Question

show that root 2 + root 3 is irrational​

Answer 1

Answer:

Step-by-step explanation: Proof by contradiction,

Suppose √2 + √3 is rational, then

            √2 + √3 is of the form of a/b where a and b are integers and b ≠ 0.

Square both sides , then- (√2 +  √3)² = a²/b²

                                          ⇒ (√2)² + 2(√2)(√3) + (√3)² = a²/b²

                                          ⇒ 2 + 2√6 + 3 = a²/b²

                                          ⇒ 2√6 = a²/b² - 5

                                          ⇒ 2√6 = a² - 5b²/b²

                                          ⇒ √6 = a² - 5b²/ 2b²

The RHS is rational when a and b are integers , so the LHS , that is √6 also has to be rational , but it is not.

So √2 + √3 is irrational .

Answer 2

• Let us assume that $\sqrt{2} \: + \: \sqrt{3}$ is a rational number.

Now,

$\implies \: \sqrt{2} \: + \: \sqrt{3} \: = \: \dfrac{a}{b}$

Here, a and b are co-prime numbers.

Squaring on both sides

$\implies \: (\sqrt{2} \: + \: \sqrt{3}) ^{2} \: = \: \bigg (\dfrac{a}{b} \bigg)^{2}$

(a + b)² = a² + b² + 2ab

So,

$\implies \: (\sqrt{2})^{2} \: + \: (\sqrt{3}) ^{2} \: + \: 2 (\sqrt{2}) (\sqrt{3} ) \: = \: \bigg(\dfrac{a^{2} }{ {b}^{2} } \bigg)$

$\implies \: 2 \: + \: 3 \: + \: 2 \sqrt{6} \: = \: \bigg(\dfrac{a^{2} }{ {b}^{2} } \bigg)$

$\implies \: 5\: + \: 2 \sqrt{6} \: = \: \bigg(\dfrac{a^{2} }{ {b}^{2} } \bigg)$

$\implies \: 2 \sqrt{6} \: = \: \bigg(\dfrac{a^{2} }{ {b}^{2} } \: - \: 5 \bigg)$

$\implies \: 2 \sqrt{6} \: = \: \bigg(\dfrac{a^{2} \: - \: 5 {b}^{2} }{ {b}^{2} } \bigg)$

$\implies \: \sqrt{6} \: = \: \bigg(\dfrac{a^{2} \: - \: 5 {b}^{2} }{2 {b}^{2} } \bigg)$

Here.. $\bigg(\dfrac{a^{2} \: - \: 5 {b}^{2} }{2 {b}^{2} } \bigg)$ is a rational number.

And $\bigg(\dfrac{a^{2} \: - \: 5 {b}^{2} }{2 {b}^{2} } \bigg)$ is equl to $\sqrt{6}$

This means that, $\sqrt{6}$ is also a rational number.

But we know that $\sqrt{6}$ is an irrational number.

This contradicts that our assumption is wrong.

$\sqrt{2} \: + \: \sqrt{3}$ is an irrational number.