Defining Elements on the Real Line?

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RE-EDIT: I'm confused again, continue on reading Bonjourno, I'm trying to work at the lectures provided on youtube by nptelhrd but I've gotten my foot stuck in a hole in the real line only 15 minutres into it

(In my head I say "such that" whenever the symbol : pops up!).

We define a set;

A : {r ∈ Q: r²<2}

Does ∃ a largest element of A in Q?

1: We seek to find some n ∈ N : $$( r \ + \ \frac{1}{n} )$$ will satisfy the conditions specified by A.

2: $$(r \ + \ \frac{1}{n} )^2 \ < \ 2$$

3: $$r^2 \ + \ \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ 2$$

4: $$\frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ 2 \ - \ r^2$$

The R.H.S. is strictly positive due to r²<2.

Okay, I understand up to here but then the lecturer starts to get confusing, he then says that It suffices only to find some n ∈ N :

$$\frac{2r}{n} \ + \ \frac{1}{n} \ < \ 2 \ - \ r^2$$

Notice the n and not n² on the bottom of the L.H.S. Fraction!

He says;

This is because;

[tex] \frac{1}{n^2} < \frac{1}{n} \ and \ this \ implies \ \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ \frac{2r}{n} \ + \ \frac{1}{n} [/itex]

I have no idea where this came from!The video is on youtube http://www.youtube.com/watch?v=0lzO...DB30C539B&playnext_from=PL&index=0&playnext=1 and I would say everything he is trying to do is described from 10:00 to 14:00.

I would extremely appreciate it if someone could take 6 minutes to watch this and correct me as I have nobody else to explain it to me.

What I think is going on is that he is trying to prove a least upper bound or something and that this will show that the real line can be continuously divided, or something.

 

[HallsofIvy]

RE-EDIT: I'm confused again, continue on reading


Bonjourno, I'm trying to work at the lectures provided on youtube by nptelhrd but I've gotten my foot stuck in a hole in the real line only 15 minutres into it

(In my head I say "such that" whenever the symbol : pops up!).

We define a set;

A : {r ∈ Q: r²<2}

Does ∃ a largest element of A in Q?

1: We seek to find some n ∈ N : $$( r \ + \ \frac{1}{n} )$$ will satisfy the conditions specified by A.

2: $$(r \ + \ \frac{1}{n} )^2 \ < \ 2$$

3: $$r^2 \ + \ \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ 2$$

4: $$\frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ 2 \ - \ r^2$$

The R.H.S. is strictly positive due to r²<2.

Okay, I understand up to here but then the lecturer starts to get confusing, he then says that It suffices only to find some n ∈ N :

$$\frac{2r}{n} \ + \ \frac{1}{n} \ < \ 2 \ - \ r^2$$

Notice the n and not n² on the bottom of the L.H.S. Fraction!

He says;

This is because;

[tex] \frac{1}{n^2} < \frac{1}{n} \ and \ this \ implies \ \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ \frac{2r}{n} \ + \ \frac{1}{n} [/itex]

I have no idea where this came from!

For any integer n> 1, $1> \frac{1}{n}$ (divide both sides of n> 1 by the positive number n) so $\frac{1}{n}> \frac{1}{n^2}$ (divide both sides by n again). Now, adding any number a to both sides, $a+ \frac{1}{n}> a+ \frac{1}{n^2}$. In particular, if $a= \frac{2r}{n}$, $\frac{2r}{n}+ \frac{1}{n}> \frac{2r}{n}+ \frac{1}{n^2}$.

The video is on youtube http://www.youtube.com/watch?v=0lzO...DB30C539B&playnext_from=PL&index=0&playnext=1 and I would say everything he is trying to do is described from 10:00 to 14:00.

I would extremely appreciate it if someone could take 6 minutes to watch this and correct me as I have nobody else to explain it to me.

What I think is going on is that he is trying to prove a least upper bound or something and that this will show that the real line can be continuously divided, or something.

 

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Thanks a lot HallsofIvy for clearing that up!