Introductory analysis question

[synkk]

I've split the proof for part a) into 3 parts:

here's what I have thus far:

1) to prove H is an upper bound consider: $H < 0$ as an upper bound for S. We take $H = -1$ and hence $H^2 = 1$ and $1 \leq 3$. 1 is in the set S and hence $H < 0$is not an upper bound which implies that $H > 0$ is an upper bound.

2) consider $H^2 < 3$is an upper bound, and for N belonging to the natural numbers $H + 1/N$ is a rational number. Consider $(H+1/N)^2 = H^2 + 2H/N + 1/N^2 < 3$for N sufficiently large. This implies that $H + 1/N$ is in the set S, but$H + 1/N > H$ hence a contradiction therefore$H^2 < 3$ is not an upper bound and $H^2 \geq 3$ is.

3)if H is a rational number and $H >0$, $H^2 \geq3$ we see $H\geq \sqrt{3}$ as $H > 0$ thus all $H\geq \sqrt{3}$ are upper bounds as $x \leq \sqrt{3}$. That is all I have for part a),

is this correct?

for part b) here is what I have:
I called $H' = H - 1/N$ and tried to prove it by contradiction, i.e. assume$H - 1/N <\sqrt{3}$ but I got to $H < \sqrt{3} + 1/N$ and we know $H \geq \sqrt{3}$ from part a), and for N sufficiently large 1/N = 0, but I don't really think this is a good enough contradiction, for instance. $\sqrt{3}\leq H < \sqrt{3} + 1/N$ is ok, if N is say 2 or so. How would I go about contradicting this?

 

[Office_Shredder]

Part 1 is not good, as all you've shown is that -1 would be in the set if negative numbers were allowed in the set (which they aren't anyway).

Part 2 and 3 are basically OK but you have some language issues (which also show up in part 1). When you say

therefore $H^2 < 3$ is not an upper bound and $H^2 \geq 3$ is

you haven't said anything meaningful. $H^2 \geq 3$ isn't an upper bound, it's a condition which upper bounds must satisfy. A correct statement would be something like 'therefore if H is an upper bound, $H^2 \geq 3$.

For part b you basically want to copy your (2) idea. You know that if H is an upper bound of S in Q then H is not equal to $\sqrt{3}$ - or you should go ahead and prove that the square root of 3 is irrational - so H² > 3 strictly. Therefore (H-1/N)² > 3 if N is small enough (you have to prove this).

 

[synkk]

Part 1 is not good, as all you've shown is that -1 would be in the set if negative numbers were allowed in the set (which they aren't anyway).

Part 2 and 3 are basically OK but you have some language issues (which also show up in part 1). When you say

you haven't said anything meaningful. $H^2 \geq 3$ isn't an upper bound, it's a condition which upper bounds must satisfy. A correct statement would be something like 'therefore if H is an upper bound, $H^2 \geq 3$.

For part b you basically want to copy your (2) idea. You know that if H is an upper bound of S in Q then H is not equal to $\sqrt{3}$ - or you should go ahead and prove that the square root of 3 is irrational - so H² > 3 strictly. Therefore (H-1/N)² > 3 if N is small enough (you have to prove this).

Thank you for correcting my language issues, I understand my presentation is not very good, but I'm trying to improve.

I don't understand what you mean by (1)? I'm trying to show that if H is an upper bound, then H>0 and H^2 > 3, with 1), I'm using contradiction by assuming H<=0 is the upper bound then going on to show that this is impossible hence H>0 must be an upper bound. Then from 2) I use a similar argument to show H^2 >= 3 must be an upper bound (by contradiction). Could you explain where I've gone wrong here?

 

[synkk]

OK for part b) here is what I got so far

Assume that $(H-1/N)^2 < 3$ then we have $H < \sqrt{3} + 1/N$ for $N \in \mathbb{N}$ as $H > \sqrt{3}$ we get $\sqrt{3} < H < \sqrt{3} + 1/N$ and for N very large (specifically for $N > \frac{1}{H-\sqrt{3}}$ ) we get $1/N$ tending to 0, hence $\sqrt{3} < H < \sqrt{3}$ which is a contradiction hence $(H-1/N)^2 > 3$

I'm not sure if this is correct, but that's all I can get so far from it, please advise.

 

[DarthRoni]

I haven't looked at a problem like this in a while. I came up with this idea.
Take $P,H\in \mathbb{Q}$ Such that $P>\sqrt{3}>H$ and assume
$$H + \frac{1}{N} \notin (H, P) \forall\ N\in\mathbb{N}\implies H = P = \sqrt{3}$$
since for all rational numbers, take $x,y \in \mathbb{N}$
$$\frac{x}{y} + \frac{1}{2y} < \frac{x+1}{y}$$
or
$$\frac{x}{y} + (\frac{1}{2})(\frac{1}{y}) < \frac{x}{y} + \frac{1}{y}$$
I believe this proves both a and b

 

[DarthRoni]

Hmm I'm not sure if this works but I believe my proof is incomplete.
Assume $P$ is the smallest rational number greater than $\sqrt{3}$
$$\implies \exists Z\in\mathbb{Q}\ and\ Z\in [H,P]\ |\ Z < \sqrt{3} \implies 3 > Z^2 > H^2$$
I believe we can make the assumption the other way in order to prove that there is an infinite amount of rational numbers across $[H,P]$
I'm not sure if this is allowed, any feedback would be great !