Question

prove that prove that root 3 + root 2 is an irrational number ​

Answer 1

Answer:

refer this as an example!

plz..extrmly sry i don't have the exact answer for ur question

Answer 2

Assume to reach the contradiction that $\displaystyle\sf {\sqrt3+\sqrt2}$ is a rational number.

By the assumption, let $\displaystyle\sf {x=\sqrt 2+\sqrt3,}$ for a rational number $\displaystyle\sf {x.}$

Then,

$\displaystyle\longrightarrow\sf {x=\sqrt2+\sqrt3}$

$\displaystyle\longrightarrow\sf {x^2=(\sqrt2+\sqrt3)^2}$

$\displaystyle\longrightarrow\sf {x^2=(\sqrt2)^2+(\sqrt3)^2+2\sqrt{2\times 3}}$

$\displaystyle\longrightarrow\sf {x^2=2+3+2\sqrt{6}}$

$\displaystyle\longrightarrow\sf {x^2=5+2\sqrt{6}}$

$\displaystyle\longrightarrow\sf {2\sqrt{6}=x^2-5}$

$\displaystyle\longrightarrow\sf {\sqrt{6}=\dfrac {x^2-5}{6}\quad\quad\dots(1)}$

(1) is not a valid equation because the irrational number $\displaystyle\sf {\sqrt6}$ in the LHS is written as a fraction in RHS which is rational (since $\displaystyle\sf {x}$ is assumed to be rational), contradicting the fact that no irrational number can write in fractional form.

Hence we've reached the contradiction and thus proved that $\displaystyle\sf {\sqrt2+\sqrt3}$ is an irrational number.