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 Olson likes strong coffee, the stronger the better. But he can’t distinguish small differences. Over the years, Mrs. Olson has discovered that if she changes the amount of coffee by more than one teaspoon in her six-cup pot, Olson can tell that she did it. But he cannot distinguish differences smaller than one teaspoon per pot. Where A and B are two different cups of coffee. Suppose that Olson is offered cups A, B, and C all brewed in the Olsons’ six-cup pot. Cup A was brewed using 14 teaspoons of coffee in the pot. Cup B was brewed using 14.75 teaspoons of coffee in the pot and cup C was brewed using 15.5 teaspoons of coffee in the pot. Q. Is Olson’s “at-least-as-good-as” relation, transitive? And, answer is NO! I can't understand why! Please help!
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 I think its like this - C has the highest amount of caffeine- but B differs from C by less than 1 tea-spoon coffee - so pakka C and B are indifferent options. Now B and A are indifferent choices - differences is again less than 1 tea-spoon coffee. she cant tell difference. But are C and A indifferent options? No, because the difference exceeds 1 tea-spoon. so C is preferred to A so, C indifferent to B , B indifferent to A, but is C indifferent to A? (by transitivity rule) no! thus transitivity violated. “Operator! Give me the number for 911!”
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 Pardon me if I am wrong, but C indifferent to B , B indifferent to A, but is C indifferent to A?(by transitivity rule) no! though transitivity is not violated as we are concerned about “at-least-as-good-as”. So, either C and A should be indifferent or C should be preferred to A. and indeed, C is preferred to A! And that makes me feel like answer given in Varian workouts for this problem is wrong. Please help! (I'm posting here the problem from Varian workouts as image as I can't type some notations.)
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 It is not transitiv as A>=B is tru,B>= C is true but A>=C is false.
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