In division by a fraction why is it that we invert and then multiply?
I posted this question so that I could explain why this works.

Question

In division by a  fraction why is it that we invert and then multiply?
I posted this question so that I could explain why this works.

Luna Angel · Accepted Answer

An alternate but similar approach added.

What does a fraction when represented as #a/b# mean? By definition it means for any two numbers #a and b#, with the condition that #b!=0# #a# divided by #b# or symbolically #=>a -: b# ........(1) We also know that #a/b# can also be written as #a" multipled by "1/ b# or symbolically #=>axx1/ b# .....(2) For expressions in lines (1) and (2) to be equal #a -: b-=axx1/ b# This is same as saying that #"division is multiplication with the reciprocal"#

Jeremiah Andrews · Answer

See the demonstration in the explanation

Solution 1 of 2
Also see my equivalent using algebra. (2 of 2)

#color(blue)("Preamble")# Consider the example #1/4 -:1/8 = 2# This is the same as:#" "(1/4color(magenta)(xx1))-:1/8=2# #(1/4color(magenta)(xx2/2))-:1/8=2# #(1xx2)/(4xx2)-:1/8=2# #2/8-:1/8=2# #color(brown)("Notice that if we just have the numerators it gives the same answer")# #2-:1=2# #color(brown)("However, for direct division of numerators to work the denominators must be the same")# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Using an example to demonstrate the principle")# Selecting numbers that are obviously different. Suppose we had: #color(brown)([2/4])color(green)(-:8/6)# Change the 4 in #color(brown)([2/4]# such that it is in #6^("ths")" instead " -> color(red)(4)xx6/4# For multiply or divide, what we do to the bottom we do to the top. #color(brown)([(2xx6/4)/(4xx6/4)]) -:8/6# Now that both the denominators are in #6^("ths")# we can ignore them and just consider the numerators. #2xx6/4 -:8" "->" "2-:8xx6/4# #2/8xx6/4" "->" "(2xx6)/(8xx4)# Swap the 8 and 4 round #(2xx6)/(4xx8)" "=" "color(brown)(2/4)color(green)(xx6/8)# #color(purple)("The "8/6" has now been inverted and the divide ")# #color(purple)("has become multiply. The principle has been demonstrated.")#

Gabriel Agustin · Answer

See below for an alternate (perhaps more abstract) explanation than the one provided by Tony.

In part this question deals with what it means to divide. In general #color(red)(a) div color(blue)(b) = color(magenta)(k)# means #color(red)(a)# is equivalent to #color(blue)(b)# "pieces" each of size #color(magenta)(k)# or#color(white)("XXX")color(red)(a)=color(magenta)k xx color(blue)(b)# When dividing by a fraction, say #color(blue)(p/q)# instead of #color(blue)(b)# we could write #color(white)("XX")color(red)(a)divcolor(blue)(p/q)=color(magenta)(k)# meaning #color(white)("XX")color(red)(a)=color(magenta)(k)xxcolor(blue)(p/q)# Provided #color(green)(q/p)!=0# we also know that we can multiply both sides of an equation by #color(green)(q/p)# and the equation will remain valid. So #color(white)("XXX")color(red)(a)xxcolor(green)(q/p) = color(magenta)(k)xxcolor(blue)(cancel(p)/cancel(q)) xx color(green)(cancel(q)/cancel(p)# and since from our original specification that #color(red)(a)divcolor(blue)(p/q)=color(magenta)(k)# it follows that #color(white)("XXX")color(red)(a) divcolor(blue)(p/q)=color(red)(a)xxcolor(green)(q/p)#

Autumn Arnold · Answer

#color(magenta)("I added this for completeness of my solution")#

#color(magenta)("It is a general solution")#
Solution 2 of 2
Also see my equivalent using numbers ( 1 of 2)

Suppose we hade #color(blue)(a/b -: c/d)# Making the denominators so that they are all #d# #color(blue)([a/bxx1]-:c/d)# #color(blue)([(axx d/b)/(bxxd/b)] -:c/d)# '................................................................................... Note that: #" "bxxd/b" " ->" " b/bxxd" " ->" " 1xxd" "=" "d# '...................................................................................... This becomes #color(blue)([(axxd/b)/d] -:c/d)# This gives the same answer as: #color(blue)(axxd/b -:c)rarr" compare to "[4/8-:2/8]=2=[4-:2]# #color(blue)(axxd/bxx1/c)# #color(blue)(a/bxxd/c)# #color(green)("Notice that the "-:c/d " has now become "xxd/c)# #color(magenta)("Thus the rule, invert and multiply is true")#

Luna Angel · Answer

The explanation is really simple... Consider first #24div3# What we are actually asking is " If I have 24 of anything, how many groups can I make with 3 in each group?" This could be shown like this:
24 = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
= (1+1+1)+(1+1+1)+(1+1+1)+(1+1+1)+(1+1+1)+(1+1+1)+(1+1 +1)+(1+1+1) We can see that 8 possible groups can be made. #24 div 3 = 8# What about #4 div 1/2?# We are asking how many groups can be made with #1/2# in each? #4 = 1/2+1/2+1/2+1/2+1/2+1/2+1/2+1/2"     "larr# there are 8 This is because each #1# has two halves in it and #4 xx2 =8# What about #6 div 3/4?# We are asking " how many groups of #3/4# can be made from 6"?
First we need to change #6# into quarters, and then group them into threes. Each 1 has four quarters in it. #6 = 6 xx4 = 24# quarters "How many groups of 3 quarters can be made from 24 quarters?" #6 = (1/4+1/4+1/4)+(1/4+1/4+1/4)+(1/4+1/4+1/4)+(1/4+1/4+1/4)+(1/4+1/4+1/4)+(1/4+1/4+1/4)+(1/4+1/4+1/4)+(1/4+1/4+1/4)# There are 8 groups with three quarters in each. #6 div 3/4 = 8# What did we do?  We changed everything into quarters by multiplying by 4 and then divided by 3 to make groups of #3/4# #6 div 3/4 = 6 xx4 div 3# # = 6 xx4/3#  #=8# In the same way. #4 div 2/5# Make everything into fifths by #xx5#, then #div2# to make groups of #2/5# #4 div 2/5 = 4 xx5 div 2# #=4 xx5/2# #=20/2# #=10#

Aurora Arias · Answer

undefined When dividing by a fraction, you invert the fraction and then multiply because it is equivalent to multiplying by the reciprocal of the fraction. This works because division by a fraction is the same as multiplying by its reciprocal, which is obtained by flipping the fraction upside down.

In division by a fraction why is it that we invert and then multiply? I posted this question so that I could explain why this works.