Question

Rational numbers ,Set theory

Answer 1

Let \(S\) be the set of rational numbers with the following properties:

\(I.\quad \dfrac{1}{2}\in S\): \(II.\quad \text{If } x\in S,\text{then both } \dfrac{1}{x+1}\in S \text{ and } \dfrac{x}{x+1} \in S\)

Which of the following is true ?

\((a)\) \(S\) contains all rational numbers in the interval \(0<x<1\)

\((b)\) \(S\) contains all rational numbers in the interval \(-1<x<1\)

\((c)\) \(S\) contains all rational numbers in the interval \(-1<x<0\)

\((d)\) \(S\) contains all rational numbers in the interval \(1<x<\infty\)

Answer 2

do you think the intervals define S? or a subset of S?

Answer 3

they define \(S\)

Answer 4

then 2 of them dont have 1/2 in them to start with

Answer 5

the other 2 differ by a negative ... make a generalization some -a/b and see what works out .. is my thought process

Answer 6

ok option \(c.\) and \(d.\) are kicked off

Answer 7

if the intervals define S infull, then c and d are simply not acceptable since 1/2 is in our domain

Answer 8

a and b differ in that -1/2 is a member of b; does -1/2 fit the setup any?

Answer 9

why did u choose \(-\dfrac{1}{2}\)

Answer 10

because its a rational number between -1 and 0 ....

Answer 11

if we are going to determine of we are just

a: rationals from 0 to 1
or
b: rationals from -1 to 1

then the difference between then is b contains rationals from -1 to 0

Answer 12

it simply a test point

Answer 13

we dont have a 'none of the above' option so one of them has to be right ...

Answer 14

\(II.\quad \text{If } -\dfrac{1}{2}\in S,\text{then both } \dfrac{1}{-1/2+1}=2\in S \text{ and } \dfrac{-1/2}{-1/2+1}=-1 \in S\)

Answer 15

so, pass or fail?

Answer 16

\(2\cancel{\in} S\) is not in option \((b.)\)

Answer 17

good, then weve got only one option to pick now :)

Answer 18

thanks

Answer 19

youre welcome