Hey guys,

I'm having a little trouble with these two questions. It would be great if you could lend me a hand

p1.

By the archimedian property, there is some n N such that,

Therefore,

(1,5)

I don't understand why

Also why does (1,5) ?

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- Apr 1st 2013, 12:19 PMrokman54Archimedian Property Question
Hey guys,

I'm having a little trouble with these two questions. It would be great if you could lend me a hand

p1.

By the archimedian property, there is some n N such that,

Therefore,

(1,5)

I don't understand why

Also why does (1,5) ? - Apr 1st 2013, 12:31 PMemakarovRe: Archimedian Property Question
I would do this problem as follows. Let x ∈ (1, 5) and let e = min(x - 1, 5 - x) > 0. Then there exists an n such that 1/n < e. Therefore, x ∈ [1 + e, 5 - e] ⊆ [1 + 1/n, 5 - 1/n] ⊆ ∪

_{k=1}^{∞}[1 + 1/k, 5 - 1/k]. The converse inclusion is obvious because [1 + 1/k, 5 - 1/k] ⊆ (1, 5) for every k. - Apr 1st 2013, 12:38 PMrokman54Re: Archimedian Property Question
Why did you set e as the minimum of (x-1 , 5-x)?

Also can you elaborate on how x ∈ [1 + e, 5 - e] ⊆ [1 + 1/n, 5 - 1/n] ⊆ ∪k=1∞[1 + 1/k, 5 - 1/k] shows that (1,5) ⊆ ∪k=1∞[1 + 1/k, 5 - 1/k]? - Apr 1st 2013, 12:38 PMPlatoRe: Archimedian Property Question
- Apr 1st 2013, 12:51 PMemakarovRe: Archimedian Property Question
Because then e ≤ x - 1

*and*e ≤ 5 - x. The first inequality implies that 1 + e ≤ x, and the second one implies that x ≤ 5 - e. Therefore, x ∈ [1 + e, 5 - e]. But since 1/n ≤ e, it follows that 1 + 1/n ≤ 1 + e and 5 - e ≤ 5 - 1/n. Thus, [1 + e, 5 - e] ⊆[1 + 1/n, 5 - 1/n].

I started by assuming x ∈ (1, 5) and concluded that x ∈ ∪_{k=1}^{∞}[1 + 1/k, 5 - 1/k]. This shows that (1, 5) ⊆ ∪_{k=1}^{∞}[1 + 1/k, 5 - 1/k].