Question

Show that `(2+3√2)/7` is not a rational number, given that √2 is an irrational number.

Answer 1

To prove: `(2+3√2)/7 "is irrational, let us assume that" (2+3√2)/7` is rational.

`(2+3√2)/7 = ab`; b ≠ 0 and a and b are integers.
 ⇒ 2b + 3√2b =7a
⇒ 3√2b = 7a − 2b
⇒ √2 = `(7a − 2b)/(3b)`
Since a and b are integers so, 7a−2b will also be an integer.
So, `(7a − 2b)/(3b)` will be rational which means √2 is also rational.
But we know √2 is irrational(given).
Thus, a contradiction has risen because of incorrect assumption. 
Thus, `(2+3√2)/7` is irrational.