00.01.E1 · precalc / numbers

Algebra and number-systems exercise pack (Lang Basic Mathematics Part I-II supplement)

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Formal definition of the pack Intermediate

Lang's Basic Mathematics Part I treats precalculus as a foundations course: the arithmetic of integers, rationals, and reals is built from the field axioms and the order axioms rather than assumed as a collection of procedures. This pack exercises that foundational layer. The problems test the field and order axioms for $R$ 00.01.01, the absolute value and the triangle inequality 00.01.02, polynomial and rational manipulation 00.01.03, the quadratic formula and discriminant 00.03.02, and the sign-chart solution of inequalities 00.04.01. A handful of elementary number-theory problems — divisibility, the Euclidean algorithm, irrationality proofs — round out the set, matching the proof-flavoured exercises Lang scatters through Part I.

The pack collects ten problems: three easy, four medium, three hard. Each carries a hint and a full solution. The conventions follow the prerequisite units: $N = {0, 1, 2, \dots}$ as in the Babel Bible number-systems unit (Lang's own text uses $N$ from $1$ , a difference flagged in the Lang notation crosswalk); $∣ x ∣$ denotes absolute value; intervals are written with parentheses for open and brackets for closed.

The problems are meant to be worked alongside the prerequisite concept units rather than as a standalone development. The recurring theme is that every algebraic manipulation a reader will later perform mechanically — clearing a denominator, completing a square, flipping an inequality when multiplying by a negative — is a consequence of a named axiom, and the exercises make that dependence explicit.

Key theorem with full solution Intermediate

We work one exercise in full as an exemplar of the format. The remaining nine follow the same structure: problem, hint, full answer in <details> blocks.

Lead exercise. Prove that $2$ is irrational, then show $2 + 3$ is irrational.

Solution. First, $2$ . Suppose for contradiction that $2 = p / q$ with $p, q$ integers, $q \neq = 0$ , and the fraction in lowest terms, so $g cd (p, q) = 1$ . Squaring gives $2 q^{2} = p^{2}$ , so $p^{2}$ is even. The square of an odd integer is odd (since $(2 k + 1)^{2} = 4 k^{2} + 4 k + 1$ is odd), so $p$ must be even: write $p = 2 m$ . Then $2 q^{2} = 4 m^{2}$ , so $q^{2} = 2 m^{2}$ , meaning $q^{2}$ is even and hence $q$ is even. But then $2 ∣ p$ and $2 ∣ q$ contradicts $g cd (p, q) = 1$ . So no such fraction exists, and $2$ is irrational.

Now $2 + 3$ . Suppose it equals a rational $r$ . Then $3 = r - 2$ . Square both sides:

3 = r^{2} - 2 r 2 + 2, so 2 r 2 = r^{2} - 1.

If $r \neq = 0$ , this gives $2 = (r^{2} - 1) / (2 r)$ , a ratio of rationals, hence rational — contradicting the first part. If $r = 0$ then $2 + 3 = 0$ , impossible since both summands are positive. Either way we reach a contradiction, so $2 + 3$ is irrational. $□$

The two-step pattern — assume rational, square to isolate a known irrational, derive a contradiction — is the workhorse of every elementary irrationality proof. It rests only on the field axioms (to manipulate the equation) and the closure of $Q$ under arithmetic (to conclude the isolated expression is rational).

Exercises Intermediate

Exercise 1 (easy, symbolic). Distributive-law expansions.

Using only the field axioms (commutativity, associativity, distributivity), expand and simplify $(a + b) (a - b)$ and $(a + b)^{2}$ , naming the axiom at each step where distributivity is invoked.

Hint

Distributivity is $x (y + z) = x y + x z$ . Apply it once to split off the first factor, then again to each resulting product.

Answer

For $(a + b) (a - b)$ : by distributivity over the second factor, $(a + b) (a - b) = (a + b) a - (a + b) b$ . Distribute again: $(a + b) a = a^{2} + ba$ and $(a + b) b = ab + b^{2}$ . So

(a + b) (a - b) = a^{2} + ba - ab - b^{2} = a^{2} - b^{2},

using commutativity $ba = ab$ to cancel the middle terms.

For $(a + b)^{2} = (a + b) (a + b)$ : distributing gives $a^{2} + ba + ab + b^{2} = a^{2} + 2 ab + b^{2}$ , again using commutativity to combine $ba + ab = 2 ab$ .

Both identities are consequences of distributivity plus commutativity of multiplication — not separate rules to memorise. This is Lang's central pedagogical point in Ch. 1.

Exercise 4 (medium, proof). The triangle inequality and its reverse.

