Linear Independence of Irrational Numbers

Lesson · Intermediate

Algebra

a+b5=0a=b=0,a,bQa+b\sqrt5=0\Longrightarrow a=b=0,\qquad a,b\in\mathbb Q
a+b5=c+d5a=c,b=d,a,b,c,dQa+b\sqrt5=c+d\sqrt5\Longrightarrow a=c,\qquad b=d,\qquad a,b,c,d\in\mathbb Q

Instead of √5, we can consider ∛5 or any other irrational number.