Standard +0.8 This is a classic proof by contradiction requiring students to understand divisibility properties and contrapositive reasoning. While the structure is standard (assume n is not divisible by 3, show n² cannot be divisible by 3), it requires careful case analysis (n = 3k+1 or 3k+2) and algebraic manipulation beyond routine exercises, making it moderately challenging for A-level students.
Prove by contradiction that if \(n^2\) is multiple of 3, then \(n\) is multiple of 3
(5)
B1 Assume that there exists number \(n\) that isn't multiple of 3 yet \(n^2\) is multiple of 3; M1 States that \(m = 3k + 1\) or \(m = 3k + 2\) and attempts to square. Alternatively exist such as \(m = 3k + 1\) or \(m = 3k - 1\); M1 States that \(m = 3k + 1\) and \(m = 3k + 2\) and attempts to square; A1 Achieves forms that can be argued as to why they are not multiple of 3, e.g. \(m^2 = (3p + 1)^2 = 9p^2 + 6p + 1 = 3(3p^2 + 2p) + 1\) and \(m^2 = (3p + 2)^2 = 9p^2 + 12p + 4 = 3(3p^2 + 4p + 1) + 1\); A1 Correct proof which requires: Correct calculations, Correct reasons (e.g. \(9p^2 + 6p + 1\) or \(9p^2 + 12p + 4\) is not multiple of 3), Minimal conclusion
GRAND TOTAL: 87 marks
Prove by contradiction that if $n^2$ is multiple of 3, then $n$ is multiple of 3 | (5) | B1 Assume that there exists number $n$ that isn't multiple of 3 yet $n^2$ is multiple of 3; M1 States that $m = 3k + 1$ or $m = 3k + 2$ and attempts to square. Alternatively exist such as $m = 3k + 1$ or $m = 3k - 1$; M1 States that $m = 3k + 1$ and $m = 3k + 2$ and attempts to square; A1 Achieves forms that can be argued as to why they are not multiple of 3, e.g. $m^2 = (3p + 1)^2 = 9p^2 + 6p + 1 = 3(3p^2 + 2p) + 1$ and $m^2 = (3p + 2)^2 = 9p^2 + 12p + 4 = 3(3p^2 + 4p + 1) + 1$; A1 Correct proof which requires: Correct calculations, Correct reasons (e.g. $9p^2 + 6p + 1$ or $9p^2 + 12p + 4$ is not multiple of 3), Minimal conclusion
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**GRAND TOTAL: 87 marks**