Prove the triangle inequality $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ for all real $x, y$ , then deduce the reverse triangle inequality $∣ x ∣ - ∣ y ∣ \leq ∣ x - y ∣$ .

Hint

For the forward inequality, square both sides (both are nonnegative) and use $x y \leq ∣ x y ∣ = ∣ x ∣∣ y ∣$ . For the reverse, apply the forward inequality to $x = (x - y) + y$ .

Answer

Forward. Both sides of $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ are nonnegative, so it suffices to compare squares. We have $∣ x + y ∣^{2} = (x + y)^{2} = x^{2} + 2 x y + y^{2}$ and $(∣ x ∣ + ∣ y ∣)^{2} = x^{2} + 2∣ x ∣∣ y ∣ + y^{2}$ . The difference is $2 (∣ x ∣∣ y ∣ - x y) \geq 0$ because $x y \leq ∣ x y ∣ = ∣ x ∣∣ y ∣$ . So $∣ x + y ∣^{2} \leq (∣ x ∣ + ∣ y ∣)^{2}$ , and taking nonnegative square roots gives the inequality.

Reverse. Write $x = (x - y) + y$ and apply the forward inequality: $∣ x ∣ \leq ∣ x - y ∣ + ∣ y ∣$ , so $∣ x ∣ - ∣ y ∣ \leq ∣ x - y ∣$ . By symmetry (swap $x, y$ and use $∣ y - x ∣ = ∣ x - y ∣$ ), $∣ y ∣ - ∣ x ∣ \leq ∣ x - y ∣$ . Together, $∣ x ∣ - ∣ y ∣ \leq ∣ x - y ∣$ . $□$

Exercise 6 (medium, proof). Order axioms: multiplying an inequality by a negative.

From the order axioms — (O1) for any $x$ , exactly one of $x > 0$ , $x = 0$ , $x < 0$ holds; (O2) $x, y > 0 \Rightarrow x + y > 0$ ; (O3) $x, y > 0 \Rightarrow x y > 0$ — prove that if $a < b$ and $c < 0$ , then $a c > b c$ .

Hint

$a < b$ means $b - a > 0$ , and $c < 0$ means $- c > 0$ . Multiply these two positives using (O3).

Answer

By definition of the ordering, $a < b$ means $b - a > 0$ , and $c < 0$ means $- c > 0$ . By (O3), the product of two positive numbers is positive:

(b - a) (- c) > 0.

Expand using distributivity: $(b - a) (- c) = - b c + a c = a c - b c$ . So $a c - b c > 0$ , which means $a c > b c$ . $□$

This is the precise reason an inequality reverses when both sides are multiplied by a negative number: the rule is not a convention but a theorem from the three order axioms. It underwrites every sign-chart manipulation in 00.04.01.

Exercise 7 (medium, symbolic). Quadratic inequality by sign chart.

Solve $x^{2} - x - 6 \geq 0$ , giving the solution as a union of intervals, and verify the answer by testing one point in each region.

Hint

Factor the quadratic to find the roots, then determine the sign of the product on each interval the roots cut the line into.

Answer

Factor: $x^{2} - x - 6 = (x - 3) (x + 2)$ , with roots $x = - 2$ and $x = 3$ . These cut the line into three regions.

On $(- \infty, - 2)$ : take $x = - 3$ , giving $(- 6) (- 1) = 6 > 0$ . ✓
On $(- 2, 3)$ : take $x = 0$ , giving $(- 3) (2) = - 6 < 0$ .
On $(3, \infty)$ : take $x = 4$ , giving $(1) (6) = 6 > 0$ . ✓

The product is $\geq 0$ on the outer two regions, including the endpoints where it equals zero. Solution set:

(- \infty, - 2] \cup [3, \infty) .

The sign-chart method 00.04.01 turns a quadratic inequality into a sign count of the linear factors.

Exercise 8 (hard, proof). AM-GM for two terms, with equality.

Prove that for nonnegative reals $a, b$ , the arithmetic-geometric-mean inequality $ab \leq \frac{a + b}{2}$ holds, and that equality holds if and only if $a = b$ .

Hint

Start from the fact that a real square is nonnegative: $(a - b)^{2} \geq 0$ . Expand and rearrange.

Answer

Since $a, b \geq 0$ , the real numbers $a$ and $b$ exist, and any real square is nonnegative:

(a - b)^{2} \geq 0.

Expand: $a - 2 a b + b \geq 0$ , i.e. $a + b \geq 2 ab$ . Dividing by $2$ (positive, so the inequality direction is preserved) gives

\frac{a + b}{2} \geq ab .

