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

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

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Answer 1 · 2015-05-15T20:32:05

The answer is : $\frac{7}{8} < \frac{22}{25} < \frac{8}{9}$ $7/8 < 22/25 < 8/9$ .

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

$\frac{22}{25} = \frac{22 \cdot 9 \cdot 8}{25 \cdot 9 \cdot 8} = \frac{1584}{1800}$ $22/25 = (22*9*8)/(25*9*8) = 1584/1800$

$\frac{8}{9} = \frac{8 \cdot 25 \cdot 8}{9 \cdot 25 \cdot 8} = \frac{1600}{1800}$ $8/9=(8*25*8)/(9*25*8) = 1600/1800$

$\frac{7}{8} = \frac{7 \cdot 25 \cdot 9}{8 \cdot 25 \cdot 9} = \frac{1575}{1800}$ $7/8=(7*25*9)/(8*25*9)=1575/1800$

Therefore, $\frac{7}{8} = \frac{1575}{1800} < \frac{22}{25} = \frac{1584}{1800} < \frac{8}{9} = \frac{1600}{1800}$ $7/8 = 1575/1800 < 22/25 = 1584/1800< 8/9=1600/1800$ .

Answer 2 · 2015-05-16T00:04:24

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

$\frac{22}{25}, \frac{8}{9}, \frac{7}{8}$ $22/25, 8/9, 7/8$

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

First the easy pair $\frac{8}{9}, \frac{7}{8}$ $8/9, 7/8$

The least common denominator is $9 \times 8$ $9xx8$ , I don't care what the number really is. It's the numerators I need to compare.

$\frac{8}{9} = \frac{8 \times 8}{9 \times 8} = \frac{64}{9 \times 8}$ $8/9 = (8xx8)/(9xx8) = 64/(9xx8)$

$\frac{7}{8} = \frac{7 \times 9}{9 \times 8} = \frac{63}{9 \times 8}$ $7/8 = (7xx9)/(9xx8) = 63/(9xx8)$

So $\frac{7}{8} < \frac{8}{9}$ $7/8 < 8/9$
(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 $\frac{7}{8}$ $7/8$ is $\frac{1}{8}$ $1/8$ less than $1$ $1$ , while $\frac{8}{9}$ $8/9$ is $\frac{1}{9}$ $1/9$ less than $1$ $1$ , so $\frac{7}{8} < \frac{8}{9}$ $7/8 <8/9$ )

$\frac{22}{25}, \frac{8}{9}$ $22/25 , 8/9$

The least common denominator is $9 \times 25$ $9xx25$ , Again, I don't care what the number really is. It's the numerators I need to compare.

$\frac{22}{25} = \frac{22 \times 9}{25 \times 9}$ $22/25 = (22xx9)/(25xx9)$

$\frac{8}{9} = \frac{25 \times 8}{25 \times 9}$ $8/9 = (25xx8)/(25xx9)$

The numerators are:

$22 \times 9$ $22xx9$ $ssssssssssssssssssss$ $color(white)"ssssssssssssssssssss"$ and $25 \times 8$ $25xx8$ , which we can rewrite as:

$22 \times (8 + 1) = 22 \times 8 + 22$ $22xx(8+1)=22xx8 + 22$ and $(22 + 3) 8 = 22 \times 8 + 24$ $(22+3)8 = 22xx8 +24$

Whatever $22 \times 8$ $22xx8$ is, adding 24 will give a bigger total than adding 22. The second numerator is greater. So

$\frac{22}{25} < \frac{8}{9}$ $22/25 < 8/9$

Third pair
$\frac{22}{25}, \frac{7}{8}$ $22/25, 7/8$ Denominator $8 \times 25$ $8xx25$ ,

Numerators:

$22 \times 8$ $22xx8$ $ssssssssssssssssssss$ $color(white)"ssssssssssssssssssss"$ and $25 \times 7$ $25xx7$

$22 \times 8 = 22 (7 + 1) = 22 (7) + 22$ $22xx8 = 22(7+1)=22(7)+22$ and $25 \times 7 = (22 + 3) 7 = 22 (7) + 21$ $25xx7=(22+3)7 = 22(7)+21$

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

$\frac{7}{8} < \frac{22}{25}$ $7/8 < 22/25$

Final Answer

$\frac{7}{8} < \frac{22}{25} < \frac{8}{9}$ $7/8 < 22/25 < 8/9$

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

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