+0  
 
0
378
5
avatar

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!

 Jan 29, 2019
edited by Guest  Jan 29, 2019
 #1
avatar+29877 
+2

Try [3, 8)   i.e. open interval on the right.

 Jan 29, 2019
 #2
avatar+770 
+3

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
edited by PartialMathematician  Jan 30, 2019
 #3
avatar+29877 
+1

True!

Alan  Jan 30, 2019
 #4
avatar+7824 
+2

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].

 Feb 1, 2019
 #5
avatar+770 
+4

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 \(\boxed{[3,9)}\).


12 Online Users

avatar
avatar
avatar