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19

In case (c), the x Electors can, as in case (a), supply s Candidates with ⁠vx/x⁠ votes apiece. But each of the (e − x) Electors can only use (m + 1 − s) of his votes: hence the (e − x) Electors can only give (m + 1 − s).(e − x) votes, thus supplying (m + 1 − s) Candidates with (e − x) votes apiece. Hence we must have

⁠vx/s⁠ > e − x;
∴ vx > se − sx;
∴ x.(s + v) > se;
∴ x > ⁠se/s + v⁠.

In case (d), the x Electors can, as in case (b), supply s Candidates with x votes apiece. And the (e − x) Electors can, as in case (c), supply (m + 1 − s) Candidates with (e − x) votes apiece. Hence we must have

x > e − x;
∴ 2x > e;
∴ x > ⁠e/2⁠.
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