Now rewrite this as $$\sqrt5={a\over b}-\sqrt2.$$ Squaring both sides of this equation we obtain $$5={a^2\over b^2}-2\sqrt2{a\over b}+2.$$ Now, carefully solve for $\sqrt2$ and obtain $$\sqrt2={-3b\over 2a}+{a\over 2b.}$$ This implies that $\sqrt2$ is a rational number which is a contradiction. Thus $$\sqrt2+\sqrt5$$ is an irrational number.

Sep 19, 2020 · Once you know that $[ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ]=4$, with basis $\{1,\sqrt{5},\sqrt{7},\sqrt{35}\}$, you can proceed as follows, without finding ...

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Jun 6, 2015 · The simple approach (take the product of $(x-\alpha)$ over all of the possible $\alpha$ obtained by applying the automorphisms in the Galois group to $\sqrt2+\sqrt3+\sqrt5$) works here because $\sqrt2$, $\sqrt3$ and $\sqrt5$ are 'mutually irreducible' over $\mathbb{Q}$ (in the sense that $\sqrt2\not\in\mathbb{Q}(\sqrt{3}, \sqrt{5})$, etc.) , so ...

Mar 22, 2015 · According to Galois, if $\sqrt3 +\sqrt5+\sqrt7$ is rational, then it must be invariant after changing the sign of any or all of the square roots. But the expression is clearly positive, so it cannot be equal to its negative.

$\begingroup$ Forgive my ignorance, but isn't it a little simpler than this? Here's my argument. Since $2^2 < (\sqrt{5})^2 < 3^2,$ and since the positive square root function is strictly increasing, thus $2 < \sqrt{5} < 3.$ Since there are not natural numbers between $2$ and $3$, this means that $\sqrt{5}$ is non-natural.

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How can I find the continued fraction expansion for the square root of 5. Do this without the use of a calculator and show all the steps.

Sep 23, 2016 · prove that $2\sqrt5 +\sqrt{11}$ is irrational. 2. Prove constructively that $\log_2 3$ is irrational. 6.

An easy way to show that $\sqrt{5}$ is not in $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is to note that $[\mathbb{Q}(\sqrt{2},\sqrt{3})\colon\mathbb{Q}]=4$, so by the Fundamental Theorem of Galois Theory it has either one or three intermediate fields of degree $2$ over $\mathbb{Q}$ (since a group of order $4$ has either a single subgroup of index $2$, when the group is cyclic, or exactly three, when it ...