What is the range of the function $$f(x,y,z):=\left \{\frac{xyz}{xy+xz+yz} \right \}$$ over all the natural numbers $x,\,y,\,z$ (Zero does not belong to the naturals.), where $\{x\}$ stands for the fractional part of a real number $x$?
Asked
Active
Viewed 61 times
2
-
are you sure you want natural numbers, as opposed to positive real numbers? In the case of the former, it will be a subset of the rationals. – Calvin Lin Apr 25 '14 at 20:16
-
Thank you for the interest to the question. I am sure that the range is a subset of $\mathbb{Q} \cap [0,1]$. The question is "What is that set?" – user64494 Apr 25 '14 at 20:40
-
Thanks for that clarification,it helps knowing that they are rational. Why must it be less than 1? E.g. take $x=y=z=10$. – Calvin Lin Apr 26 '14 at 00:03
1 Answers
2
The range is all rationals in $(0,1).$ We have for $t\ge 1$ $$f(2k,2k,tk)=\frac{kt}{t+1}.\tag{1}$$ We first claim that for each $n>1$ the rational $\frac1n$ is in the range. To see this put $n=t+1$ so that $\gcd(n,t)=1$ and one may find $k$ for which $kt$ is $1$ mod $n.$ That is, $kt=1+nw$ for some positive integer $w$, and then the right side of $(1)$ becomes $$\frac{1+nw}{n}=\frac{1}{n}+w,$$ which has fractional part $\frac1n$ as desired.
Once we have obtained a value with fractional part $\frac1n$ we may go from the relation $f(x,y,z)=m+\frac1n$ to the relation $f(kx,ky,kz)=km+\frac{k}{n}$ and thus we have also $\frac kn$ in the range, for $1 \le k < n.$
Added: For an explicit triple to cover $k/n$ where $n\ge 2$ and $0<k<n$ the above can be worked out to get $$f(2k(n-1),2k(n-1),k(n-1)^2)=\frac{k}{n}+k(n-2)$$ which has fractional part $\frac kn.$ Note also that using $k=n$ here gives a fractional part $0$, so that the actual range of $f$ is the set of rationals in $[0,1).$ [Thanks to @user64494 for pointing out $0$ is in the range.]
coffeemath
- 29,884
- 2
- 31
- 52
-
1
-
@user64494 Yes, it can achieve zero since it is fractional part and one can achieve some positive integers. +1 on comment. – coffeemath Apr 26 '14 at 17:50