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As title say, which is the degree of the algebraic number $\frac{1+\sqrt{2}+\sqrt{3}}{5}$ ?

Obviously it is not degree 1 (rationals). It seems also not degree 2 (I do not known any way to convert form $\frac{a+b\sqrt{c}}{d}$).

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    With all respect, what is wrong in this question, that deserves a down vote?. If there are something wrong, I will delete it. – 2017-01-01
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    Complete lack of personal input, most probably. – 2017-01-01
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    There are few things I can add. Question seems to be clear. And answer or method to find the answer is unknown to me. – 2017-01-01
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    "Question seems to be clear." Clear questions with no personal input are offtopic on this site (should I say "should be"... :-)). "It is also not degree 2 (I do not known any way to convert form (a+b√c)/d." Why do you assert the degree is not 2? Is it an opinion or do you have a proof? – 2017-01-01
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    It is in fact part of the question – 2017-01-01
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    So your thoughts on this are that the degree is not 1, nothing else? My impression is that you could try to **prove** the degree is not 2 either without killing yourself at the task... – 2017-01-01
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    First of all show its degree is the same as the degree of $\sqrt{2}+\sqrt{3}$, which is easier to compute, because it is rather straightforward to see that $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$. – 2017-01-01
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    @pasabaporaqui : dont be taken back by down votes when you want to know something, number of votes is not about how good or bad a question is, it is about how others think it is. Sometimes you might get down voted first then when people realise what you asking then vote ups pour in. Just ask and dont be bothered with up or down. – 2017-01-01
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    @pasabaporaqui this is strange, this question has got over 100 views and 4 downvotes, so many views means it is interesting, – 2017-01-01
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    @Arjang Your advice (invoking some rather mysterious and ad hoc mechanism according to which upvotes should follow downvotes, probably like the sun follows the rain) to a relatively new user to neglect the remarks explaining how their question does not fit the rules of the site, is not the most constructive contribution one can imagine, to stay polite. – 2017-01-01
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    @Arjang "so many views means it is interesting" Huh? What are you talking about? – 2017-01-01
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    @Did : Down or Upvotes are not the only measure, considering the number of views this question has been getting within a short time and number of votes it has received, very often a question with very few views gets downvoted to oblivion, I have also seen number of times where a downvoted question has ended up being upvoted. In many occasions a question that was against the guidlines, e.g. it looked like a homework with no effort shown has been upvoted, most likely by the friends of the poster, or whoever that did not give a damn about guidlines, Yes, people should take comments into account. – 2017-01-01
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    Problem appears when there are several downvotes and no comments. There are no way to improve or delete the question. – 2017-01-01
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    @Arjang Your rather fascinated and exclusive concern for votes (despite your claims of the contrary) is odd, it seems to make you miss that the intrinsic characteristics of a question are, at least for some of us, coming first. – 2017-01-01
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    @pasabaporaqui "when there are several downvotes and no comments" But you got comments, specific ones even, right? – 2017-01-01

3 Answers 3

4

Some hints:

  1. the degree of $\dfrac{1+\sqrt{2}+\sqrt{3}}{5}$ is the same as the degree of $\sqrt{2}+\sqrt{3}$

  2. the degree of $\sqrt{2}+\sqrt{3}$ is the dimension of $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ as a vector space over $\mathbb{Q}$

  3. $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$

  4. dimension formula for finite field extensions

It's basic knowledge about algebraic numbers that, if $b\in\mathbb{C}$ is algebraic over $\mathbb{Q}$, with minimal polynomial of degree $n$, then $\{1,b,b^2,\dots,b^{n-1}\}$ is a basis of $\mathbb{Q}(b)$ over $\mathbb{Q}$, where $\mathbb{Q}(b)$ is the least subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and $b$.

Check also “Degree of a field extension” on Wikipedia or your lecture notes.