Equality. The single inequality used was $(a - b)^{2} \geq 0$ , which is an equality exactly when $a - b = 0$ , i.e. $a = b$ , i.e. $a = b$ . So equality in AM-GM holds iff $a = b$ . $□$

The proof needs only the order axiom that squares are nonnegative — itself a consequence of (O3) applied to $x$ and $x$ , or to $- x$ and $- x$ . Lang presents AM-GM as a worked example precisely to show how much follows from the ordering alone.

Exercise 9 (hard, proof). Density of the rationals.

Prove that between any two distinct reals $a < b$ there is a rational number. You may use the Archimedean property: for every real $t > 0$ there is a positive integer $n$ with $1/ n < t$ .

Hint

Choose $n$ so large that $1/ n < b - a$ , so steps of size $1/ n$ cannot jump over the interval. Then take the smallest integer $m$ with $m / n > a$ and show $m / n < b$ .

Answer

Since $b - a > 0$ , the Archimedean property gives a positive integer $n$ with $1/ n < b - a$ , equivalently $n (b - a) > 1$ .

Consider the integers $m$ with $m / n > a$ , i.e. $m > na$ . Such integers exist (Archimedean property again, applied to bound $na$ above), and they are bounded below, so by well-ordering there is a least one; call it $m$ . Then $m / n > a$ , while minimality gives $(m - 1) / n \leq a$ , i.e. $m \leq na + 1$ .

Now estimate $m / n$ from above:

\frac{m}{n} \leq \frac{na + 1}{n} = a + \frac{1}{n} < a + (b - a) = b .

So $a < m / n < b$ , and $m / n$ is rational. $□$

The argument is the precalculus seed of the density statement in 00.01.01; it is the reason the rationals can approximate any real to arbitrary precision, which Apostol's epsilon-delta limits later exploit.

Exercise 10 (hard, numeric). Rational-roots test.

Find all rational roots of $p (x) = 2 x^{3} - 3 x^{2} - 11 x + 6$ , then factor $p$ completely over $R$ .

Hint

The rational-root theorem says any rational root $p / q$ in lowest terms has $p$ dividing the constant term and $q$ dividing the leading coefficient. Test the finite list of candidates, then divide out and solve the remaining quadratic.

Answer

Constant term $6$ , leading coefficient $2$ . Candidate rational roots $p / q$ have $p \in {\pm 1, \pm 2, \pm 3, \pm 6}$ and $q \in {1, 2}$ , giving candidates $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$ .

Test $x = 3$ : $2 \cdot 27 - 3 \cdot 9 - 11 \cdot 3 + 6 = 54 - 27 - 33 + 6 = 0$ . So $x = 3$ is a root and $(x - 3)$ divides $p$ .

Divide: $2 x^{3} - 3 x^{2} - 11 x + 6 = (x - 3) (2 x^{2} + 3 x - 2)$ . Factor the quadratic: $2 x^{2} + 3 x - 2 = (2 x - 1) (x + 2)$ , with roots $x = \frac{1}{2}$ and $x = - 2$ .

So the rational roots are $x = 3, \frac{1}{2}, - 2$ , and the complete real factorisation is

p (x) = (x - 3) (2 x - 1) (x + 2) = 2 (x - 3) (x - \frac{1}{2}) (x + 2) .

The rational-root test plus polynomial division 00.01.03 reduces a cubic to a quadratic, which the quadratic formula 00.03.02 finishes.

Exercise pack for Lang, Basic Mathematics, Part I-II. Algebra and number systems: field and order axioms, absolute value, inequalities, polynomials, and elementary number theory. Distribution: 3 easy / 4 medium / 3 hard.

Prerequisites

00.01.01 pending
00.01.02 pending
00.01.03 pending
00.03.02 pending
00.04.01 pending

Tier anchors

beginner
intermediate: Lang, Basic Mathematics, Part I (Algebra, Ch. 1-5) and Part II exercise sets: integer/rational arithmetic, ordering axioms, absolute value, inequalities, and elementary number theory
master

References

Lang — Basic Mathematics (Springer, 1988) · Part I (Ch. 1 number systems and field axioms, Ch. 2 order and absolute value, Ch. 3 quadratics, Ch. 4 inequalities, Ch. 5 exponents); exercise sets at chapter ends
Apostol — Calculus, Volume 1 · §I.3 (the field axioms), §I.4.1-I.4.10 (the order axioms and absolute value exercises), §I 3.5 (least-upper-bound and irrationality)
Niven-Zuckerman-Montgomery — An Introduction to the Theory of Numbers · §1.1-1.3 (divisibility, the Euclidean algorithm, the irrationality of square roots)

Estimated time

intermediate: 4h