The set S contains some real numbers, according to the following three rules. (i) 1/1 is in S. (ii) If a/b is in S, where a/b is written in lowest terms (that is, a and b have highest common factor 1), then b/2a is in S. (iii) If a/b and c/d are in S, where they are written in lowest terms, then a+c/b+d is in S. These rules are exhaustive: if these rules do not imply that a number is in S, then that number is not in S. Can you describe which numbers are in S? For example, by (i), 1/1 is in S. By (ii), since 1/1 is in S, 1/2·1 is in S. Since both 1/1 and 1/2 are in S, (iii) tells us 1+1/1+2 is in S. What I have so far: Claim: Set S in contained in interval [½, 1] for a/b where 0
The set S contains some real numbers, according to the following three rules.
(i) 1/1 is in S.
(ii) If a/b is in S, where a/b is written in lowest terms (that is, a and b have highest common factor 1), then b/2a
is in S.
(iii) If a/b and c/d are in S, where they are written in lowest terms, then a+c/b+d is in S.
These rules are exhaustive: if these rules do not imply that a number is in S, then that number is not in S. Can you describe which numbers are in S? For example, by (i), 1/1 is in S. By (ii), since 1/1 is in S, 1/2·1
is in S. Since both 1/1 and 1/2 are in S, (iii) tells us 1+1/1+2 is in S.
What I have so far:
Claim: Set S in contained in interval [½, 1] for a/b where 0<a≤b≤2a
The reason is that 1/1 has this form and transformations preserve the property of being in this interval
If a≤b≤2a, then b/2a obeys the requirement, since b≤2a≤2b
And if a/b and c/d obey the requirement, then so does (a+c)/(b+d), since a+c≤b+d≤2a+2c=2(a+c)
However, I feel there is still more to this question, and also I need a rigorous proof that supports my claims
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