# Showing rational numbers are algebraic

A polynomial with integer coefficients is an expression of the form: $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

where $a_n$, $a_{n-1}, \ldots, a_1, a_0$ are integers and $a_n$ is not equal to $0$.

a zero of the polynomial is a $c \in \mathbb{R}$ such that $f(c)=0$

A real number is said to be algebraic if it is a zero polynomial with integer coefficients

1) Show that every rational number is algebraic

2) Show that if $a$, $b$ and $k$ are positive integers, then the $k$-th root of $a/b$ is algebraic

I don't even know where to start on this. What is a zero of a polynomial with integer coefficients?

• What is the simplest polynomial equation with integer coefficients that has $2/3$ as a root?
– lhf
Commented Oct 15, 2013 at 19:04
• Consider the polynomial x-1, 2x-1, x-2 as a few examples of what are the zeroes for these and what can this tell you about generating a polynomial for a given fraction? Commented Oct 15, 2013 at 19:07
• A zero of a polynomial with integer coefficients is just what you said it was in your question, when you typed "a zero of the polynomial is a, C is in R such that f(c)=0". Except that you shouldn't be changing notation in midstream from capital $C$ to lower-case $c$. Commented Oct 15, 2013 at 19:09
• What have you tried? Do you understand what an algebraic number is? Do you know how to show that $\frac{3}{5}$ is algebraic? Commented Oct 16, 2013 at 0:58
• Do you know what a zero of a polynomial is (AKA, a root of a polynomial)? Commented Oct 16, 2013 at 1:07

The rational number $5/7$ is a zero of the polynomial $7x+(-5)$. We have $n=1$, $a_1=7$, $a_0=-5$.

So try showing that works with every rational number.

• How would I do that? Would I call 5 = to a and 7 = to b and say that it is true for a+1/b+1? Commented Oct 15, 2013 at 19:21
• It seems to me you are confused about far more than this one problem. A rational number is a number that can be written as an integer over another integer, i.e. $a/b$, where $a$ and $b$ are integers. If $x=a/b$, i.e. if $x$ is rational, then $bx+(-a)$ is $0$, so $x$ is a zero of the polynomial $bx+(-a)$, and $b$ and $-a$ are integer coefficients. Commented Oct 15, 2013 at 22:36

Hint: The polynomials you need will have very simple forms.

Bigger hint:

In particular, they will all have exactly two terms.

• Of course this is true, but how does it help?
– Pedro
Commented Oct 16, 2013 at 1:13
• @PedroTamaroff, this was my best attempt at a hint that might be helpful but wouldn't give it all away. Commented Oct 16, 2013 at 1:17
• I would rather exhibit one particular polynomial in each case, and ask to generalize. As it stands, you're saying "The solution will not be complicated." which I don't see as something helpful, rather as a motivational comment.
– Pedro
Commented Oct 16, 2013 at 1:20
• @PedroTamaroff, I don't remember specific examples, but I definitely remember times when being told that the solution is simple actually helped me find it. Commented Oct 16, 2013 at 1:23
• It might be helpful to look at the duplicate question (posed by OP), and see that an explicit example was given, but OP didn't understand it. Commented Oct 16, 2013 at 1:27

Sometimes it helps to read the definition over and over.

A complex number is said to be algebraic if there exists $$n\in\mathbb{N}$$ and integers $$a_0,a_1,\dots,a_n$$ with $$a_n\neq0$$ that $$\sum_{j=0}^na_jz^j=0$$.

To show that every rational is algebraic one needs two terms and make a simple observation on how to choose the $$a$$'s, as the definition says integers $$a_j$$.