Question

prove that root 2 is irrational and hence prove that 4 + 3 root 2 is irrational​

Answer 1

Let's prove that √2 is irrational.

Assume that √2 is rational.

$\sf \sqrt{2}\ =\ \dfrac{a}{b},\ were\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ is\ \ne\ 0.$

$\sf \sqrt{2}\ =\ \dfrac{a}{b}\\\\\\\bf Squaring\ both\ sides,\\\\\\\sf (\sqrt{2})^2\ =\ \bigg(\dfrac{a}{b}\bigg)^2\\\\\\2\ =\ \dfrac{a^2}{b^2}\\\\\\a^2\ =\ 2b^2\\\\\\\implies\ 2\ divides\ a^2.\\\\\\\implies\ 2\ divides\ a.\ [If\ p\ divides\ a^2,\ p\ divides\ a.]$

$\bf Let\ a\ =\ 2c\ for\ some\ integer\ c.\\\\\\Substituting\ a\ =\ 2c\ in\ a^2\ =\ 2b^2,\\\\\\\sf (2c)^2\ =\ 2b^2\\\\\\4c^2\ =\ 2b^2\\\\\\2c^2\ =\ b^2\\\\\\\implies\ 2\ divides\ b^2.\\\\\\\implies\ 2\ divides\ b.\\\\\\\bf Hence,\ 2\ is\ a\factor\ of\ both\ 'a'\ and\ 'b'.\\\\\\This\ contradicts\ the\ fact\ that\ 'a'\ and\ 'b'\ are\ co\ -\ primes.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\Therefore,\ our\ assumption\ is\ wrong.$

$\bf \sqrt{2}\ is\ irrational.$

Now, let's prove that 4 + 3√2 is irrational.

Assume 4 + 3√2 is irrational, again.

$\sf 4\ +\ 3\sqrt{2}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ primes\ and\ b\ \ne\ 0.$

$\sf 4\ +\ 3\sqrt{2}\ =\ \dfrac{a}{b}\\\\\\3\sqrt{2}\ =\ \dfrac{a}{b}\ -\ 4\\\\\\3\sqrt{2}\ =\ \dfrac{a\ -\ 4b}{b}\\\\\\\sqrt{2}\ =\ \dfrac{a\ -\ 4b}{3b}\\\\\\\bf Since\ 'a'\ and\ 'b'\ are\ both\ integers\ and\ \dfrac{a\ -\ 4b}{3b}\ is\ rational,\\\\\\\implies\ \sqrt{2}\ is\ rational\ as\ well.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{2}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\Therefore,\ our\ assumption\ is\ wrong.$

$\bf 4\ +\ 3\sqrt{2}\ is\ irrational.$