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For every positive integer $n$, prove that $$\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$$

Hence or otherwise, prove that $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$, where $[x]$ denotes the greatest integer not exceeding $x$.

This question was posed to me in class by my teacher. Since we are discussing quadratic equations. I can only imagine that this question is related to that topic.

Actually, by squaring the terms on both sides of the inequality, the first part of the question is solved easily. It is the second half that is causing me trouble.

Clearly we have to show, that if $x <\sqrt{4n+1} < x+1$ where $x$ is a natural number, then $x <\sqrt{n} + \sqrt{n+1} < x+1$, but how?

I am in high school, so please use techniques appropriate for my level.

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Clearly $\lfloor\sqrt{4n+1}\rfloor\le\lfloor\sqrt n+\sqrt{n+1}\rfloor\le \lfloor\sqrt{4n+2}\rfloor$. Thus the claim could only be wrong if $\lfloor\sqrt{4n+1}\rfloor< \lfloor\sqrt{4n+2}\rfloor$, i.e. if there exist an ineger $m$ with $\sqrt {4n+1}<m\le\sqrt {4n+2}$, equivalently: with $4n+1<m^2\le 4n+2$. But we cannot have $m^2=4n+2$ because the latter is even but not divisible by $4$, whereas $m^2$ is either odd or divisible by $4$. Thus the integer $m^2$ is between $4n+1$ and $4n+2$, which is absurd.

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Observe that $$\lfloor\sqrt{4n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$$

unless $4n+2$ is perfect square

But any square $\equiv0,1\pmod4$

$$\implies\lfloor\sqrt{4n+1}\rfloor=\lfloor\sqrt n+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$$

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  • $\begingroup$ That is true. But I already figured that out as I have mentioned. But how to prove that [\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}] $\endgroup$ – user144220 Apr 19 '14 at 12:03
  • $\begingroup$ @user144220, Please find the edited version $\endgroup$ – lab bhattacharjee Apr 19 '14 at 12:17
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$$\sqrt n+\sqrt{n+1}>\sqrt{4n+1}\\n+2\sqrt{n(n+1)}+n+1>4n+1\\\sqrt{n(n+1)}>n\\n^2+n>n^2\\n>0\\\sqrt{4n+2}>\sqrt n+\sqrt{n+1}\\4n+2>2n+1+2\sqrt{n(n+1)}\\2n+1>2\sqrt{n(n+1)}\\4n^2+4n+1>4n^2+4n\\1>0$$

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