What fraction is between #1/3# and #1/2# ?

Question

Julian Avila · Accepted Answer

#2/5# #1/2=0.5# #1/3=0.333333# So #2/5=0.4#

Luna Angel · Answer

There are infinitely many possibilities, but the midpoint between #1/3# and #1/2# is #5/12# For any two numbers #a# and #b#, the average #1/2(a+b)# lies between them. So to find a fraction between #1/3# and #1/2# we can form the average: #1/2(1/3+1/2) = 1/2(2/6+3/6) = 1/2(5/6) = 5/12#

Jeremiah Andrews · Answer

#5/12#

First, make the denominators equal:

#(1/2) * (3/3) = 3/6#

#(1/3) * (2/2) = 2/6#

Now figure out the middle number between the numerators:

#2/6 < x < 3/6#

#x = 2.5/6#

Since #2.5# is a decimal, multiply both the numerator and denominator to make the #2.5# into a whole number:

#(2.5/6) * (2/2) = 5/12#

Check the answer:

#(1/3) < (5/12) < (1/2) " "?#

#1/2# is the same thing as #6/12#

#1/3# is the same thing as #4/12#

so

#(4/12) < (5/12) < (6/12)# is true!

Cameron Arnold · Answer

There are many ... but #5/12# is exactly half-way between them.

There are many fractions between #1/2 and 1/3# If you use the LCD you end up with #3/6 and 2/6#. The values between #2 and 3# are all fractions. However, use a larger value in the denominator: #6/12 and 4/12# Now it is easy so see that a fraction exactly between them is #5/12# Consider an even bigger value in the denominator: #12/24 and 8/24# Now we have the fractions: #9/24, 10/24, 11/24# lying between them. You can continue in this using larger and larger values and each time you will find more and more fractions between #1/2 and 1/3# There are infinitely many equivalent fractions.

Genesis Amaya · Answer

#2/5# is one of them.

The fraction #(a+c)/(b+d)# is between #color(red)(a/b)# and #color(blue)(c/d)# for any integers #a,c# and positive integers #b,d#. Proof: Assuming that #a/b< c/d# #color(red)(a/b)=color(red)(a/b)*(b+d)/(b+d)# #=color(red)(a/b)*b/(b+d)+color(red)(a/b)*d/(b+d)<# #< a/b*b/(b+d)+c/d*d/(b+d)<# #< color(blue)(c/d)*b/(b+d)+color(blue)(c/d)*d/(b+d)# #=color(blue)(c/d)*(b+d)/(b+d)=color(blue)(c/d)# The middle term #a/cancel(b)*cancel(b)/(b+d)+c/cancel(d)*cancel(d)/(b+d)=a/(b+d)+c/(b+d)=(a+c)/(b+d)# That means #a/b< (a+c)/(b+d)< c/d# In our case the fraction between #1/3# and #1/2# is #(1+1)/(3+2)=2/5# It's a nice trick if you need a fraction between fractions quick. Ford's circles are closely related to that.

Roman Alvarez · Answer

undefined The fraction that is between $ \frac{1}{3} $ and $ \frac{1}{2} $ is $ \frac{2}{5} $.