How do you subtract two rational numbers with different denominators?

1 Answer
Aug 3, 2018

Answer:

Generic explanation given using algebra. Followed by a numeric example.

Explanation:

#color(blue)("Initial concept")#
A fractions construct is such that we have:

#("numerator")/("denominator") -> ("count")/("size indicator of what is being counted")#

This may be further considered as:

#" count" color(white)("d")ubrace(" of ")color(white)("ddd")1/("size indicator of what is being counted")#
#color(white)("ddddddd.")darr#

#"count "color(white)("d.")xxcolor(white)("ddd")ubrace(1/("size indicator of what is being counted"))#
#color(white)("dddddddddddddddddddddddddd.")darr#

#"count "color(white)("d.")xxcolor(white)("dddddddddd")" unit of measurment"#

#color(red)("To directly add counts the units of measurement must be the same")#
#color(green)("To directly add numerators the denominators must be the same")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

Let the first rational number be #a/b#
Let the second rational number be #c/g#

Consider the context:
#a/b-c/g larr" denominators are not the same"#

I opted for #g# instead of #d# as its difference in look to #b# if very obvious. #b and d# are easily confused with each other.

Multiply by 1 and you do not change the actual value. However, 1 comes in many forms.

#color(green)(a/b-c/g color(white)("dddd")->color(white)("dddd")[a/bcolor(red)(xx1)]-[c/gcolor(red)(xx1)] )#

#color(green)(color(white)("dddddddddd")->color(white)("dddd")[a/b color(red)(xxg/g)]-[c/g color(red)(xxb/b)] )#

#color(green)(color(white)("dddddddddd")->color(white)("dddddd")[(ag)/(bg)]color(white)("d")-color(white)("d")[(bc)/(bg)] )#

#color(green)(color(white)("dddddddddd")->color(white)("ddddddddd")(ag-bc)/(bg))#

So basically you change the denominators to the same value and then directly subtract the suitably adjusted numerators.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Example")#

#3/5-7/10#

#[3/5xx2/2]-7/10#

#[(3xx2)/(5xx2)]-7/10#

#color(green)(6/10color(red)(-7/10))#

But #color(red)(-7/10)# may be split into #color(red)(-6/10-1/10)# giving:

#color(green)(ubrace(6/10color(red)(-6/10))color(red)(-1/10))#
#color(white)("d.d")darr#
#color(white)("ddd")0color(white)("dd")-1/10#

So #3/5-7/10= -1/10#