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I was requested to prove $\mathbb{R}$ is a vector space over $\mathbb{Q}$. However, I do not see how this problem is not completely trivial. We know $(\mathbb{R}, +)$ is an abelian group. Since $\mathbb{Q} \subset \mathbb{R}$ and $(\mathbb{R}, +, \cdot)$ is a field on itself, then we know

  • $\frac{a}{b}\cdot (x+y)=\frac{a}{b}x+\frac{a}{b}y$
  • $\frac{a}{b} \cdot (x \cdot y)= (\frac{a}{b} \cdot x) \cdot y$
  • $1 \cdot x=x$, with $1 \in \mathbb{Q}$,

for any $\frac{a}{b} \in \mathbb{Q}, x \in \mathbb{R}$.

What seems trivial to me is that the three properties immediately follow from the fact that $(\mathbb{R}, +, \cdot)$ is a field and $\mathbb{Q} \subset \mathbb{R}$. Then, for example, we already know $\cdot$ is distributive over $+$, is associative, etc.

lafinur
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    You are right, it is completely trivial. Many questions are when you are being taught fundamentals. – JMoravitz Dec 07 '22 at 14:38
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    Thank you. I was just worried I was missing something. It is also the case that many questions seem trivial when you don't understand them correctly. – lafinur Dec 07 '22 at 14:39
  • For what its worth, this example is an important one. Consider exploring it further in depth with things such as talking about a basis for this space or looking at $\Bbb R/\Bbb Q$ which is useful in defining things like Vitali Sets. – JMoravitz Dec 07 '22 at 15:16

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