Proof and algebraic division flashcards

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Proof
A logical argument showing that a statement is true in every case it claims to cover: assumptions → logical steps → conclusion.
Proof by deduction
Starting from known facts or a general algebraic form and reaching the result through a chain of logical steps.
Proof by exhaustion
Splitting all possibilities into a finite set of cases (or checking every member of a finite set) and proving the result in each case.
Counterexample
One specific example that meets the conditions of a statement but makes it false. One is enough to disprove a statement about 'all' cases.
Identity (≡)
A statement true for every value of the variable, e.g. (x + 1)² ≡ x² + 2x + 1.
Rational number
A number that can be written as p/q, where p and q are integers and q ≠ 0.
Prime number
A positive integer with exactly two factors, 1 and itself (so 1 is not prime).
Quotient and remainder
In f(x) ≡ (ax + b) q(x) + r, q(x) is the quotient (one degree lower than f) and the constant r is the remainder.
⇒ and ⇔
A ⇒ B: if A is true then B is true. A ⇔ B: A is true if and only if B is true (each implies the other).
General even, odd and consecutive integers
Even 2n; odd 2n + 1 (or 2n − 1); consecutive n, n + 1, n + 2; two different odd numbers 2m + 1 and 2n + 1. (No — learn it)
Division identity for a linear divisor
f(x) ≡ (ax − b) q(x) + r, with r a constant. (No — learn it)
Remainder theorem
The remainder when f(x) is divided by (ax − b) is f(b/a). (No — learn it)
Factor theorem
(ax − b) is a factor of f(x) ⇔ f(b/a) = 0. (No — learn it)
Completed square of x² + bx + c
(x + b/2)² + c − b²/4; positive for all real x when c − b²/4 > 0. (No — derive it)
Mean of two positive numbers vs. root of their product
(a + b)/2 ≥ √(ab) for a, b > 0, from (√a − √b)² ≥ 0; equality when a = b. (No — derive it)
Graph of y = (x − p)² + q with q > 0
A U-shaped parabola with minimum point (p, q) above the x-axis — so the expression is always positive.
What the remainder looks like on the graph of y = f(x)
The remainder on dividing by (ax − b) is the height of the curve at x = b/a, i.e. the point (b/a, f(b/a)).
Graph of a cubic with factors (x − α)(x − β)(x − γ)
When α, β and γ are all different, it crosses the x-axis at x = α, β and γ; the y-intercept is the value of the cubic at x = 0.
Graph at a repeated factor (x − α)²
The curve touches the x-axis at x = α without crossing it.
Table of n² ÷ 4 remainders for n = 1, 2, 3, …
Only 0 (n even) and 1 (n odd) appear — the table suggests the result; the even/odd cases prove it.
Why checking many examples is not a proof
A later case could fail (n² + n + 11 is prime for n = 1 to 9 but 121 = 11² at n = 10). Examples prove a result only if they cover every case.
Why the remainder theorem works
Put x = b/a in f(x) ≡ (ax − b) q(x) + r: the bracket is 0, so f(b/a) = r.
Why a linear divisor leaves a constant remainder
The remainder always has lower degree than the divisor; a degree-1 divisor leaves degree 0, a constant.
Three parts every written proof needs
Clear assumptions (what the letters stand for), logical steps that each follow, and a concluding sentence linked to the claim.
How to make cases exhaustive for integers
Use forms that cover every integer: 2k and 2k + 1, or 3k, 3k + 1 and 3k + 2, and say so in the proof.
Why one counterexample is a full disproof
A statement about all cases is false if any single case fails; no further cases are needed.
Prove: the sum of three consecutive odd numbers is a multiple of 3.
(2n − 1) + (2n + 1) + (2n + 3) = 6n + 3 = 3(2n + 1), a multiple of 3.
Remainder when x³ − 2x² + 5 is divided by (x + 2)?
f(−2) = −8 − 8 + 5 = −11.
Divide 4x³ + 4x² − 5x + 3 by (2x + 1).
Quotient 2x² + x − 3, remainder 6; check f(−½) = 6.
Factorise 3x³ + 7x² − 22x − 8 completely, given f(2) = 0.
(x − 2)(3x² + 13x + 4) = (x − 2)(3x + 1)(x + 4).
(2x − 3) is a factor of 2x³ − x² + kx − 6. Find k.
f(3/2) = 27/4 − 9/4 + 3k/2 − 6 = 0 ⇒ k = 1.
Disprove: n² + n + 11 is prime for every positive integer n.
n = 10: 100 + 10 + 11 = 121 = 11 × 11, not prime, so the statement is false.
Prove x² − 6x + 11 > 0 for all real x.
x² − 6x + 11 = (x − 3)² + 2; (x − 3)² ≥ 0, so the expression ≥ 2 > 0.
ax³ + bx² − 11x + 6 has factor (x + 2) and remainder −6 on division by (x − 1). Find a and b.
f(−2) = 0 ⇒ 2a − b = 7; f(1) = −6 ⇒ a + b = −1; so a = 2, b = −3.
f(3) = 0, so which factor?
(x − 3), not (x + 3): f(p) = 0 ⇔ (x − p) is a factor.
Remainder on dividing by (2x − 1)
Use f(½), not f(1) or f(2): substitute the value that makes the divisor zero.
'Factorise completely' after finding one factor
Keep going: factorise the quadratic factor too, or show it has no real roots (discriminant < 0).
Using (…)² > 0 to prove an expression is positive
A square can be 0: write (…)² ≥ 0 and let the positive constant make the whole expression > 0.
Finishing a proof
End with a sentence in the words of the question, e.g. 'so the sum is always a multiple of 3' or 'so the statement is false'.

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