Question

prove that root3 + root8 is an irrational no.​

Answer 1

substituting values

roots of both numbers are an irrational numbers

Thus there addition will also give an irrational number.

Answer 2

To prove $\sqrt{3}+\sqrt{8}$ is an irrational number

• Let us prove this by contradiction method.
• Let $\sqrt{3}+\sqrt{8}$ be a rational number. Then

                     $\sqrt{3}+\sqrt{8}=\frac{p}{q}$        

                              [ p and q are co primes and q ≠ 0]

• Squaring on both sides

                    $(\sqrt{3}+\sqrt{8})^{2} =(\frac{p}{q})^{2}$

                     $3+8+2\sqrt{24}=\frac{p^{2} }{q^{2} }$

                     $2(2\sqrt{6})=\frac{p^{2} }{q^{2}}-11$

                     $\sqrt{6}=\frac{p^{2} }{4q^{2}}-\frac{11}{4}$

• This is a contradiction because $\sqrt{6}$ is an irrational number and $\frac{p^{2} }{4q^{2}}-\frac{11}{4}$ is rational.
• Therefore, our assumption that $\sqrt{3}+\sqrt{8}$ is a rational number is wrong.
• Hence, $\sqrt{3}+\sqrt{8}$ is an irrational number.