Consider the set of all fractions \(\frac{x}{y}\)

Where x and y are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by 1, the value of the fraction is increased by 10%?

Guest Sep 10, 2017

2+0 Answers


Here is one such fraction: 5/11.  [5+1] / [11+1] =6/12 = 1/2 which is 10% > 5/11.

Guest Sep 10, 2017

Cantor is spinning in his grave.....  I am willing to stand corrected but -

I think the answer is that there is an infinite number of co-primes that satisfy this. I can write the numerator x  as the product of factors

(a1*a2*a3....an)  and the denominator y  as the product (b1*b2*b3...bm)   where no a's or b's are equal,to guarantee that I have co-primes.  So

x/y =(a1*a2*a3...an) / (b1*b2*b3...bm)              {  n  and m to provide that  x and y don't need to have the same number of factors}             Clearly I can keep choosing more and more a's  and b's to generate more and more co-primes  ie an infinite set.        What about (x+1)/(y+1)  ?     This is

(a1*a2*a3...an  + 1) /(b1*b2*b3....bm  +1)   and we are requiring from these infinite sets that


(a1*a2*a3....an  +1)/(b1*b2*b3.....bm +1)   =(1.1) (a1*a2*a3....an)/(b1*b2*b3....bm) .   Again,every time I reach a solution,I can choose more a's and b's  until I find another one   ad infinitum.   

frasinscotland  Sep 10, 2017

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