How do you order these fractions smallest to largest: 22/25, 8/9, 7/8?

The answer is : #7/8 < 22/25 < 8/9#.

In order to see which one is larger than another, you need to put them all to the same denominators :

#22/25 = (22*9*8)/(25*9*8) = 1584/1800#

#8/9=(8*25*8)/(9*25*8) = 1600/1800#

#7/8=(7*25*9)/(8*25*9)=1575/1800#

Therefore, #7/8 = 1575/1800 < 22/25 = 1584/1800< 8/9=1600/1800#.

Same idea, but less actual arithmetic. (Is it easier? Probably not, but the arithmetic is easier, because we don't finish much of it.)

#22/25, 8/9, 7/8#

We can order them, pairwise (two at a time).

First the easy pair # 8/9, 7/8#

The least common denominator is #9xx8#, I don't care what the number really is. It's the numerators I need to compare.

#8/9 = (8xx8)/(9xx8) = 64/(9xx8)#

#7/8 = (7xx9)/(9xx8) = 63/(9xx8)#

So #7/8 < 8/9#
(At the end, we won't need this as a separate step, but it's not difficult to do.)

Second Pair
(Note: it is even quicker to observe that #7/8# is #1/8# less than #1#, while #8/9# is #1/9# less than #1#, so #7/8 <8/9#)

#22/25 , 8/9#

The least common denominator is #9xx25#, Again, I don't care what the number really is. It's the numerators I need to compare.

#22/25 = (22xx9)/(25xx9)#

#8/9 = (25xx8)/(25xx9)#

The numerators are:

#22xx9# #color(white)"ssssssssssssssssssss"#and #25xx8#, which we can rewrite as:

#22xx(8+1)=22xx8 + 22# and #(22+3)8 = 22xx8 +24#

Whatever #22xx8# is, adding 24 will give a bigger total than adding 22. The second numerator is greater. So

#22/25 < 8/9#

Third pair
#22/25, 7/8# Denominator #8xx25#,

Numerators:

#22xx8# #color(white)"ssssssssssssssssssss"#and #25xx7#

#22xx8 = 22(7+1)=22(7)+22# and #25xx7=(22+3)7 = 22(7)+21#

Adding #22# will give a greater total than adding #21#, so the first number is greater:

#7/8 < 22/25#

Final Answer

#7/8 < 22/25 < 8/9#