A beautifoul thing!

[fu^2]

proof that $sum_(k=0)^(n-1)[alpha+k/n]=[nalpha]$ where $alphainRR^+$, $ninn$ and $alpha,n>0$.

NB $[gamma]$ is the max integer minor of $gamma$, for example $[2.89]=2$.

proof this using the inductive's method:

i) if n=1 $[alpha]=[alpha]$

ii) i suppose true for $n-1$.

iii) show the fact for n

$sum_(k=0)^n[alpha+k/(n+1)]= sum_(k=0)^(n-1)[alpha+k/n]+[alpha+n/(n+1)] =$ using the induction's hp $=[nalpha]+[alpha+n/(n+1)]$

$alpha$ is real numerd, so i can write $alpha$ as $alpha=C_0+C_1/10+C_2/100+...$

so

$nalpha=nC_0+nC_1/10+nC_2/100+...$ and $alpha+n/(n+1)=n/(n+1)+C_0+C_1/10+C_2/100+...=((n+1)C_0+n)/(n+1)+C_1/10+C_2/100+...

i can suppose that $nC_1/10<1$ for simplicity (and because is equal: if this number is >9, i must change the integer component).

now i can white $[nalpha]=nC_0$ (because i've just supposed that $nC_1/10<1$) and $[alpha+n/(n+1)]=((n+1)C_0+n)/(n+1)

so

$[nalpha]+[alpha+n/(n+1)]=nC_0+((n+1)C_0+n)/(n+1)=(C_0(n+1)^2+n)/(n+1)=C_0(n+1)+n/(n+1)$

but $n/(n+1)<1$ so the integer part of the result is $C_0(n+1)=[alpha(n+1)]$, because i'd supposed that $nC_1/10<1$).

cvd

[mathefan]

beautifoul?!
sorry.. what is this??

BEAUTIFUL is right.

[TomSawyer1]

https://www.matematicamente.it/f/viewtop ... 933#109933

[fu^2]

"luca.barletta":[quote="fu^2"] $ninnn$

I suppose $n in NN$[/quote]

yes, of course!

[_luca.barletta]

"fu^2": $ninnn$

I suppose $n in NN$