Find the supremum of
$$\sqrt{n+1} - \sqrt{n} $$
where $n$ is in the set of natural numbers. It's a silly question I am stuck at. Is the answer $1$ or is it $\sqrt{2} -1$? There's a bit of confusion. Please help out.
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2is it $\sqrt{n+1-\sqrt{n}}$ or $\sqrt{n+1}-\sqrt{n}$? – VIVID Aug 17 '20 at 12:42
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6Is $0$ a natural number or not? – lulu Aug 17 '20 at 12:43
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@VIVID It looks like $\sqrt{n}+1-\sqrt{n}=1$, but, well, ...then the answer would be $1$ at least. – Dietrich Burde Aug 17 '20 at 12:44
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1Error in the title, different error in the question, and error when you say that $n$ is a set. Make an effort, Shatabdi! But there is certainly confusion about what "natural number" means $-$ does it include $0$ or not? (See my answer here). If the question doesn't specify either way, then it is flawed. – TonyK Aug 17 '20 at 12:53
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@TonyK: But, in axiomatic set theory, natural numbers are sets! – Bernard Aug 17 '20 at 12:58
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@Bernard: Very clever. But $n$ is not the set of natural numbers. – TonyK Aug 17 '20 at 13:18
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Certainly– $\mathbf N$ is the union of all finite ordinals. – Bernard Aug 17 '20 at 13:30
2 Answers
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$$ \sqrt{n+1}-\sqrt{n} =\left(\sqrt{n+1}-\sqrt{n}\right) \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} =\frac{1}{\sqrt{n+1}+\sqrt{n}} = f(n)$$ Since $f$ is decreasing, the supremum is $f(0)=1$ if $0 \in \mathbb N$ and $f(1)= \sqrt 2 - 1$ if $0 \notin \mathbb N$.
lhf
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This sequence is strict descenting sequence so then if you are dealing with natural numbers the maximum or supremum of this sequence are the same and happens when $n = 1$ and equals to $\sqrt 2 - 1$.
