Show that 6 + √2 is an irrational number.

1.	Show that 6 + √2 is an irrational number.
Answer» Let’s assume on the contrary that 6+√2 is a rational number. Then, there exist co prime positive integers a and b such that 6 + √2 = \(\frac{a}{b}\) ⇒ √2 = \(\frac{a}{b – 6}\) ⇒ √2 = \(\frac{(a-6b)}{b }\) ⇒ √2 is rational [∵ a and b are integers ∴ \(\frac{(a-6b)}{b }\)is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 6 + √2 is an irrational number.

Answer»

Let’s assume on the contrary that 6+√2 is a rational number.

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

6 + √2 = \(\frac{a}{b}\)

⇒ √2 = \(\frac{a}{b – 6}\)

⇒ √2 = \(\frac{(a-6b)}{b }\)

⇒ √2 is rational [∵ a and b are integers ∴ \(\frac{(a-6b)}{b }\)is a rational number]

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

Hence, 6 + √2 is an irrational number.

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