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Prove that (5 − √3) is irrational. |
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Answer» Let us assume contrary that (5 – √3) is a rational number. ∴ 5 – √3 = \(\frac{p}{q}\), q ≠ 0 & p, q ∈ I (Because every rational number can be written in \(\frac{p}{q}\) form where q ≠0 & p, q ∈ I ) ⇒ 5 − \(\frac{p}{q}\) = √3. ⇒ \(\frac{5q-p}{q}\) = √3 … (1) L.H.S = \(\frac{5q-p}{q}\), (q ≠ 0 & , 5q − p ∈ I) is a rational number. (Because, every number which can be written in \(\frac{p}{q}\) form is a rational number) R.H.S = √3 is an irrational number. (∵\(\sqrt{prime \,number}\) is always a irrational number) But rational ≠ irrational which is contradiction of equation (1). Hence, Our assumption is wrong. ∴ (5 – √3) is an irrational number. |
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