How do you divide #3/8 div 1/4#?

1 Answer
Oct 19, 2016

Answer:

#3/2#
Both shortcut and first principle methods shown. Followed by what is actually happening in the shortcut approach. This is not always explained!

Explanation:

#color(blue)("Shortcut method")#

#color(brown)(3/8-:color(green)(1/4))color(green)(" "vec( "invert and multiply")" " color(brown)(3/8)xx4/1)#

#color(brown)(3/8xxcolor(green)(4/1)" the same as "3/(color(green)(1))xx(color(green)(4))/8)" " =" " 3xx1/2" "=" "3/2#
.............................. or ..............................................................

#3/8xx4/1 " "->" "3/(cancel(8)^2)xx(cancel(4)^1)/1 = color(red)(3/2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("First principles")" "larr" shortcut based on this method" #

Method: make the denominators (bottom numbers) the same then just divide the numerators (top numbers). This way you are just dividing the counts and ignoring the indicator of size (bottom number).

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

#color(green)(3/8 -:[1/4color(magenta)(xx1)]#

But #2/2=1#

#color(green)(3/8 -:[1/4color(magenta)(xx2/2)]#

#3/8-:2/8#

Now that the denominators (bottom numbers) are both 8 we can just divide the numerators (top numbers# -># counts).

#3/8-:2/8" "=" "3-:2 =color(red)( 3/2)#
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

#color(blue)("What is actually happening in the shortcut")#

A fraction consists of two parts.

Their function is: #" "("count")/("size indicator of what you are counting")#
#color(white)(.)#

#" "(color(red)(3))/(color(blue)(8))-:(color(red)(1))/(color(blue)(4))#

Invert and multiply #-> (color(red)(3))/(color(blue)(8))xx(color(blue)(4))/(color(red)(1))" "->" "color(red)(3/1)xxcolor(blue)(4/8)#

#color(white)(2/2)#

Where #" "color(red)(3/1)" "xx" "color(blue)(4/8)#
#color(white)(.)#
#" "color(red)(uarr)" "color(blue)(uarr)#
#color(red)("division of counts")" "color(blue)("conversion factor")#