Let p(x) be defined on \(2 \le x \le 10\) such that

{ x + 1 if \(\left \lfloor{x}\right \rfloor \) is prime

p(x) = { p(y) + (x + 1 - \(\left \lfloor{x}\right \rfloor \)) otherwise

where Y is the greatest prime factor of \(\left \lfloor{x}\right \rfloor \). Express the range of \(p\) in interval notation.

I tried [3, 8] because when x = 2, p(x) = 3, and when x = 7, p(x) = 8. I thought that was the range, but it is apparently not.

Please help, thanks in advance!

Guest Jan 29, 2019

edited by
Guest
Jan 29, 2019

#1

#2**+1 **

I tried that for guest, and I am pretty sure it is incorrect. I think p(x) can equal 8 because if x = 7, p(x) = 7+1 = 8.

PartialMathematician
Jan 30, 2019

#4**+1 **

If you only consider the integer values, then the range is [3,8].

Consider a number infinitesimally close to 8. (7.9999999999999999999999...)

p(7.9999999999999999999...) = x + 1 = 8.999999999999999999999... = 9

So the range is actually [3,9].

MaxWong Feb 1, 2019

#5**+1 **

Oh, I see what you are saying. I thought the first condition with x+1 if floor x is prime was saying: If floor x is prime, then p(x) = floor x + 1.

That explains a lot. But again, the number can be infinitely close to 8, such as 7.99999..., but the number can never be 8, so the answer range is \(\boxed{[3,9)}\).

PartialMathematician
Feb 2, 2019