Moderate -0.8 This is a straightforward proof by contradiction requiring only basic logic about integers. Students need to assume a greatest odd integer exists, then show n+2 is larger and odd, which is a direct contradiction with minimal steps. The technique is standard and the argument is simpler than typical proof questions.
States the largest odd number and an odd number greater than it. E.g. odd number \(n\) and \(n+2\)
M1
Allowable starts: "\(n\)" with "\(n+2\)"; "\(n\)" with "\(n^2\)"; "\(2k+1\)" with "\(2k+3\)"; "\(2k+1\)" with "\((2k+1)^3\)"; "\(2k+1\)" with "\(2k+1+2k\)". Note: stating \(n=2k\) even when "\(n\)" is odd is M0
Fully correct proof by contradiction including: assumption there exists greatest odd number \(n\); correct statement showing second odd number is greater; minimal conclusion "this is a contradiction, hence proven"
A1*
Must include: (1) assumption e.g. "there exists greatest odd number \(n\)"; (2) minimal statement showing second number is greater e.g. \(2k+3>2k+1\); (3) minimal conclusion. Any incorrect algebra scores A0 e.g. \((2k+1)^2=4k^2+2k+1\) would be A0. Can ignore spurious info e.g. \(n>0\)
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| States the largest odd number and an odd number greater than it. E.g. odd number $n$ and $n+2$ | M1 | Allowable starts: "$n$" with "$n+2$"; "$n$" with "$n^2$"; "$2k+1$" with "$2k+3$"; "$2k+1$" with "$(2k+1)^3$"; "$2k+1$" with "$2k+1+2k$". Note: stating $n=2k$ even when "$n$" is odd is M0 |
| Fully correct proof by contradiction including: assumption there exists greatest odd number $n$; correct statement showing second odd number is greater; minimal conclusion "this is a contradiction, hence proven" | A1* | Must include: (1) assumption e.g. "there exists greatest odd number $n$"; (2) minimal statement showing second number is greater e.g. $2k+3>2k+1$; (3) minimal conclusion. Any incorrect algebra scores A0 e.g. $(2k+1)^2=4k^2+2k+1$ would be A0. Can ignore spurious info e.g. $n>0$ |