Put the $>, < o r =$ signs without changing into equivalent fractions. Apply the rule.
$\frac{1}{7} □ \frac{1}{9}$

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Solution

LHS $= \frac{1}{7} = \frac{9}{63}$
RHS $\frac{1}{9} = \frac{7}{63}$
On observing LHS > RHS So, $\frac{1}{7} > \frac{1}{9}$

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>, < o r =

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Question 2(b) :

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