Concise Introduction To Pure: Mathematics Solutions Manual

Let remainder be (ax+b). Write (x^100 = (x^2-1)Q(x) + ax+b). Set (x=1): (1 = a+b). Set (x=-1): (1 = -a+b). Solve: adding → (2=2b \Rightarrow b=1,\ a=0). Remainder = 1. Chapter 7 – Relations and Functions Exercise 7.2 Define relation (R) on (\mathbbZ) by (aRb) if (a-b) is even. Prove (R) is an equivalence relation.

Work mod 7: (2^1\equiv 2,\ 2^2\equiv 4,\ 2^3\equiv 1 \pmod7) (since (8\equiv 1)). Thus (2^3k\equiv 1). Write (100 = 3\cdot 33 + 1). (2^100 = (2^3)^33\cdot 2^1 \equiv 1^33\cdot 2 \equiv 2 \pmod7). Remainder = 2.

Prove by contradiction: (\sqrt2) is irrational.

[ A\cup B = 1,2,3,4,\quad A\cap B = 2,3 ] [ A\setminus B = 1,\quad B\setminus A = 4 ] Remark : Set difference removes elements of the second set from the first. Concise Introduction To Pure Mathematics Solutions Manual

Inverse of 3 mod 11: (3\times 4 = 12\equiv 1), so inverse is 4. Multiply both sides by 4: (x \equiv 20 \equiv 9 \pmod11). Check: (3\times 9=27\equiv 5) ✓. Chapter 4 – Real Numbers Exercise 4.1 Prove: if (x) is real and (x^2 < 1), then (-1 < x < 1).

Assume (\sqrt2 = p/q) in lowest terms ((p,q\in\mathbbZ), (\gcd(p,q)=1)). Squaring: (2q^2 = p^2 \Rightarrow p^2) even (\Rightarrow p) even. Write (p=2k). Then (2q^2 = 4k^2 \Rightarrow q^2 = 2k^2 \Rightarrow q) even. Contradiction since (\gcd(p,q)\ge 2). Hence (\sqrt2) irrational. Chapter 2 – Natural Numbers and Induction Exercise 2.3 Prove by induction: (1 + 2 + \dots + n = \fracn(n+1)2) for all (n\in\mathbbN).

Subcase A: first digit is even. Then first digit ∈ 2,4,6,8 (4 ways), other even digit ∈ 0,2,4,6,8 \ first digit choice? Wait, repetition allowed? Usually yes unless stated. Let’s assume repetition allowed unless “exactly two even digits” means count of even digits =2, not positions. Then easier: Let remainder be (ax+b)

Case 1: first digit odd (4 choices: 1,3,5,7,9? Actually 5 odds, but careful: first digit ≠0, so even allowed but handled separately). Better systematic: Choose positions for the two even digits: (\binom42=6) ways.

Choose 2 positions for evens: (\binom42=6). Fill evens: (5^2) ways (0–8 evens). Fill odds: (5^2) ways. Total = (6 \times 25 \times 25 = 3750).

Assume (\sqrt3=p/q) in lowest terms. Then (3q^2=p^2). So 3 divides (p^2) ⇒ 3 divides (p) (since 3 prime). Write (p=3k). Then (3q^2=9k^2\Rightarrow q^2=3k^2) ⇒ 3 divides (q). Contradiction ((\gcd(p,q)\ge 3)). Chapter 5 – Complex Numbers Exercise 5.2 Find ((2+3i)/(1-i)) in (a+bi) form. Set (x=-1): (1 = -a+b)

(x^2 < 1 \Rightarrow x^2 -1 < 0 \Rightarrow (x-1)(x+1) < 0). Product negative iff one factor positive, the other negative. Case 1: (x-1<0) and (x+1>0) → (x<1) and (x>-1) → (-1<x<1). Case 2: (x-1>0) and (x+1<0) impossible (would require (x>1) and (x<-1)). Thus (-1<x<1).

Let (y=x^2): (y^2-5y+4=(y-1)(y-4)=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)).

Solve (3x \equiv 5 \pmod11).

Find all cube roots of (-8).