Primarily, let the improper fraction be written as a/b, where a > b, and both a and b are positive integers. Also, let c be the value that we wish to subtract and divide from the fraction. Using this notation, we can form the equation:
(a/b) - c = (a/b) / c
Finding a common denominator in the LHS, and evaluating the RHS, we can simplify the above equation into:
(a/b) - (bc/b) = a/(bc); or (a - bc)/b = a/(bc)
We can now cross multiply to get rid of the fractions:
(a - bc)(bc) = ab; or expanding the brackets, abc - (b^2)(c^2) = ab
Now, as all of the terms in the equation have a factor of b, we can divide both the RHS and LHS by b, leaving us with a quadratic equation:
ac - bc^2 = a; or by rearranging bc^2 - ac + a = 0
Just for simplicity, let a = y, b = w and c = x,(so we do not get muddled up when we use the quadratic formula next):
wx^2 - yx +y = 0
We can now use the quadratic formula to find x in terms of w and y (the quadratic formula being x = [-b +- sqrt(b^2 - 4ac)]/2a ):
x = [y +- sqrt(y^2 - 4wy)]/2w; or converting back to a, b and c:
c = [a +- sqrt(a^2 - 4ab)]/2b
HOWEVER, a c will ONLY be real iff (IF AND ONLY IF) a^2 - 4ab >= 0 (as the square root of a negative number is impossible in the real plane). Thus, we can solve the above above inequality:
a^2 - 4ab >= 0; Factorise out a
a(a - 4b) >= 0
The expression a(a - 4b) has roots at a, and at 4b, thus, we find that either:
a <= 0, which is not accepted as a is a positive integer, or
a >= 4b, which we can allow.
Thus, we have proven that for ANY 2 positive integers a and b (where a >= 4b), there exists a value c (which is [a +- sqrt(a^2 - 4ab)]/2b)* such that (a/b) - c = (a/b) / c, or in other words, if Aisha's mixed fraction is converted into an improper fraction, and the numerator is greater than or equal to 4 times the denominator, there is a number which when divided or subtracted from the fraction gives the same result either way. QED.
* The value of c is not necessarily an integer: but for it to be rational, a^2 - 4ab must be a square number and for it to be an integer, 2b must be a factor of a + sqrt(a^2 - 4ab)