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MathDude
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I've been having trouble partitioning number systems into sets. The complex and rational number systems blow me away, so I'll stick with all reals, integers, and naturals for now.
1a) With five sets of infinitely many positive integers, partition the set of all real numbers.
1b) With five infinite sets, partition the set of all real numbers.
2a) With infinitely many infinite sets, partition the set of all real numbers/
2b) With infinitely many infinite sets, partition the set of integers.
(see above)
1a)
A1 = { k \in R>0 : k \equiv 0(mod 5) }
A2 = { k \in R>0 : k \equiv 1(mod 5) }
A3 = { k \in R>0 : k \equiv 2(mod 5) }
A4 = { k \in R>0 : k \equiv 3(mod 5) }
A5 = { k \in R>0 : k \equiv 4(mod 5) }
1b) Same as above except k \in R.
2a) Using Fundamental Thm of Arithmetic,
Sp = {pn : n \in N, p is prime)
Let L, Sp, ... partition R.
where L = R - \bigcupSp
2b) Same as above except instead of L = R -..., R is instead Z (set of integers).
Homework Statement
1a) With five sets of infinitely many positive integers, partition the set of all real numbers.
1b) With five infinite sets, partition the set of all real numbers.
2a) With infinitely many infinite sets, partition the set of all real numbers/
2b) With infinitely many infinite sets, partition the set of integers.
Homework Equations
(see above)
The Attempt at a Solution
1a)
A1 = { k \in R>0 : k \equiv 0(mod 5) }
A2 = { k \in R>0 : k \equiv 1(mod 5) }
A3 = { k \in R>0 : k \equiv 2(mod 5) }
A4 = { k \in R>0 : k \equiv 3(mod 5) }
A5 = { k \in R>0 : k \equiv 4(mod 5) }
1b) Same as above except k \in R.
2a) Using Fundamental Thm of Arithmetic,
Sp = {pn : n \in N, p is prime)
Let L, Sp, ... partition R.
where L = R - \bigcupSp
2b) Same as above except instead of L = R -..., R is instead Z (set of integers).