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    Could you add a link to the definitions of "dimension of Q(xxx) over Q" and "Q(xxx)"? Thanks a lot. – 2017-01-01
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    @pasabaporaqui Those are basic tools when you talk about degree of algebraic elements. A field extension $F$ of $\mathbb{Q}$ is a vector space over $\mathbb{Q}$ and it's the dimension as vector spaces I'm dealing with. – 2017-01-01
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    nothing in googling for "dimension of Q". Any hint? – 2017-01-01
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    @pasabaporaqui Any lecture notes on algebraic numbers. I don't know how you're supposed to solve your homework, but just checking with degree of polynomials surely is the wrong way. – 2017-01-01
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    No homework. Writing a program submodule (in a tool to guide math learning in secondary school) for calculations over Z, Q, A(restricted to integer operations +,-,...), R and C. Z, Q, R, C are easy, A is not. Question under this is one: is it possible to "simplify" the form of this value (or someone similar) is expressed? Under previous: what can be defined as the result of a calculation on A? It is not possible write a program if their correct result is undefined. However, after several disappointing experiences, I try to ask simple, concrete and clear questions and continue by myself. – 2017-01-01
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    @pasabaporaqui It's easy to find that $(1+\sqrt{2}+\sqrt{3})/5$ satisfies a degree $4$ polynomial; quite more difficult is to show this polynomial is irreducible and that's where considerations about intermediate fields help. – 2017-01-01
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Set $ x $ equal to the number and work until you have a polynomial in $ x $ that's equal to zero.

$$ \begin{array}{r c l} x & = & \frac{1 + \sqrt{2} + \sqrt{3}}{5} \\ 5x & = & 1 + \sqrt{2} + \sqrt{3} \\ 5x - 1 & = & \sqrt{2} + \sqrt{3} \\ (5x - 1)^2 & = & \left(\sqrt{2} + \sqrt{3}\right)^2 \\ 25x^2 - 10x + 1 & = & 2 + 2 \sqrt{6} + 3 \\ \frac{25}{2} x^2 - 5x - 2 & = & \sqrt{6} \\ \left(\frac{25}{2}x^2 - 5x - 2\right)^2 & = & \left(\sqrt{6}\right)^2 \\ \frac{625}{4}x^4 - 125x^3 - 25x^2 + 20x + 4 & = & 6 \\ \frac{625}{4}x^4 - 125x^3 - 25x^2 + 20x - 2 & = & 0 \\ 625x^4 - 500x^3 - 100x^2 + 80x - 8 & = & 0 \end{array} $$ $$ \bbox[5px,border:2px solid black]{\mbox{The degree of the minimal polynomial is 4.}} $$

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    You have not shown that $625x^4 - 500x^3 - 100x^2 + 80x - 8$ is minimal. – 2017-01-01
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    "The degree of the minimal polynomial is 4" Is this a conjecture, a wish, or a fact your answer would have proven? – 2017-01-01
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Let $x=\frac{1+\sqrt{2}+\sqrt{3}}{5}$. Then, $x$ is a root of the following polynomial,

$$625x^4-500x^3-100x^2+80x-8$$. So,the degree of $x$ is 4, if you think over quotient number.

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    I don't think that x is a root of that polynomial. – 2017-01-01
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    Thanks, but can you explain how to find this polynomial? – 2017-01-01
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    "Let y=5x−1, then y=√2+√3" Sure but this is only the beginning of the determination of the polynomial. This answer is also incomplete on a second count, which is that you do not explain why lesser degree polynomials would not fit. (Of course arguing that incomplete answers fit questions with no context is tempting... but slightly too paradoxical for my taste.) – 2017-01-01
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    Let $y=5x-1$, then$y=\sqrt{2}+\sqrt{3}$. $(y-\sqrt{2})^2=3$,$(y^2-1)^2=(2\sqrt{2}y)^2$,$y^4-10y^2+1=0$. Substitute for this formula. – 2017-01-01
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    Quote: `explain why lesser degree polynomials would not fit`. – 2017-01-01
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    Sorry, I forgot it. We can find all of the root of the polynomial are $\frac{1\pm \sqrt{2}\pm \sqrt{3}}{5}$ . So, you can check any cases whether lesser degree polynomials would fit or not. – 2017-01-01
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    "you can check any cases whether lesser degree polynomials would fit or not" Sorry but how do you suggest to do that exactly? – 2017-01-01