Just picking values at random, suppose we had: #2.35/2.65#
#color(blue)("Making the numbers more straight forward")#
Multiply by 1 and you do not change the value. However, 1 comes in many forms.
#color(green)( [2.35/2.65color(red)(xx1)] color(white)("ddd") -> color(white)("ddd")[2.35/2.65color(red)(xx100/100)] )#
#color(green)(color(white)("dddddddddddd")->color(white)("ddddddd")235/265 = 47/53)#
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#color(blue)("Conducting the division")#
What follows is one method of many.
Note that #47# is the same as #470xx1/10#
I will use this adjustment approach to make division itself more straight forward. We will need to apply the appropriate adjustment at the end.
So for #47-:53# we have #470-:53color(magenta)(xx1/10)#
#color(white)("ddddddd")470 color(magenta)(xx1/10)#
#color(magenta)(8)(53) ->ul( 424 larr" Subtract")#
#color(white)("dddddddd")46 larr" less than 53 so we adjust again"#
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#color(white)("ddddddd")460 color(magenta)(xx1/10)#
#color(magenta)(8)(53)->ul(424larr" Subtract")#
#color(white)("dddddddd") 36larr" less than 53 so we adjust again"#
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#color(white)("ddddddd")360 color(magenta)(xx1/10)#
#color(magenta)(6)(53)->ul(318larr" Subtract")#
#color(white)("dddddddd")42larr" less than 53 so we adjust again"#
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#color(magenta)("AND SO THE CYCLE CONTINUES:")#
So far we have:
#color(magenta)(886xx1/10xx1/10xx1/10 = 0.886)#
The full printout from my calculator is: 0.88679245283
