Quick check: prove a 2 + b 2 ≥ 2ab for all real a, b.

Since (a − b) 2 ≥ 0, expanding gives a 2 − 2ab + b 2 ≥ 0, so a 2 + b 2 ≥ 2ab, with equality iff a = b.

Outline proving the inequality 2 n > n 2 for integers n > 4 by induction.

Base: n = 5 gives 32 > 25. Step: assume 2 k > k 2 . Then 2^(k+1) = 2·2 k > 2k 2 ≥ (k+1) 2 , since 2k 2 − (k+1) 2 = k 2 − 2k − 1 ≥ 0 for k ≥ 5.

Free HSC Mathematics Extension 2 Flashcard Review

Review 20 public cards with spaced repetition. Flip a card, then grade your recall. Cards you miss return today. Your schedule stays in this browser.

Browse the free cards

State the difference between an implication (=>) and an equivalence (<=>).

P ⇒ Q means 'if P then Q' (P is sufficient for Q). P <=> Q means 'P if and only if Q', i.e. both P ⇒ Q and Q ⇒ P hold, so each is necessary and sufficient for the other.

Define the converse, negation and contrapositive of the statement 'P ⇒ Q'.

Converse: Q ⇒ P. Negation: P and not-Q (P is true but Q is false). Contrapositive: (not Q) ⇒ (not P).

What do the quantifiers ∀ and ∃ mean, and how is 'iff' used?

∀ means 'for all' (holds for every element); ∃ means 'there exists' (at least one element satisfies it). 'iff' abbreviates 'if and only if', denoting equivalence <=>.

Outline the structure of a proof by contradiction.

Assume the statement is false (assume its negation). Use valid steps to derive a contradiction. Conclude the assumption is false, so the original statement must be true.

Sketch the proof that √2 is irrational.

Assume √2 = a/b in lowest terms. Then a² = 2b², so a is even, a = 2k; then b² = 2k², so b is even too — contradicting 'lowest terms'. Hence √2 is irrational.

Sketch the proof that log₂ 5 is irrational.

Suppose log₂ 5 = p/q with positive integers p, q. Then 2^p = 5^q, but the left side is even and the right side is odd — impossible. Hence log₂ 5 is irrational.

State the definition of a > b for real numbers a, b used in inequality proofs.

For real a, b, a > b means a − b > 0, i.e. a − b is a positive real number. Many inequality proofs reduce to showing the relevant difference is positive.

State the triangle inequality for real numbers and its geometric meaning.

For all real x, y: |x + y| ≤ |x| + |y|. Geometrically the distance for x + y is at most the sum of the separate distances; equality holds iff x and y have the same sign (or one is zero).

State the AM–GM inequality for two non-negative numbers and prove it.

For a, b ≥ 0, (a + b)/2 ≥ √(ab), equality iff a = b. Proof: (√a − √b)² ≥ 0 gives a − 2√(ab) + b ≥ 0, so a + b ≥ 2√(ab).

Quick check: prove a² + b² ≥ 2ab for all real a, b.

Since (a − b)² ≥ 0, expanding gives a² − 2ab + b² ≥ 0, so a² + b² ≥ 2ab, with equality iff a = b.

In a proof by mathematical induction, what are the two essential steps?

Base case: verify the statement at the initial value of n. Inductive step: assume it holds for n = k (the hypothesis) and prove it then holds for n = k + 1.

How is induction adapted when the initial value exceeds 1 or n does not increase by 1?

The base case is checked at the actual starting value (e.g. n = 5, or the first even integer), and the inductive step proves P(k) ⇒ P(next value), e.g. P(k) ⇒ P(k + 2) for even n only.

What does sigma notation Σ_r=1ⁿ a_r represent, and why is it useful in induction?

It denotes a₁ + a₂ + … + a_n. In the inductive step Σ_r=1^k+1 a_r = (Σ_r=1^k a_r) + a_k+1, so the hypothesis is substituted for the first part.

Outline proving the inequality 2ⁿ > n² for integers n > 4 by induction.

Base: n = 5 gives 32 > 25. Step: assume 2^k > k². Then 2^(k+1) = 2·2^k > 2k² ≥ (k+1)², since 2k² − (k+1)² = k² − 2k − 1 ≥ 0 for k ≥ 5.

Outline proving 3^(2n+4) − 2^(2n) is divisible by 5 for all positive integers n.

Base: n = 1 gives 729 − 4 = 725 = 5·145. Step: assume 3^(2k+4) − 2^(2k) = 5M; then 3^(2k+6) − 2^(2k+2) = 9·3^(2k+4) − 4·2^(2k) = 45M + 5·2^(2k), divisible by 5.

How is induction used to prove d/dx(xⁿ) = n·x^(n−1) for positive integers n?

Base: n = 1, d/dx(x) = 1 = 1·x⁰. Step: assume d/dx(x^k) = k·x^(k−1); by the product rule on x^(k+1) = x·x^k, d/dx = x^k + x·k·x^(k−1) = (k+1)x^k.

State the binomial theorem that can be proven by induction.

(x + a)ⁿ = Σ_r=0ⁿ (nCr) x^(n−r) a^r, where nCr = n!/[r!(n−r)!]. The inductive step uses Pascal's rule (nCr) + (nC(r−1)) = (n+1 C r).

State the geometric result about exterior angles provable by induction.

The sum of the exterior angles of any n-sided convex polygon (one per vertex) is 360°. The inductive step adds a vertex while the total remains 360°.

State the standard unit vectors in three dimensions and the component form of a vector.

i = (1,0,0), j = (0,1,0), k = (0,0,1). A 3D vector is v = xi + yj + zk, equivalently the ordered triple (x, y, z) or column vector [x; y; z].

State the magnitude of a 3D vector and how to form a unit vector.

For v = (x, y, z), |v| = √(x² + y² + z²). For non-zero v the unit vector in its direction is v̂ = v/|v|, which has magnitude 1.