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**Extra resources for Algebraic Numbers (Pure & Applied Mathematics Monograph)**

**Example text**

Hence, for any κ with 0 ≤ κ ≤ ∞, there are infinitely many proportional points ( p1 , p2 ) ∈ IPS p −1 with p2 − 21 = κ. This establishes part bi. 1 2 The proof for bii is the same except that here we must not allow κ = 0 or κ = ∞, since this would correspond to a point ( p1 , p2 ) with either p1 = 12 or p2 = 12 , and such a point is not strongly proportional. 2. The line segment in the figure contains the point ( 12 , 12 ) and has slope κ. Every point along the solid part of this line segment, including ( 12 , 12 ), is proportional, and every point along this part of the line segment, not including ( 12 , 12 ), is strongly proportional.

The horizontal line through the origin) intersects the outer and inner boundaries at the same point, namely (1, 0). , the vertical line through the origin) intersects the outer and inner boundaries at the same point, namely (0, 1). The points (1, 0) and (0, 1) are each Pareto maximal and Pareto minimal. 4. It is clear that these theorems, which involved proportionality, strong proportionality, and the corresponding chores notions, 36 3. What the IPS Tells Us About Fairness and Efficiency need to use the point ( 12 , 12 ) rather than the origin.

The IPS has no points that are both strongly proportional and Pareto maximal. b. If m 1 = m 2 , then i. the IPS has infinitely many points that are both proportional and Pareto maximal. In particular, for any κ with 0 ≤ κ ≤ ∞, there is a point ( p1 , p2 ) ∈ IPS that is both proportional and Pareto maximal and is such p −1 that p2 − 21 = κ (where we set λ0 = ∞ for any real number λ = 0). 1 2 ii. the IPS has infinitely many points that are both strongly proportional and Pareto maximal. In particular, for any κ with 0 < κ < ∞, there is a point ( p1 , p2 ) ∈ IPS that is both strongly proportional and Pareto maximal and p −1 is such that p2 − 21 = κ.