Logical Question:
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Let D be the set of all points in the real plane such that |x| + |y| <= 1,
where |x| (respectively |y|) denotes the absolute value of x (respectively y).
Prove that amongst every 5 points in D, there exist two points whose distance
from one another is at most 1.

Answer:

Yes,D is a square with ends at (0,1),(1,0),(-1,0),(0,-1), and also we could draw a round which will go through these four points. Its equation is x^2+y^2=1. we could see the square is totally contained within the round. Also we will notice that any points in the round will have distance less than 1 from each other. So in conclusion, every points in D have distance at most 1 from each other .

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The letters A, B, C, D, E, F and G, not necessarily
in that order,
stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F

1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F

2. A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E

3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16		Seventy five years ago pharmacists weighed medicine on balances like the ones you have been using. The mass pieces were very expensive, so a pharmacist would buy as few mass pieces as possible. If a pharmacist had 1-g, 3-g, and 9-g mass pieces, he or she could weight out any number of grams from 1 g to 13 g. Show how could measure all of the masses from 1 g to 13 g using only the three mass pieces given?