
12. Check the triangle inequality for x=−7, y=3. Compute both sides.
13. Prove by induction that 7 divides 8ⁿ−1 for n∈ℕ.
14. For M={1/n:n∈ℕ}, give its supremum and infimum. Say which are attained.
15. Compute C(7,3), showing the cancellation.
16. Convert [10.11]₂ to base ten.
17. Solve |2x+1|≥5 by cases and check endpoints.
18. In an ordered field, if a<b, what can you conclude about a+c and b+c? What if you multiply by c<0?
19. Use Pascal’s identity to evaluate C(6,2)+C(6,3).
20. Use Bernoulli’s inequality to give a lower bound for (1.1)⁵.
21. What is the principal real fourth root of 16?
22. Evaluate ∑_k=1ⁿ1/(k(k+1)) by writing each term as 1/k−1/(k+1).
23. Evaluate ∏_k=2⁴k.
24. Prove by induction that 1+3+…+(2n−1)=n².
25. Find the coefficient of x²y³ in (x+y)⁵.

Exam-style problems

Allow about 20 minutes for each. Work every part before checking the answer. Afterward, note where you lost a step and revisit the cited sections.

1 · Bounds, distance and counting

Let M={x∈ℝ: |x−2|≤3} and let S be the set of 2-element subsets of {1,2,3,4,5}.

1. Write M as an interval and give its minimum, maximum, infimum and supremum.
2. Find all x∈M for which |x|>2.
3. How many members does S have? Explain why order does not matter.
4. Find the coefficient of x³y² in (x+y)⁵.

2 · Prove the binomial theorem

Let P(n) be the claim (a+b)ⁿ=∑_k=0ⁿC(n,k)a^n−kb^k, for non-zero real a,b and n∈ℕ.

1. Verify P(1).
2. Assume P(n). Multiply its right-hand side by (a+b) to obtain the expression for (a+b)ⁿ⁺¹.
3. Group adjacent terms and use Pascal’s identity to prove P(n+1).
4. Use the theorem to expand (x+2)⁴.

Slips in the notes

These are source-checked issues and the corresponding precise statement used in this guide. Printed page numbers refer to the lecture notes.

1. Printed page 12, missing condition: the remark says that for arbitrary rational a,b, the equation ax=b always has a solution x∈ℚ. This fails when a=0 and b≠0. The statement is correct when a≠0; for a=0 it has no solution if b≠0 and every rational x is a solution if b=0.
2. Printed page 18, error: the telescoping rule is printed as ∑_k=mⁿ(a_k−a_k+1)=a_m−a_n. Writing out the terms leaves a_m−a_n+1, and the product rule likewise gives a_m/a_n+1. For m=n the printed form gives 0 instead of a_n−a_n+1. This guide uses the corrected rules.
3. Printed page 19, missing condition: Theorem 2.13(iii) is stated for k=0,…,n, but at k=n it needs C(n,n+1), which the notes do not define. This guide states Pascal’s identity for 0≤k≤n−1 (or set C(n,k)=0 for k>n).
4. Printed page 15, loose wording: “for ℝ² and ℂ the order axioms do not hold” means that no ordering of ℂ satisfies all of (A10)–(A13) together with the field operations. ℂ can be totally ordered, for example lexicographically, but such an order is not compatible with multiplication (A13).

Self-check

Tick a skill only in a later session, at least one day after learning it, when you can show it without notes. Use the problem numbers as evidence.

• Classify values in ℕ, ℤ, ℚ and ℝ and distinguish irrational numbers. (1, 7)
• Convert binary integers and fixed-point numbers. (11, 16)
• Rebuild the contradiction proof for √2. (7)
• Explain why ℚ is an ordered field but not complete. (9, 18, 21)
• Find bounds, suprema, infima, maxima and minima. (5, 14)
• Solve absolute-value inequalities with correct endpoints. (2, 8, 17)
• Apply the triangle inequality and check it numerically. (12)
• Evaluate and interpret summation and product notation, including telescoping sums. (6, 22, 23)
• Use factorials and binomial coefficients for counting. (10, 15)
• Use Pascal’s identity and expand a binomial. (4, 19, 25)
• Write complete induction proofs from the correct base index. (3, 13, 24)
• Apply Bernoulli’s inequality with its stated condition. (20)
• Rebuild the listed derivations without notes.
• Solve both multi-part problems and justify every step.