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Toolkit 114

Linear Independence of Irrational Numbers

a+b5=0⟹a=b=0,a,b∈Qa+b\sqrt5=0\Longrightarrow a=b=0,\qquad a,b\in\mathbb{Q}
a+b5=c+d5⟹a=c,b=d,a,b,c,d∈Qa+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.

Proof

Proof (Part 1)

a+b5=0a+b\sqrt5=0
b5=−ab\sqrt5=-a

Assume b≠0b\neq0.

5=−ab∈Q\sqrt5=-\frac{a}{b}\in\mathbb{Q}

This is a contradiction because 5\sqrt5 is irrational.

⇒b=0\Rightarrow b=0
a+b5=0⟹a=0a+b\sqrt5=0\Longrightarrow a=0
a=b=0a=b=0

Proof (Part 2)

a+b5=c+d5a+b\sqrt5=c+d\sqrt5
a−c+(b−d)5=0a-c+(b-d)\sqrt5=0
(a−c)+(b−d)5=0(a-c)+(b-d)\sqrt5=0

By the first property,

a−c=0,b−d=0a-c=0,\qquad b-d=0
a=c,b=da=c,\qquad b=d
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