What is #\frac { \frac { 1} { 3} } { \frac { 3} { 9} }#?

3 Answers
Dec 6, 2016

The answer is #1#.

Explanation:

There are a couple of ways to go about simplifying this number. The first way is to remember that "dividing" is the same thing as "multiplying by the reciprocal". (The reciprocal is the number you get when you take a fraction and swap the numerator and denominator.) For example: dividing by #2# is the same as multiplying by #1/2#.
Using this, we get

#(1/3)/(color(white)(.)3/9color(white)(.))=1/3-:3/9=1/3times9/3=(1times9)/(3times3)=9/9=1.#

We could also recognize that the denominator #3/9# can be simplified itself, because #3# and #9# are both multiples of 3. So we can get

#(1/3)/(color(white)(.)3/9color(white)(.))=(1/3)/(color(white)(.)(1xxcancel3)/(3xxcancel3)color(white)(.))=(1/3)/(color(white)(.)1/3color(white)(.))#

And any nonzero number divided by itself will be 1, so we get

#(1/3)/(color(white)(.)1/3color(white)(.))=1#.

Dec 6, 2016

A different approach. Tends not to be used as the shortcut method is very much faster. Explained why the shortcut method works.

#1/3-:3/9=1#

Explanation:

A fractions structure is such that you have:

#("count")/("size indicator")" "->" "("numerator")/("denominator")#

You can not #ul("directly divide the counts")# unless the size indicators are the same.

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The shortcut method adopts a sort of 'indirect division' in that first it divides the counts then multiplies by a conversion factor.
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#color(blue)("Using first principles")#

Initial condition: #-> 1/3-:3/9#

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

#color(green)( [1/3 color(red)(xx1)] -:3/9" "->" "[1/3color(red)(xx3/3)]-:3/9 )#

#" "[3/9]-:3/9 #

Now the 'size indicators' (denominators) are the same you can directly divide the counts;

#" "3-:3=1#

#color(green)(" Try this with different values. It really does work!")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Short cut method")#

#" "(color(green)(1))/(color(brown)(3))-:(color(green)(3))/(color(brown)(9))#

Turn the divisor upside down and multiply:

#" "(color(green)(1))/(color(brown)(3))xx(color(brown)(9))/(color(green)(3))#

This is the same as:

#" "color(green)(1/3)xxcolor(brown)(9/3)#

#" "color(green)(1/3)" "xx" "color(brown)(9/3)#

#" "uarr" "uarr#

#"Division of counts conversion factor"#

#" "=1#

Jan 18, 2017

1

Explanation:

What the problem is asking you is what is how many #3/9#'s are in #1/3#? #3/9# is the same thing as #1/3#, so how many #1/3#'s are in one #1/3#? That would be 1.