Being x_n a real number sequence, i'm really interested in the fact that in order to show that x_n is unbounded we can either show that " for all M in R, there exists k with x_k > M." or that "for all n in N, x_n >= n "

Maybe the latter implies the former but the former doesnt imply the later ?

To prove that inductively defined sequences are unbounded it seems like only one of these definitions work ( the last one ) , the other doesn't make sense

Why ?

@nerdy The second statement is too much to ask, in general. Suppose $x_n=\ln n$.

It's just that i want to prove that inductively defined sequences are unbounded, but since it is gonna have to be proof by induction it doesn't make sense to try and prove the first statement ..

So, i guess i only have the second statement to prove ?

what is an "inductively defined sequence"

In proving that a inductively defined sequence satisfies the second statement, am i proving that it is unbounded ?

like $(x_(n+1) = 3.(x_n)^2$ with $x_1$ = 1

oh, you mean a particular sequence

not in general

gotcha

no, i don't mean a particular sequence

i just mean sequences that are recursively defined

for example $x_n = n^2 $ isn't

while the one i gave you is

then I can hand you a recursively defined sequence that is not unbounded

I'm trying to understand the steps i need to take to prove a inductively defined is unbounded

since trying to prove the negation of the definition of boundedness doesnt work for them

the negation of definition of boundedness would be my first statement : " for all M in R, there exists k with x_k > M "

Hi @Mike ...

@nerdy: Suppose $x_{n+1}=x_n/2$.

Yes, @nerdy, your first statement is correct.

oops, I've been beaten to the punch

hello @Ted

@MikeMiller I think I have it. $\Bbb F_p(X)^\times$ has the infinite cyclic group $\langle X \rangle \cong \Bbb Z$ sitting inside it.

and?

@MikeMiller And, $\Bbb F_p(X)^\times$ is abelian.

Aaaand?

Someone unaccepted my answer. Now I can't even remember which question it was for.

@BalarkaSen I think you may have forgotten the question

You're right. Where were we?

"show that $F^\times$ is finitely generated iff $F$ is finite"

That $F^\times$ is finite iff $F$ is finite... follows from the fact that $F^\times = F \setminus \{0\}$ as sets :)

OK, so we need to prove $\Bbb F_p(X)^\times$ is not finitely generated.

aye

Well, it has infinitely many $\Bbb Z$s sitting inside it.

$\langle X + a \rangle$

prove it

err

what's a, here?

and... oh noes. $a \in \Bbb F_p$. rats.

would you like a hint

there are only finitely many such $a$s.

@Mike No, no, no.

never never never

ok

Besides, you can't help me by giving out hints.

cute

@BalarkaSen He should not wear a tie.

@MikeMiller how about $\langle f(X) \rangle$ where $f(X)$s are irreducible polynomials on $\Bbb F_p[X]$? There are infinitely many of them (they are all prime ideals of $\Bbb F_p[X]$, so we can apply Euclid's proof here)

@BalarkaSen You've got it. The hint I was going to give was: "How did you prove $\Bbb Q^\times$ wasn't finitely generated?" (By picking irreds $p_i$ in $\Bbb Z$ and considering $1/p_i$... :) )

YES

Thus all such fields are algebraic extensions of $\Bbb F_p$, hence finite.

Mmm. Careful. There are infinite algebraic extensions of $F_p$.

Did we deal with those already?

Right, we didn't.

Well, lemme know, @BalarkaSen

hello @mixedmath

Coincidentally, I have a question about moderation: this user just "defaced" his own closed question (it was a duplicate) by switching nouns to, uh, "not noun". Since it's closed, do we just ignore that?