If you add the aforementioned cardinal to the numerator and denominator of a able fraction, the amount of the able atom increases.

If you add the aforementioned cardinal to the numerator and denominator of an abnormal fraction, the amount of the abnormal atom decreases.

Note: You can bethink this by befitting in apperception that,

1/2 < 2/3 < 3/4 < 4/5 …

and

3/2 > 4/3 > 5/4 > 6/5 …

Example,

Arrange the afterward in accretion order: 117/229, 128/239, 223/449.

Let’s aboriginal analyze 117/229 & 128/239.

If we added 11 to the numerator and the denominator of the aboriginal able fraction, the consistent able atom would be 128/240, which will be bigger in amount than the aboriginal (as per Funda 2).

We apperceive that 128/240 is abate than 128/239, as the closing has a lower base.

So, 117/229 < 128/240 < 128/239

→ 117/229 < 128/239

Now let’s analyze 117/229 and 223/449.

If we added 11 to the numerator and the denominator of the additional able fraction, the consistent able atom would be 234/460, which will be bigger in amount than the original.

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