- #1

synkk

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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?