Question

prove that root 11 minus root 6 is irregular number​

Answer 1

Step-by-step explanation:

i hope that it will helps you.....

Answer 2

Answer:

assume that root 11- root 6 is rational number

root 11- root 6=a/b

root 11=a/b + root 6

squaring both sides

11=(a/b)^2 + 2(a/b) root 6 +6

2(a/b) root 6= (a/b)^2 -5

root 6= [(a/b)^2 -5 ]÷ 2(a/b)

root 6 is is rational no

but this contradicts the fact that root 6 is irrational

our assumption is wrong

so root11 - root 6 is irrational