# Proving that $\frac{1+\sqrt{3}}{2}$ is not an algebraic integer

How can I prove that $$\frac{1+\sqrt{3}}{2}$$ is not algebraic integer in $$\mathbb{Z}$$?

I understand algebraic integer in a commutative ring $$R$$ as any element $$r\in R$$ which satisfies equation $$P(r)=0$$ where $$P$$ is a nontrivial polynomial whose coefficients are multiplies of $$1_{R}$$ and the top degree coefficient is $$1_{R}$$.

• Well, what is the minimal polynomial for that algebraic number?
– lulu
Oct 23, 2019 at 18:28
• You probably want to say that it's not an algebraic integer. Oct 23, 2019 at 18:29
• If it were, then so too would $2\times \frac{1+\sqrt{3}}{2} = 1+\sqrt{3}$ be an integer since integers are closed under multiplication. Further, so too would $1+\sqrt{3}-1=\sqrt{3}$ since integers are closed under subtraction. Now, can you prove the much simpler problem of showing that $\sqrt{3}$ is not an integer? Oct 23, 2019 at 18:36
• It’s between $1$ and $2$ and there is no such element in $\Bbb Z$ ;-) Oct 23, 2019 at 18:40
• See algebraic integer and integral element for the correct definitions. Oct 23, 2019 at 21:03

HINT. I assume you meant algebraic integer. Find the minimal polynomial for $$\dfrac{1+\sqrt{3}}{2}$$. If you are having trouble with this, follow the idea from this answer. Now examine that minimal polynomial. Why does this being the minimal polynomial imply that $$\dfrac{1+\sqrt{3}}{2}$$ is not an (algebraic) integer?

If you meant actual integer, then this is much easier. If you take two integers, then their product is an integer. And if you sum/difference two integers, you get an integer. How do you use these to show that $$\dfrac{1+\sqrt{3}}{2}$$ is an integer if and only if $$\sqrt{3}$$ is an integer? Now show that $$\sqrt{3}$$ is not even rational (nevertheless an integer!).

• Though it is easy to show $\sqrt{3}$ is not rational, it is even easier to show it is not an integer: $1^2<3<2^2$ therefore $1<\sqrt{3}<2$. Oct 23, 2019 at 18:41

If $$\alpha$$ is an algebraic integer then so too is $$\alpha' = 1-\alpha$$ hence so too is $$\alpha\alpha' = -1/2,\,$$ contradiction.

Remark  More conceptually let's recall one motivation for the definition of algebraic integers. Suppose that we desire to consider as "integers" some subring $$\:\mathbb I\:$$ of the field of all algebraic numbers. To be a purely algebraic notion, it cannot distinguish between conjugate roots, so if $$\rm\:\alpha,\alpha'$$ are roots of the same polynomial irreducible over $$\rm\:\mathbb Q,\:$$ then $$\rm\:\alpha\in\mathbb I\iff \alpha'\in\mathbb I.\:$$ Also we desire $$\rm\:\mathbb I\cap \mathbb Q = \mathbb Z\$$ so that our notion of algebraic integer is a faithful extension of the notion of a rational integer. Now suppose that $$\rm\:f(x)\:$$ is the monic minimal polynomial over $$\rm\:\mathbb Q\:$$ of an algebraic "integer" $$\rm\:\alpha\in \mathbb I.\:$$ Then $$\rm\:f(x) = (x-\alpha)\:(x-\alpha')\:(x-\alpha'')\:\cdots\:$$ has coefficients in $$\rm\:\mathbb I\cap \mathbb Q = \mathbb Z.\:$$ Therefore the monic minimal polynomial of elements $$\in\mathbb I\:$$ must have coefficients $$\in\mathbb Z.\,$$ In particular a quadratic irrational $$\,\alpha\in\Bbb I\iff (x\!-\!\alpha)(x\!-\!\alpha') = x^2\!-(\alpha\!+\!\alpha') x + \alpha\alpha'\in\Bbb Z[x],\,$$ i.e. iff $$\alpha$$ has trace and norm $$\in \Bbb Z,\,$$ which fails in the OP since $$\,\alpha\alpha' = -1/2$$.

Conversely, one easily shows that the set of all such algebraic numbers contains $$1$$ and is closed under both difference and multiplication, so it forms a ring.

Hence a few natural hypotheses on the notion of an algebraic integer imply the standard criterion in terms of minimal polynomials.

Hint:

Find a polynomial equation satisfied by this number, and apply the rational roots theorem.

I would be inclined to prove it is not rational (and so cannot be an integer) using the "rational root theorem". The rational root theorem say that "if $$\frac{m}{n}$$ is a root of the polynomial equation $$a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$$, with all coefficients integers, then the denominator, n, must divide the leading coefficient, $$a_n$$, and the numerator, m, must divide the constant term, $$a_0$$".

If $$x= \frac{1+ \sqrt{3}}{2}$$ then it satisfies the equation $$\left(x-\frac{1+ \sqrt{3}}{2}\right)\left(x+ \frac{1- \sqrt{3}}{2}\right)= \left(-\frac{1}{2}- \frac{\sqrt{3}}{2}\right)\left(x-\frac{1}{2}+ \frac{\sqrt{3}}{2}\right)= \left(x- \frac{1}{2}\right)^2- \frac{3}{4}= x^2- x- \frac{1}{2}= 0$$. Multiplying by 2, that is equivalent to $$2x^2- 2x- 1= 0$$.

So any rational roots of that must be of the form $$\frac{m}{n}$$ where m divides -1 (so can only be 1 or -1) and n divides 2 (so can only be 1, -1, 2, and -2). That is the only possible rational roots of that equation are 1, -1, 1/2, or -1/2. The only integer roots are 1 and -1. Since $$\frac{1+ \sqrt{3}}{2}$$ is not equal to either 1 or -1, it is not an integer.