I have this question:
Let $f$ be a Polynomial so that $\mathbb{Q}[x] \ni f=ax+b$.
With the initial information: $f(\sqrt{2})=0$,
How can someone prove that $f=0$?
Thanks.
I have this question:
Let $f$ be a Polynomial so that $\mathbb{Q}[x] \ni f=ax+b$.
With the initial information: $f(\sqrt{2})=0$,
How can someone prove that $f=0$?
Thanks.
Suppose $f(\sqrt{2})=a\sqrt{2}+b=0$, where $a$ and $b$ are rational numbers. If $a\neq 0$, then after rearranging, we have $\sqrt{2}=\frac{-b}{a}$; but $\frac{-b}{a}$ is rational (since $a$ and $b$ are) and $\sqrt{2}$ is not, which is a contradiction. Thus, $a=0$, and we have $0\cdot\sqrt{2}+b=b=0$, so $b=0$. Thus $f=0\cdot x+0=0$.