Prove that 7 + 3 root 2 is not a rational number

Prove by contradiction.

See below.

Step-by-step explanation:

Assume it is rational, then:

Let x = 7 + 3√2 = p/q where p and q are integers and p/q is in simplest form.

x^2 = 49 + 42√2 + 18 = p^2/q^2

67q^2 = p^2 - 42√2q^2

67q^2 - p^2 = - 42√2 q^2

67q^2 - p^2 is rational and q^2 is rational so that makes - 42√2 rational.

but we know that - 42√2 is irrational, so our original assumption is not true.

Step-by-step explanation:

let 7 + 3√2 be an rational number where

7+3√2 = a/b [ a and b are coprime and b is not equal to zero]

3√2 = a/b-7

3√2 = (a-7b) / b

√2 = (a-7b) / 3b ... (i)

Now, from equation (i), we get that √2 is rational but we know that √2 is irrational. so actually 7 + 3√2 is irrational not rational. thus our assumption is wrong. The number is irrational.