Question

A rational number such that its denominator is greater then its numerator by a constant k

Answer 1

Answer:

Step-by-step explanation:

or this could be called a trick question.

n= the numerator

d=n%2BK= the denominator

2%2Ad=2%28n%2BK%29= the denominator, doubled

 

The original number is n%2Fd .

The number with the denominator doubled (just the denominator) is n%2F2d .

For both of those numbers to be equal, the only solution is n=0 .

That would make the original number 0%2FK=0 ,

and the number with the denominator doubled would be 0%2F2K=0.