Look at these calculations:

i) Take some more fractions equal to
and form fractions by multiplying the numerators and denominators by 3 and 4 and adding.
Do you get fractions equal to ![]()
ii) Take some other pairs of equal fractions and check this
iii) In all these, instead of multiplying numerators and denominators by 3 and 4, multiply by some other numbers and add. Do you still get equal fractions?
iv) Explain why, if the fraction
is equal to the fraction
then for any pair of natural numbers m and n, the fractions
is equal to ![]()
i) ![]()
![]()
We observe that, in these cases also, we obtain ![]()
ii) ![]()
![]()
![]()
We observe in these cases also, we obtain equal fraction, ![]()
iii) let us now take 2 and 5 instead of 3 and 4 in part(ii).
![]()
![]()
![]()
So, again we get equal fractions.
iv) We know that,
When
then, aq = pb ………. (1)
∴ for ![]()
(ma + np) × b = a × (mb + nq) →this must be satisfied
∴ (ma + np) × b = mab + npb ………… (2)
And a × (mb + nq) = mab + nqa
= mab + npb ………… using (1)
∴ a × (mb + nq) = mab + npb ………….. (3)
Since, (2) is equal to (3)
∴ ![]()
Couldn't generate an explanation.
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