Question

Prove that 4-5 square root 3 is irrational

Answer 1

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

Answer 2

         

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.

$\frac{p}{q}=4-5\sqrt{3} \\ \\ \\ (\frac{p}{q})^2=(4-5\sqrt{3})^2 \\ \\ \\ \frac{p^2}{q^2}=91-40\sqrt{3} \\ \\ \\ 40\sqrt{3}=91-\frac{p^2}{q^2}$

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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