- #1
glebovg
- 164
- 1
How to show that these sets are nonempty ([itex]\mid [/itex] means "divides")?
Here N is an arbitrary large integer and q is some fixed integer.
[itex]{R_{k,q}} = \{ k \in {\mathbb N}:(kN\mid k!) \wedge ((k - 1)N\mid k!) \wedge \cdots \wedge (N\mid k!) \wedge (k > Nq)\}[/itex]
[itex]{S_{k,q}} = \{ k \in {\mathbb N}:({(2k - 1)^2}N\mid k!) \wedge ({(2k - 3)^2}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\}[/itex]
[itex]{T_{k,q}} = \{ k \in {\mathbb N}:({k^5}N\mid k!) \wedge ({(k - 1)^5}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\}[/itex]
They exist by the axiom schema of separation, but how do I determine which [itex]k[/itex] to choose so that it satisfies all the properties? Is there a general approach?
Here N is an arbitrary large integer and q is some fixed integer.
[itex]{R_{k,q}} = \{ k \in {\mathbb N}:(kN\mid k!) \wedge ((k - 1)N\mid k!) \wedge \cdots \wedge (N\mid k!) \wedge (k > Nq)\}[/itex]
[itex]{S_{k,q}} = \{ k \in {\mathbb N}:({(2k - 1)^2}N\mid k!) \wedge ({(2k - 3)^2}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\}[/itex]
[itex]{T_{k,q}} = \{ k \in {\mathbb N}:({k^5}N\mid k!) \wedge ({(k - 1)^5}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\}[/itex]
They exist by the axiom schema of separation, but how do I determine which [itex]k[/itex] to choose so that it satisfies all the properties? Is there a general approach?
Last edited by a moderator: