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I am trying to deduce whether or not

$$K=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\text{ and }ab<\sqrt{2}\}$$

is a subfield of $\mathbb{C}$. I feel like I'm getting stuck. If I subtract two numbers $a+b\sqrt{2}$ and $c+d\sqrt{2}$, then we have to show that

$$(a-c)(b-d)=ab+cd-(ad+bc)<2\sqrt{2}-(ad+bc)<\sqrt{2}$$

Thus I have $ad+bc>\sqrt{2}$ and I'm not sure how to procede here. It I look at the multiplicative inverse of $a+b\sqrt{2}$,

$$(a+b\sqrt{2})^{-1}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}$$

and so the conditions must hold for

$$\frac{-ab}{(a^2-2b^2)^2}<\frac{-\sqrt{2}}{(a^2-2b^2)^2}<\sqrt{2}$$

$$\frac{1}{(a^2-2b^2)^2}>-1$$

Which is true for all $a,b$. I'm stuck....

  • 2
    How about $1+\sqrt 2 \in K$ but $2 + 2\sqrt 2 = (1+ \sqrt 2) + (1 + \sqrt 2) \notin K$? – 2017-02-05
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    So how do you know what numbers to use when constructing a counderexample? – 2017-02-05
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    Trial and error. Any time you're trying to prove or disprove something, you need to consider both possibilities: If it's true, try to prove it, if it's false, look for a counterexample. – 2017-02-05
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    Also for your case $ 4 - 2\sqrt{2} - ( 3 - 5\sqrt{2}) = 1 + 3 \sqrt{2} \not \in K$ – 2017-02-05
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    Okay I see. Thank you. – 2017-02-05

2 Answers 2

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No, as clearly $1,\sqrt{2}\in K$ we have since this is a field that $a+b\sqrt{2}\in K$ for all $a,b\in\Bbb Z$. In particular you can choose $a=1, b=2$ to quickly come to a counterexample.

You can also see this in other ways, since $\Bbb Q(\sqrt 2)$ has degree $2$ over $\Bbb Q$, any proper subfield is simply the rational numbers, but then clearly this is not equal to $K$ as $1+\sqrt 2\in K$ is not rational.

Finally if you assumed $K$ was a field, clearly it is not simply the rational numbers so it has some degree $n>1$ over $\Bbb Q$ which must be finite because $K$ contains only algebraic numbers. But then its ring of integers is a rank-$n$ lattice in $\Bbb R\otimes K\subseteq\Bbb R^n$, which in particular means it has a vector, $\mathbf{v}$, with all coordinate not equal to zero, so $n\cdot \mathbf{v}$ has arbitrarily large coordinates, and as $\sqrt{2}$ is linearly independent from $1$ over $\Bbb Z$ and is in $K$, we can choose it as one of the basis vectors for the vector space, implying we can make $ab$ as large as we like. (This one motivated by the geometric consideration that your inequalities define the space under a hyperbola, and so in particular would imply a bound in a particular direction for the integers of $K$, which is impossible by the full rank description of the integer ring.)

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The key here is observing that, for $a,b,c,d\in\mathbb{Q}$, $$ a+b\sqrt{2}=c+d\sqrt{2} \quad\text{if and only if}\quad a=c\text{ and }b=d $$ which is easy, but should be noted before going on. One direction is obvious; for the reverse, if $b\ne d$, then $$ \sqrt{2}=\frac{a-c}{d-b} $$ which is a contradiction. Therefore $b=d$ and consequantly also $a=c$.

Now $2=2+0\sqrt{2}$ and $2\sqrt{2}=0+2\sqrt{2}$ both belong to $K$. However $2+2\sqrt{2}$ doesn't satisfy $2\cdot2<\sqrt{2}$ and there cannot be other rational numbers $a,b$ with $a+b\sqrt{2}=2+\sqrt{2}$. Thus $K$ is not closed under addition.