MHB Can you prove this floor function challenge involving square roots?

AI Thread Summary
The discussion centers on proving the equality $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor$ for all positive integers $n$. Participants share various approaches to demonstrate this mathematical statement, emphasizing the use of properties of square roots and floor functions. The conversation highlights the importance of rigorous proof techniques in number theory. Contributions include detailed calculations and logical reasoning to validate the equality. Overall, the challenge fosters engagement and collaboration among members interested in mathematical proofs.
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Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor$ for any positive integer $n$.
 
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anemone said:
Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor$ for any positive integer $n$.

LHS = $\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor $
= $\lfloor\sqrt{(\sqrt{n}+\sqrt{n+1})^2}\rfloor $
= $\lfloor\sqrt{n+n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{(n+\dfrac{1}{2})^2 + \dfrac{3}{4}}}\rfloor $
= $\lfloor\sqrt{2n+1+\sqrt{(2n+1)^2 + 3}}\rfloor $

now realising that integral part of square root of x and square root of x + t where t is less than 1 are sameso we need to find the integral part of $\sqrt{(2n+1)^2 + 3}$

$(2n+1)^2 + 3\gt(2n+1)^2$
and $(2n+1)^2 + 3\lt(2n+2)^2$ as $(2n+2)^2-(2n+1)^2 = 4n + 3$so integral part of $\sqrt{(2n+1)^2 + 3}$ = (2n + 1)

so LHS = $\lfloor\sqrt{2n+1+2n+1}\rfloor $
= $\lfloor\sqrt{4n+2}\rfloor $
= RHS
 
Last edited:
kaliprasad said:
LHS = $\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor $
= $\lfloor\sqrt{(\sqrt{n}+\sqrt{n+1})^2}\rfloor $
= $\lfloor\sqrt{n+n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{(n+\dfrac{1}{2})^2 + \dfrac{3}{4}}}\rfloor $
= $\lfloor\sqrt{2n+1+\sqrt{(2n+1)^2 + 3}}\rfloor $

now realising that integral part of square root of x and square root of x + t where t is less than 1 are sameso we need to find the integral part of $\sqrt{(2n+1)^2 + 3}$

$(2n+1)^2 + 3\gt(2n+1)^2$
and $(2n+1)^2 + 3\lt(2n+2)^2$ as $(2n+2)^2-(2n+1)^2 = 4n + 3$so integral part of $\sqrt{(2n+1)^2 + 3}$ = (2n + 1)

so LHS = $\lfloor\sqrt{2n+1+2n+1}\rfloor $
= $\lfloor\sqrt{4n+2}\rfloor $
= RHS

Well done and thanks for participating, kaliprasad!(Yes)

Here's another solution that I want to share with MHB:

It's easy to verify that

$\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+3}$ where $n\in N$

Neither $4n+2$ nor $4n+3$ are squares, so $\left\lfloor{\sqrt{4n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor=\left\lfloor{\sqrt{4n+3}}\right\rfloor$ and the result follows.
 
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