Before I start with the solution, I would like to just clarify that all my mixed fractions will be written in the form a#b# where a is the whole number and b is the rational number between the # symbol. I have done this since I am unable to type a mixed fraction in properly on my laptop.
Aisha would get the same answers for (5#1/3#)-4 and (5#1/3#)÷4 = 4/3, and for (7#1/5#)-6 and (7#1/5#)÷6.
Generalising this, the division of a mixed fraction and a positive integer and the difference between the two will work when the expression is in the form ((n+2)#1/n#)-(n+1), or ((n+2)#1/n#)÷(n+1), either is fine since they will give the same answer anyway.
Proof of this:
((n+2)#1/n#)-(n+1) = ((n+2)#1/n#)÷(n+1)
((n^2 +2n + 1)/n) - (n+1) = ((n^2 +2n + 1)/n) * 1/(n+1)
(n^2 + 2n +1 - n(n+1))/n = ((n^2 +2n + 1)/n) * 1/(n+1)
(n^2 + 2n + 1 - n^2 - n)/n = ((n^2 +2n + 1)/n(n+1))
(n+1)/n = ((n+1)^2)/n(n+1)
(n+1)/n = (n+1)/n
LHS = RHS
QED.
This means that if we pick any positive integer n, the subtraction of a mixed fraction and a constant in the form given above will equal to their division.
However, using this expression doesn't work with negative integers.
Example: Let n = -5
Subtraction:
-3#1/-5# -(-4)
=16/5 + 4
=36/5
Division:
-3#1/-5# ÷ (-4)
= 16/5 ÷-4
=16/-20 = -4/5
36/5 ≠-4/5
We can match the division with the addition though;
-3#1/-5# + (-4)
= 16/5 - 4
=-4/5
This shows that the rule would only stand with the absolute values of integers.