sequences?

[Sarahisme]

Hey all, yet another question

 

Suppose an → a, bn → b and an ≤ bn ultimately. Prove that a ≤ b.

 

i am not sure how to do this.... if anyone has some advice or could show me how, it would be greatly appriciated

 

Sarah

[Dapthar]

Suppose an → a' date=' bn → b and an ≤ bn ultimately. Prove that a ≤ b.

[/quote']Here's a small hint. From it you should be able to work out the details of the proof.

 

Try a proof by contradicition. Assume the following:

 

[math]\lim_{n\to \infty}a_n = a[/math], [math]\lim_{n\to \infty}b_n = b[/math], and that [math]\forall n \in \mathbb{Z},[/math][math] a_n \leq b_n[/math]. However, assume that [math] a > b[/math].

 

Now, recall the definition of a limit of a sequence, and think carefully about exactly why if the following two statements are true:

• [math]\lim_{n\to \infty}a_n = a[/math], [math]\lim_{n\to \infty}b_n = b[/math]

 

• [math]\forall n \in \mathbb{Z},[/math][math] a_n \leq b_n[/math]

 

Then [math]a[/math] cannot be greater than [math]b[/math].

 

When you figure that out, you have answered your question.

 

If you have any more questions, feel free to ask.

[Sarahisme]

ok, i have to go to bed now, but tomorrow i will have a good think about it, but just reading it then even put a few ideas into my head

 

Thanks Dap

[Sarahisme]

but arn't we tring to prove an<=bn? if so, haven't we assumed that to begin with?

[Dapthar]

but arn't we tring to prove an<=bn?

Nope, we're trying to prove that [math]a \leq b[/math].

 

Suppose you wish to prove something of the form [math]a\implies b[/math] (in your problem, [math]a[/math] is "[math]\lim_{n\to \infty}a_n = a[/math], [math]\lim_{n\to \infty}b_n = b[/math], and that [math]\forall n \in \mathbb{Z},[/math][math] a_n \leq b_n[/math]", and [math]b[/math] is "[math] a \leq b[/math]").

 

Proofs by contradiction 'assume' that [math]\neg (a \implies b)[/math] is true, which is equivalent to assuming that [math]a \wedge \neg b[/math] is true (I use the [math]\neg[/math] symbol for 'not', and [math]\wedge[/math] for 'and'.). Thus, if by assuming [math]a[/math] is true and [math]b[/math] is false, we can show that [math]a \wedge \neg b[/math] is false, then that implies that [math]\neg (a \implies b)[/math] is false, which means that [math]a \implies b[/math] is true, which is what we originally wanted.

[Sarahisme]

oh right i see now thanks dap