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So we have the following given to us this weekend just on a handout:

Considering $\mathbb R$ as a vector space over the field $\mathbb Q$, show that $\sqrt 3$ is not an element of the span of $(1, \sqrt 2)$. His hint was to assume $\sqrt 3$ and $\sqrt 2$ are not elements of $\mathbb Q$ and $\sqrt 3$ is not in the span of $\sqrt 2$.

Assuming that he will want something short and sweet since on his handout, and I know the span means that I have to show that no linear combination of 1 and $\sqrt 2$ will ever equal $\sqrt 3$. Yet for some reason I seem to be going in circles with this problem. Do I need to even worrying about $\mathbb Q$ or not. Anybody with hints or ideas?

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andemw01
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    "I don't know why its not letting type the square root symbols today" See this page for information on how to type with MathJax and $\LaTeX$ here. – JMoravitz Sep 24 '17 at 21:49
  • Related general case, but not the same https://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of – Widawensen Jun 15 '18 at 12:19

1 Answers1

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Suppose we could find some $a,b$ in the rationals with $$ a+b\sqrt{2}=\sqrt{3} $$ If $a=0$, then we have that $$ b=\frac{\sqrt{3}}{\sqrt{2}} $$ Put $b=p/q$ is into lowest terms and mimic the proof for the irrationality of $\sqrt{2}$, namely $$ 2p^2=3q^2 $$ which means $p$ is divisible by $3$ and $q$ by $2$, and thus $p^2$ is divisible by $9$ and $q^2$ by $4$, thus $q$ is divisible by $3$ and $p$ by $2$ s well, contradicting our putting the fraction in lowest terms, since both $gcd(p,q)\geq 6$.

If b is zero, then we have that $\sqrt{3}$ is rational.

So, this then means that $$ a^2+2ab\sqrt{2}+2b^2=3\implies\sqrt{2}=\frac{3-2b^2-a^2}{2ab} $$ a contradiction of the irrationality of $\sqrt{2}$.

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