Question

prove that √2+√3 is irrational​

Answer 1

Q .GIVEN,

TO PROVE √2+√3 IS IRRATIONAL...

ANS. LET √2+√3 RATIONAL ,

SO,√2+√3=a/b, where a and b are integers

SQARING ON BOTH SIDES,

(√2+√3)^2= (a/b)^2

(√2)^2 + (√3)^2+ 2 (√2) (√3)=a^2/b^2

2+ 3 +2√6 =a^2/b^2

5+2√6 =a^2/b^2

2√6 =a^2/b^2 - 5

2√6 = a^2- 5b^2/b^2

√6 = a^2- 5b^2/2b^2

AS,a and b are integers,

a^2- 5b^2/2b^2 is a rational number...

SO ,√6 IS ALSO A RATIONAL NUMBER...

BUT THIS CONTRADICTS THAT THE

FACT THAT√6 IS AN IRRATIONAL

NUMBER....

THERE FORE,

√2+√3 IS AN IRRATIONAL NUMBER...

Answer 2

Answer:

suppose root 2+root 3 is a rational number

root 2+root 3 = a/b ( where a and b are integers )

squaring both sides

( root 2+root 3 )^2= ( a/b )^2

( root 2 )^2 + 2(root 2)(root 3)+(root 3)^2 = a^2/ b^2

2+2 root 6 + 3 = a^2/b^2

5+2 root 6 = a^2/b^2

2 root 6 = a^2/b^2 ‐ 5/1

2 root 6 = a^2‐5b^2/b^2

root 6 = a^2‐ 5b^2/2b^2

here a^2‐ 5b^2/2b^2 is rational as a and. b are integers

root 6 is also rational.

but actually root 6 is irrational

so we can say root 2+root 3 is irrational