How do you compare each pair of fractions with <, > or = given #5/4, 7/6#?

1 Answer
Dec 2, 2016

#color(white)"XXX"5/4?7/6"                "# (#4times3=12=6times2#)

#3/3times5/4?7/6times2/2"       "# (multiply both sides by "one")

#"       "15/12?14/12"              "# (multiply across)

#"       "15>14"              "# (reduce)

#:.5/4>7/6#.

Explanation:

The easiest way to compare two fractions is to rewrite one (or both) of them so that their denominators match.

You can think about the denominator of a fraction like a unit, similar to inches, centimeters, feet, etc. If you were asked to compare 5 inches to 7 centimeters, you'd first need to convert one value (or both) so that their units matched.

In fractions, the "unit" is the denominator. The first value is 5 "fourths", and the second value is 7 "sixths". These aren't in the same unit, so we can't just compare 5 to 7. But, we can convert the numbers to the same unit.

The best way to do this is to find the lowest multiple that 4 and 6 have in common. Without too much effort, this is found to be 12 #(4times3=12"# and #"6times2=12)#. Any common multiple will do, but the lowest one is the easiest to work with.

So we need to convert #5/4# into twelfths. Looking at the equation #4times3=12#, if we move the 12 to the left and the 4 to the right, we get #3/12=1/4#. So there are 3 #"12"^"th""s"# in a #"4"^"th".# (Just like how there are 3 feet in a yard.)

For every fourth, there are 3 twelfths. We have 5 fourths, so we have:

#5/4=5(1/4)=5(3/12)=(5times3)/12=15/12#.

This was just for explanation. There's a much more concise way to write this change of "unit":

#5/4=(5times3)/(4times3)=15/12#

What we're doing in this shorter method is multiplying #5/4# by #3/3#. Remember, #3/3=1#, so the value represented by #5/4# remains the same, but the unit (denominator) changes.

Similarly, if we have 7 sixths, and there are 2 twelfths in a sixth, then we have 14 twelfths:

#7/6=(7times2)/(6times2)=14/12#

Now the values represented by #5/4# and #7/6# are in the same unit. Comparing #5/4# to #7/6# is the same as comparing #15/12# to #14/12#. And since #15>14#, it follows that #15/12>14/12,# and thus:

#5/4>7/6.#

Bonus:

There's a trick to comparing fractions where the numerator is 1 more than the denominator:

#2/1>3/2>4/3>5/4>6/5>7/6>...>(n+1)/n>...#

This works because the numerators are getting proportionally closer to their denominators.

Think of a really big number. Now consider the fraction #("your number"+1)/("your number")#. This ratio is greater than 1, and the bigger your number, the smaller the relative difference between it and the number that's one more than it. So this fraction gets closer to #1# as your number gets larger. In fact, the limit #(lim)# of this fraction, as your number #(n)# grows forever, is equal to #1#. In math language, we write:

#lim_(n->oo)(n+1)/n=1#

This is a little taste of calculus for you. It's not so bad. ;)