If r > 0, then for all p and q such that pq does not equal 0 and such that pr > qr, we have:
- –p > –q
- –p > q
- 1 > q/p
- 1 < q/p
- none of these
Since r is not equal to 0, we can divide by r, and since r > 0 we know pr > qr implies p > q. We have three possibilities to consider, so let’s try specific values.
Note: p < 0 and q > 0 is impossible since p > q.
| p | q | p | q | a | b | c | d | |||||
| –p > –q | –p > q | 1 > q/p | 1 < q/p | |||||||||
| + | + | 6 | 3 | F | F | T | F | |||||
| + | – | 6 | -3 | F | F | T | F | |||||
| – | – | -3 | -6 | F | T | F | T |
Since none of the conditions (a), (b), (c), or (d) holds for every possibility of p and q, we conclude that (e) is the correct response.