Question

route 2 is a irrational number. prove that 5+3(route 2) is a irrational number

Answer 1

Answer:

let it be rational for contradiction.

therefore. we can write them in the form of A/b and diving with all its cafactors give us p/q

5+3√2 = p/q

then √2 = p+5q/3

here although p+5q/3 is rational √2 should also be rational but it's not true this condition occur because our contradiction that 5+3√2 is rational was wrong hence it is irrational.....

proved...

hope it helps.... mark brainliest

Answer 2

$Given \sqrt{2}\ is \ an \ irrational\ number.\\\\now, suppose\ that \ the \ given \ number\ is\  also\ rational.\\So,  let\ 5+3\sqrt{2} \   be \ rational\ equal \ to\  \frac{b}{c} \ where\ b\ and \ c\ are\ co-prime \natural\  numbers.\\\\5+3\sqrt 2= \frac{b}{c} \\3\sqrt{2} =\frac{b}{c} -5\\\sqrt{2} =\frac{1}{3}(\frac{b}{c} -5)\\\\On \ RHS\ we \ observe\ that\ 1/3,b,c,5 \ all\ are\ rational.\\So,\ RHS \ is\ rational.\\But LHS= \sqrt{2}\ is \ irrational\ (given)$

This is not possible because RHS  and LHS  are not same.

SO, our assumption of  taking 5+3\sqrt{2} \  rational is wrong  .

So  the  given  expression  is  IRRATIONAL.