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

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor$ for any positive integer $n$.
 
Mathematics news on Phys.org
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top