Makes a valid conclusion. B1 2.4 1, 0 are not both positive integers, which is a x = y = contradiction to the opening statement. Therefore there do not exist positive integers and such that x y (5 marks) Notes Pearson Progression Step Q Scheme Marks AOs and Progress descriptor 9a Begins the proof by assuming the opposite is true. B1 3.1 7th there exists a number such that is even and Complete proofs n n2 n ‘Assumption: is odd.’ using proof by contradiction. Defines an odd number (choice of variable is not important) and M1 2.2a successfully calculates n2 Let 2k 1 be an odd number. + Factors the expression and concludes that this number must be M1 1.1b odd. , so is odd. n2 Makes a valid conclusion. B1 2.4 This contradicts the assumption is even. Therefore if is n2 n2 even, must be even. n (4) 9b Begins the by assuming the opposite is true. B1 3.1 a rational number.’ ‘Assumption:is Defines the rational number: M1 2.2a for some integers and where and have no common a b, a b factors. Squares both sides and concludes that is even: 1.1b M1 a
Makes a valid conclusion. x = 1, y = 0 are not both positive integers, which is a contradiction to the opening statement. Therefore there do not exist positive integers x and y such that B1 2.4 (5 marks) Notes Q Scheme Marks AOs Pearson Progression Step and Progress descriptor 9a Begins the proof by assuming the opposite is true. ‘Assumption: there exists a number n such that n2 is even and n is odd.’ B1 3.1 7th Complete proofs using proof by contradiction. Defines an odd number (choice of variable is not important) and successfully calculates n2 Let 2k + 1 be an odd number. M1 2.2a Factors the expression and concludes that this number must be odd. , so n2 is odd. M1 1.1b Makes a valid conclusion. This contradicts the assumption n2 is even. Therefore if n2 is even, n must be even. B1 2.4 (4) 9b Begins the by assuming the opposite is true. ‘Assumption:is a rational number.’ B1 3.1 Defines the rational number: for some integers a and b, where a and b have no common factors. M1 2.2a Squares both sides and concludes that a is even: M1 1.1b
al_maths_pure_unit_1_ms.docx