Show that 5 − 2√3 is an irrational number.

1.	Show that 5 − 2√3 is an irrational number.
Answer» Let’s assume on the contrary that 5 – 2√3 is a rational number. Then, there exist co prime positive integers a and b such that 5 – 2√3 = \(\frac{a}{b}\) ⇒ 2√3 = 5 – \(\frac{a}{b}\) ⇒ √2 = \(\frac{(5b – a)}{(2b)}\) ⇒ √2 is rational [∵ 2, a and b are integers ∴ \(\frac{(5b – a)}{(2b)}\) is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 5 – 2√3 is an irrational number.

Answer»

Let’s assume on the contrary that 5 – 2√3 is a rational number.

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

5 – 2√3 = \(\frac{a}{b}\)

⇒ 2√3 = 5 – \(\frac{a}{b}\)

⇒ √2 = \(\frac{(5b – a)}{(2b)}\)

⇒ √2 is rational [∵ 2, a and b are integers ∴ \(\frac{(5b – a)}{(2b)}\) is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 5 – 2√3 is an irrational number.

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