7. Prove that 3+5root2) is an irrational number, given that root 2 is an irrational number ​

1 thought on “7. Prove that 3+5root2) is an irrational number, given that root 2 is an irrational number ​”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that

   [tex]\rm :\longmapsto\: \sqrt{2} \: is \: irrational.[/tex]

   [tex]\rm :\longmapsto\:\: Let \: assume \: that \: 3 + 5 \sqrt{2} \: is \: not \: irrational.[/tex]

   So,

   [tex]\rm :\implies\:3 + 5 \sqrt{2} \: is \: rational.[/tex]

   [tex]\rm :\longmapsto\:Let \: 3 + 5 \sqrt{2} \: = \dfrac{x}{y} [/tex]

   where,

   [tex] \red{ \sf \: x \: and \: y \: are \: integers \: such \: that \: y \: \ne \: 0 \: and \: hcf(x,y) = 1}[/tex]

   [tex]\rm :\longmapsto\:\: 5 \sqrt{2} \: = \dfrac{x}{y} – 3[/tex]

   [tex]\rm :\longmapsto\:\: 5 \sqrt{2} \: = \dfrac{x – 3y}{y}[/tex]

   [tex]\rm :\longmapsto\:\: \sqrt{2} \: = \dfrac{x – 3y}{5y}[/tex]

   As x and y are integers,

   So, x – 3y and 5y are also integers.

   [tex]\rm :\longmapsto\:\: \dfrac{x – 3y}{5y} \: is \: rational[/tex]

   [tex]\bf\implies \: \sqrt{2} \: is \: rational.[/tex]

   which is contradiction to the fact as it is given that

   [tex]\rm :\longmapsto\: \: \sqrt{2} \: is \: irrational.[/tex]

   [tex]\rm :\longmapsto\:Hence, \: our \: assumption \: is \: wrong.[/tex]

   [tex]\bf\implies \:3 + 5 \sqrt{2} \: is \: irrational.[/tex]

   Hence, Proved

   Additional Information :-

   Irrational numbers

   • Irrational numbers are those numbers whom decimal representation is non terminating and non repeating. Basically, square root of prime numbers are always Irrational.

   Rational numbers

   • Rational numbers are those numbers whom decimal representation is either terminating or non – terminating but repeating.

   Reply