p - adic norm not equivalent to usual norm!

I know that the two norms: $$p$$ - adic norm and the usual norm ($$\left| \cdot \right|$$) defined on $$\mathbb{Q}$$ are not quivalent. This is clearly because the $$p$$ adic norms staisfies the strong triangle inequality while the usual norm does not. However, it contradicts a theorem that I have studied.

"On a finite dimensional vector space, any two norms are equivalent."

We know that a field over itself is a one - dimensional vector space and therefore, $$\mathbb{Q}$$ is a vector space. From the above statement $$p$$ adic norm and usual norm should be equivalent.

I know that there is something hiding behing the "$$p$$ - adic" norm, rather numbers. But I am not able to figure it out. I would like some help regarding the same.

The theorem you quote carries a bunch of implicit assumptions, more specifically, it only considers special kinds of norm. Specifically, the norms it concerns are norms $$\|\cdot\|$$ on a finite-dimensional vector space $$V$$ over real numbers, which further satisfies the homogeneity property $$\|av\|=|a|\|v\|$$ for all $$v\in V,a\in\mathbb R$$.
As you see, the example of the standard and of the $$p$$-adic norm show that this is not true much more generally. In fact, both requirements are necessary: the norm must both be homogeneous (otherwise there are wild counterexamples even on $$\mathbb R$$) and the vector space must be over $$\mathbb R$$ (we can find homogeneous examples over $$\mathbb Q$$. Actually, some other fields will work in place of $$\mathbb R$$, for instance $$p$$-adic numbers $$\mathbb Q_p$$, but it's an exception more than a rule).