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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.

1 Answers 1

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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 $f=0\cdot x+0=0$.

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    oh, thanks Zev. I thought that I need to use some linear-algebra theorems.2011-03-08
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    @Nir. You should click the up arrow in order to show you find Zev's answer useful. Maybe also check it as your accepted answer (though in general it's wise to wait a couple of days to see if other answers appear).2011-03-08
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    I clicked before you commented, or during it. thanks for the remainder any way.2011-03-08
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    @Nir: I guess you "only" upvoted. What Agustí Roig was referring to was that you can accept the answer as the accepted answer by clicking on the tick mark.2011-03-08