Prove that 4-5 square root 3 is irrational
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Answered by
12
Answer:
4-5²√3
4-(5²)√3=a/b
25√3=4/b-4b/b
√3=4/25b-4b/25b
where L.H.S is rational and R.H..S is rational , butit contraditcs that √3 is irrational
∴4-5²√3 is irrational
Answered by
15
Assume that 4 - 5√3 is rational.
So that 4 - 5√3 can be written as p/q, where p, q are coprime integers and q ≠ 0.
Here it contradicts our earlier assumption that 4 - 5√3 is rational.
Because, the RHS, p²/q² subtracted from 91, is rational, where p²/q² = (p/q)² is rational as it is square of a rational number p/q, but while the LHS, 40√3, is irrational.
∴ 4 - 5√3 is irrational.
Hence proved!
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