reserve p,q for Rational;
reserve g,m,m1,m2,n,n1,n2 for Nat;
reserve i,i1,i2,j,j1,j2 for Integer;

theorem Th15:
  m <> 0 implies
  denominator(i/m) = m div ( i gcd m ) &
  numerator(i/m) = i div ( i gcd m )
  proof
    set p = i/m;
    set g = i gcd m;
    set d = m div g;
    assume
A1: 0 <> m;
    then
A2: g <> 0 by INT_2:5;
    g divides m by INT_2:def 2;
    then g <= m by A1,INT_2:27;
    then
A3: d <> 0 by A2,PRE_FF:3;
A4: p = (i div g) / d by Th10;
    now
      given w being Nat such that
A5:   1 < w and
A6:   ex i1,m1 st i div g = i1*w & m div g = m1*w;
      consider i1,m1 such that
A7:   i div g = i1*w and
A8:   d = m1*w by A6;
      w divides i div g & w divides m div g by A7,A8;
      then
A9:   w divides (i div g) gcd (m div g) by INT_2:def 2;
      (i div g) gcd (m div g) = 1 by A1,Th14;
      hence contradiction by A5,A9,WSIERP_1:15;
    end;
    hence thesis by A3,A4,RAT_1:30;
  end;
