Prove that $ 4 + 3 \sqrt 2 $ is irrational

Question

I am trying to prove by contradiction that $ 4 + 3 \sqrt 2 $ is irrational.
(I may use the fact that $ \sqrt 2 $ is irrational, and must use the definition of rational numbers.)

Proofs by contradiction always trip me up, and feel incomplete for some reason. Is the following correct? specifically is the last statement sufficient to contradict the assumption?

Suppose not:
Suppose $ 4 + 3 \sqrt 2 $ is rational. $$ ( 4 + 3 \sqrt 2 ) ^ 2 = \left( \frac p q \right ) ^ 2 $$ $$ 34 + 24 \sqrt 2 = \frac { p ^ 2 } { q ^ 2 } $$ $$ 24 \sqrt 2 = \frac { p ^ 2 } { q ^ 2 } - 34 $$ $$ \sqrt 2 = \frac 1 { 24 } \left( \frac { p ^ 2 } { q ^ 2 } - 34 \right) $$ which is a contradiction because irrational does not equal a rational number.

Thank you.

Answer 1

Hints:

Suppose there are $\;a,b\in\Bbb Z\;,\;\;b\neq0\;$ , s.t.

$$4+3\sqrt2=\frac ab \implies16+24\sqrt2+18=\frac{a^2}{b^2}\implies\ldots$$

Fill in the details in the above,,,and complete the very small ammount of work that is left to reach a contradiction.

Further Hint: there is a much easier, and direct, approach.

Answer 2

As @DietrichBurde hints, $4+3\sqrt{2}=\frac{a}{b}\iff\sqrt{2}=\frac{a-4b}{3b}$.

Answer 3

You can also use this well-known method:

$$x_1=4+3\sqrt 2 ~~~\text{and}~~~ x_2=4-3\sqrt 2$$

Then, you get

$$x^2-8x-2=0$$

By Rational root theorem we deduce that $x=±1,±2$ are not the roots of the quadratic equation. This means, both $x_1=4+3\sqrt 2 $ and $x_1=4-3\sqrt 2 $ are irrational.