Prove that 2√3 − 1 is an irrational number.

1.	Prove that 2√3 − 1 is an irrational number.
Answer» Let’s assume on the contrary that 2√3 – 1 is a rational number. Then, there exist co prime positive integers a and b such that 2√3 – 1 = \(\frac{a}{b}\) ⇒ 2√3 = \(\frac{a}{b}\) – 1 ⇒ √3 = \(\frac{(a – b)}{(2b)}\) ⇒ √3 is rational [∵ 2, a and b are integers ∴ \(\frac{(a – b)}{(2b)}\) is a rational number] This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, 2√3 – 1 is an irrational number.

Answer»

Let’s assume on the contrary that 2√3 – 1 is a rational number.

Then, there exist co prime positive integers a and b such that

2√3 – 1 = \(\frac{a}{b}\)

⇒ 2√3 = \(\frac{a}{b}\) – 1

⇒ √3 = \(\frac{(a – b)}{(2b)}\)

⇒ √3 is rational [∵ 2, a and b are integers ∴ \(\frac{(a – b)}{(2b)}\) is a rational number]

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, 2√3 – 1 is an irrational number.

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