Proof by contradiction and partial fractions flashcards

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Proof by contradiction
Assume the statement is false, reason correctly from that assumption to something impossible, and conclude that the statement is true.
Irrational number
A real number that cannot be written as a/b with a and b integers and b ≠ 0.
Simplest form of a fraction a/b
a and b have no common factor other than 1. Every rational number can be written this way.
Negation of 'if P then Q'
P is true and Q is false (keep the condition, deny the result).
Partial fractions
Writing one algebraic fraction as a sum of simpler fractions whose denominators are the factors of the original denominator.
Proper algebraic fraction
The degree of the numerator is less than the degree of the denominator.
Improper algebraic fraction
The degree of the numerator is greater than or equal to the degree of the denominator — divide before splitting.
Identity (≡) in partial fractions
The numerator equation, e.g. 7x − 1 ≡ A(2x + 1) + B(x − 1), is true for every x, so any value may be substituted.
Form for f(x)/((ax + b)(cx + d)), proper
A/(ax + b) + B/(cx + d) (No — learn it)
Form for f(x)/((ax + b)²(cx + d)), proper
A/(ax + b) + B/(ax + b)² + C/(cx + d) (No — learn it)
Form for f(x)/((x − p)(x − q)) when f has degree 2
A + B/(x − p) + C/(x − q): the constant A comes from dividing first. (No — learn it)
Division step for an improper fraction N(x)/D(x)
N(x)/D(x) = Q(x) + R(x)/D(x), where the degree of R is less than the degree of D. (No — learn it)
The new number in the proof of infinitely many primes
N = p₁p₂…pₙ + 1, which leaves remainder 1 on division by every pᵢ. (No — learn it)
Odd a squared
(2k + 1)² = 2(2k² + 2k) + 1, which is odd — so a² even ⇒ a even. (No — derive it)
Vertical asymptotes of y = f(x)/((x − p)(x − q))
The lines x = p and x = q, one for each linear factor of the denominator.
Graph near a repeated factor (x − p)² in the denominator (B > 0 in B/(x − p)²)
The curve goes to +∞ on both sides of x = p: the sign does not change there.
Graph of an improper fraction with a linear quotient ax + b
Far from the origin the curve approaches the sloping line y = ax + b, because the leftover fractions tend to 0.
Curve y = A/(x − p) + B/(x − q) and its two partial fractions
At every x the curve's height is the sum of the heights of the two simpler curves.
x² − y² = 10 drawn over integer grid points
The curve passes between the grid points without meeting one — illustrating (not proving) that there are no integer solutions.
Why the √2 proof needs 'simplest form'
The contradiction is that a and b turn out both even. That only contradicts something if a/b was assumed to have no common factor.
Why p₁p₂…pₙ + 1 need not be prime
2 × 3 × 5 × 7 × 11 × 13 + 1 = 30 031 = 59 × 509. The proof only needs a prime factor that is not in the list.
Why substitution finds the constants
The numerator equation is an identity; substituting a root of one factor makes the other terms zero and isolates one constant.
Why a repeated factor needs two terms
The numerator has three coefficients to match, so three constants are needed: A/(x − p), B/(x − p)² and C/(x − q).
Why x − y and x + y have the same parity
Their difference is 2y, which is even — so both are odd or both are even.
Express (7x − 1)/((x − 1)(2x + 1)) in partial fractions.
x = 1: 6 = 3A, A = 2; x = −½: −9/2 = −(3/2)B, B = 3. So 2/(x − 1) + 3/(2x + 1).
Express (3x² + 6)/((x − 1)²(x + 2)) in partial fractions.
x = 1: B = 3; x = −2: C = 2; x² terms: A + C = 3, A = 1. So 1/(x − 1) + 3/(x − 1)² + 2/(x + 2).
Express (2x³ + 3x² − 2x + 6)/((x − 1)(x + 2)) as Ax + B + C/(x − 1) + D/(x + 2).
Divide by x² + x − 2: quotient 2x + 1, remainder x + 8; then 3/(x − 1) − 2/(x + 2). So 2x + 1 + 3/(x − 1) − 2/(x + 2).
Express 5/((x − 1)(x + 4)) in partial fractions.
1/(x − 1) − 1/(x + 4).
√2 = a/b in simplest form. Show a and b are both even.
a² = 2b², so a² even, a even; a = 2c gives 4c² = 2b², b² = 2c², so b even — contradicting simplest form.
Prove there are no integers x, y with x² − y² = 10.
(x − y)(x + y) = 10; the factors have the same parity. Both odd: product odd. Both even: product a multiple of 4. 10 is neither — contradiction.
Prove there is no greatest even integer.
Assume N is the greatest. N + 2 is even and larger — contradiction.
Express (x² + 1)/((x − 1)(x + 1)) in partial fractions.
Improper: 1 + 2/(x² − 1) = 1 + 1/(x − 1) − 1/(x + 1).
First and last lines of a proof by contradiction
Start 'Assume, for contradiction, that …' and finish 'This is a contradiction, so …' in the words of the question.
Substituting for the factor (2x + 1)
Use x = −½, not x = −1 or x = −2.
Constant left over after substituting the roots (repeated factor)
Compare coefficients of x², or substitute a spare value such as x = 0.
Check before choosing a partial-fraction form
Compare degrees: if the numerator's degree is not lower, divide first.
Checking your partial fractions
Substitute one more value (often x = 0) into the identity; both sides must agree.

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