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Mohit Sharma 8 years, 2 months ago
let 2∧a=3∧b=6∧c=d
so,2∧a=d
⇒d∧1/a=2----------------eq 1.
and 3∧b=d
⇒d∧1/b=3----------------eq 2.
and also 6∧c=d
⇒d∧1/c=6-----------------eq 3.
we know that 6=2×3-----------------eq 4.
now put value of 2, 3 and 6 from eg. 1,2 and 3
we get d∧1/a × d∧1/b = d∧1/c
d(1/a=1/b) = d∧1/c
so, 1/c = 1/a + 1/b ⇒1/c = (b+a) / (ab)
⇒c=ab / a+b hence proved
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