Recall that dividing by #5# is the same as multiplying by #1/5#. We can show this by remembering that we multiply fractions by taking the product of the numerators and the product of the denominators for the resulting numerator and denominator, respectively. So, for any number #x#
#x xx 1/5 = x/1 xx 1/5 = (x xx 1)/(1 xx 5) = x/5 = x -: 5#
The last thing we need is that multiplication is associative, that is, it doesn't matter what order we do it in. If we have #axxbxxc#, it doesn't matter if we perform the operations as #(axxb)xxc# or #axx(bxxc)#.
Now, let's look at the problem. Taking a number #x#, multiplying it by #4#, and then dividing it by #5# can be written as
#(x xx 4) -: 5#
But dividing by #5# is the same as multiplying by #1/5#, so
#(x xx 4) -: 5 = (x xx 4) xx 1/5#
Because multiplication is associative, we can write that as
#(x xx 4) xx (1/5) = x xx (4 xx 1/5)#
Because #4 = 4/1#, we multiply across as above, getting #4xx1/5 = 4/5#. So our final result becomes
#(x xx 4) -: 5 = x xx 4/5#
