Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to show that these sets are nonempty (here $\mid $ means "divides")?

Here N is an arbitrary large integer and q is some fixed integer.

$R = \lbrace k \in {\mathbb N}:(kN\mid k!) \wedge ((k - 1)N\mid k!) \wedge \cdots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

$S = \lbrace k \in {\mathbb N}:({(2k - 1)^2}N\mid k!) \wedge ({(2k - 3)^2}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

$T = \lbrace k \in {\mathbb N}:({k^5}N\mid k!) \wedge ({(k - 1)^5}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

They exist by the axiom schema of separation, but how do I determine which $k$ to choose so that it satisfies all the properties? Is there a general approach?

share|improve this question
1  
I don't think it is true for $S$. Let $p$ be a prime between $k$ and $2k$. (Such a $p$ exists by Betrand's Postulate.) Then $p\not\mid k!$, so definitely $p^2N\not\mid k!$ –  Thomas Andrews Jul 23 '12 at 19:10
    
Similarly, I don't think it is true that $T$ is non-empty, since if $p$ is a prime between $k/2$ and $k$ then $p^2\not\mid k!$ so definitely $p^5N\not\mid k!$ –  Thomas Andrews Jul 23 '12 at 19:13
    
That does not mean ${p^2}\nmid k!$, does it? –  glebovg Jul 23 '12 at 19:14
    
What does not mean it? If $p\not\mid k!$ then $p^2\not\mid k!$. @glebovg –  Thomas Andrews Jul 23 '12 at 19:16
    
Choose $k > {p^2}$, ${p^2}$ is an integer so ${p^2}\mid k!$. –  glebovg Jul 23 '12 at 19:19
show 15 more comments

1 Answer 1

up vote 2 down vote accepted

For example, for $R$, you want $k!/(k-j)$ to be a multiple of $N$ for each $j$ from $0$ to $k-1$. That will certainly be true if $k \ge 2N$.

share|improve this answer
    
Could you explain your reasoning in more detail? How would I apply the same reasoning to $S$ and $T$? Also, I think you mean $k>2N$ because $q$ is an arbitrary integer. –  glebovg Jul 23 '12 at 19:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.