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tmua.me Medium Set B Paper 2 · Q20

Paper 2Logic & proof

Let a1,…,ana_1,\ldots,a_n be real numbers, where n≥2n\geq2 and a1+⋯+an>0a_1+\cdots+a_n>0.

A student wants to prove that some cyclic rearrangement of the list has every initial sum strictly positive. Put S0=0S_0=0 and Sk=a1+⋯+akS_k=a_1+\cdots+a_k for 1≤k≤n1\leq k\leq n.

The student proposes the following proof.

  1. Choose an index jj from 0,…,n−10,\ldots,n-1 by a rule still to be specified. Use the list

    aj+1,…,an,a1,…,aj.\displaystyle a_{j+1},\ldots,a_n,a_1,\ldots,a_j.

  2. An initial sum ending at aka_k before reaching ana_n is Sk−SjS_k-S_j, where j<k<nj<k<n. It is strictly positive by the choice of jj.

  3. The sum ending at ana_n is Sn−Sj>0S_n-S_j>0.

  4. An initial sum continuing past ana_n and ending at aka_k, where 1≤k≤j1\leq k\leq j, is EE. It is strictly positive by the choice of jj and the fact that Sn>0S_n>0.

Which rule for choosing jj, together with which expression for EE, makes this proof valid? In each option, minima and maxima are taken over S0,…,Sn−1S_0,\ldots,S_{n-1}.

AChoose the smallest jj at which SjS_j is minimal; E=Sn−Sj+SkE=S_n-S_j+S_k
BChoose the largest jj at which SjS_j is maximal; E=Sn−Sj+SkE=S_n-S_j+S_k
CChoose the largest jj at which SjS_j is minimal; E=Sn+Sj−SkE=S_n+S_j-S_k
DChoose the largest jj at which SjS_j is minimal; E=Sn+SkE=S_n+S_k
EChoose j=0j=0; E=Sn−Sj+SkE=S_n-S_j+S_k
FChoose the largest jj at which SjS_j is minimal; E=Sn−Sj+SkE=S_n-S_j+S_k

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