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Introductory analysis question

  1. Oct 15, 2013 #1
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    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: [itex]H < 0[/itex] as an upper bound for S. We take [itex]H = -1[/itex] and hence [itex]H^2 = 1[/itex] and [itex]1 \leq 3[/itex]. 1 is in the set S and hence [itex]H < 0 [/itex]is not an upper bound which implies that [itex]H > 0[/itex] is an upper bound.

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

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

    is this correct?

    for part b) here is what I have:
    I called [itex]H' = H - 1/N[/itex] and tried to prove it by contradiction, i.e. assume[itex]H - 1/N <\sqrt{3}[/itex] but I got to [itex]H < \sqrt{3} + 1/N[/itex] and we know [itex]H \geq \sqrt{3}[/itex] 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. [itex]\sqrt{3}\leq H < \sqrt{3} + 1/N[/itex] is ok, if N is say 2 or so. How would I go about contradicting this?
     
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  3. Oct 15, 2013 #2

    Office_Shredder

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    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. [itex] H^2 \geq 3 [/itex] 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, [itex] H^2 \geq 3[/itex].

    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 [itex] \sqrt{3}[/itex] - 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).
     
  4. Oct 15, 2013 #3
    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?
     
  5. Oct 15, 2013 #4
    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.
     
  6. Oct 15, 2013 #5
    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
     
    Last edited: Oct 15, 2013
  7. Oct 15, 2013 #6
    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 !
     
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