Olympiad Algebra and Analysis

Common Sets in Algebra (Part 2)

In the last part of the article, I explained about the set of natural numbers, integers, rational numbers, real numbers, and complex numbers. These are the common sets we will enocunter in Olympiad Problems. However, there is a set called the set of algebraic numbers which I will explain in further detail below.

Algebraic Numbers

So, what are algebraic numbers? An algebraic number is a number that is a root of a non-zero polynomial in one variable with integer coefficients. An algebraic number is called an algebraic integer if the described polynomial is monic, i.e. has leading coefficient $1$.

Real numbers that aren't algebraic but are called transcendental number. A famous example of a transcendental number is $\pi$.

Note that some complex numbers are algebraic by definition. (e.g. $i$, the root of $x^2 + 1$)

The set of algebraic numbers

Now, we define the set of algebraic numbers (often denoted as $\mathbb{A}$), as, well, the set of all algebraic numbers (which are described above.)

Some important properties of $\mathbb{A}$.

1. The algebraic numbers form a field, i.e. The sum, difference, product and quotient of two algebraic numbers is again algebraic.

2. Given an algebraic number, there is a unique monic polynomial (with rational coefficients) of least degree that has the number as a root. This polynomial is called its minimal polynomial. If its minimal polynomial has degree $n$, then the algebraic number is said to be of degree $n$. An algebraic number of degree $1$ is a rational number. A real algebraic number of degree $2$ is a quadratic irrational.

That's all on Algebraic Numbers.

Check out the practice!

Hello!

This is sort of a Tricki-style experiment, but not exactly...

Please/feel free to contribute any ("algebraic") heuristic/trick/strategy here, no matter what level, or how broad/specific.

Thank you!

Here are some (not necessarily good) examples, to get us started...

1. A super-specific strategy: $\sin(x-y)\sin(x+y) = \sin(x)^2 - \sin(y)^2$ can sometimes help in simplifying trigonometric expressions. (What other functions satisfy $f(x-y)f(x+y) = f(x)^2 - f(y)^2$? See Putnam 1963 B3 for a partial answer.)

2. Another specific strategy: it can sometimes help to work modulo $4$ in Diophantine equations with squares. (And a more general related strategy: sometimes it helps to take a Diophantine equation modulo a specific number $m$, where the choice of $m$ depends on the context. For example, modulo $7$ helps to deal with cubes, since $7 \equiv 1\pmod3$ means that $a^3$ only takes on $\frac{7-1}{3} = 2$ nonzero values modulo $7$. This generalizes to mod $p$ to deal with $d$th powers, when $p\equiv1\pmod{d}$.)

3. A broader/more general strategy: deal with inverses by undoing inverse (radicals, logs, inverse trig, even basic stuff like $(x+1)-1$ to solve linear equations).

4. An algorithm for solving any system of polynomial equations: Buchberger's algorithm (thanks to Terry Tao for mentioning this, but unfortunately, as he said, I don't think we can teach this...)

This is a stub article; the stuff on transformations and completing the square should be expanded.

The goal of this post is to show (algebraically) that all quadratics have the "same basic form", or equivalently (geometrically), that all parabolas have the "same basic shape".

Specifically, we'll show that any quadratic $aX^2+bX+c$ can be easily transformed to a "simpler" quadratic $X^2 - d$ (which has roots at $\pm\sqrt{d}$, (almost) by definition of the square root of $d$, at least if $d\ge0$). So in order to understand the "entire family" of quadratics, it's "enough" to understand the "simpler family" of quadratics of the form $X^2 - d$. As long as we keep track of how the more "complicated" quadratics are related to their simpler basic forms, we'll usually be good to go.

For example, the quadratic formula is a special case of this idea: we can relate the roots of a "complicated" quadratic to the roots of a "simpler" quadratic $X^2 - d$. In some sense, that's all the quadratic formula really says; we're just "solving for the roots of a general quadratic in terms of the roots of simpler quadratics $X^2 - d$ (which are $\pm\sqrt{d}$, essentially by definition)".

The quadratic formula gives the roots of a quadratic, at least when the discriminant is nonnegative. But some quadratics have no real roots, so in order to "have solutions" we need to introduce imaginary numbers. Once we do that, the quadratic formula extends to real quadratics with negative discriminant as well, as discussed later.

Stub explanation

The quadratic discriminant $b^2-4ac$ of a real quadratic $P(x) = ax^2+bx+c$ has a nice interpretation in terms of the roots $r_1,r_2$ (possibly equal) of $P$. The story starts with factored form of $P$: $$P(x) = a(x-r_1)(x-r_2) = ax^2 - a(r_1+r_2)x + a(r_1r_2).$$ By the method of undetermined coefficients (or more precisely, Vieta's formulas for quadratics), we have $b = -a(r_1+r_2)$ and $c = a(r_1r_2)$. Substituting, we find the discriminant of $P$ is $$b^2 - 4ac = [-a(r_1+r_2)]^2 - 4a[a(r_1r_2)] = a^2[(r_1+r_2)^2 - 4r_1r_2],$$ where we have factored out $a^2$ in the second step. But we may expand to get $(r_1+r_2)^2 = r_1^2+2r_1r_2+r_2^2$ (the $n=2$ case of the binomial theorem), so our discriminant expression simplifies to $$a^2[r_1^2-2r_1r_2+r_2^2] = a^2(r_1 - r_2)^2.$$ So $b^2 - 4ac = a^2(r_1-r_2)^2$ (which you can also check using the quadratic formula for $r_1,r_2$).

This "squares of differences of roots" type expression can also be thought of as an alternative definition of the quadratic discriminant. In fact, it generalizes to discriminants of higher degree polynomials as well, like the cubic discriminant.

Recall that the complex numbers are numbers of the form $a+bi$ where $a, b \in \mathbb{R}$, i.e. $a$ and $b$ are real numbers, and the symbol $i$ satisfies the equation $i^2 + 1 = 0$. The complex numbers form a field, i.e. we can do all the usual operations we expect to be able to do with numbers, like addition, subtraction, multiplication and division.

Another property of the complex numbers that allows us to call them a field is that multiplication is commutative, i.e. $ab = ba$ for all complex numbers $a$ and $b$.

Finally, we note that the complex numbers satisfy the distributive rule, i.e. $a(b+c) = ab+ac$ for all complex numbers $a$, $b$ and $c$.

In fancy mathematical language, we can say that the complex numbers are isomorphic to the field $\mathbb{R}[X]/(X^2+1)$, the quotient of the polynomial ring $\mathbb{R}[X]$ by the maximal ideal $(X^2+1)$. But you don't have to understand what this means at this point.

Notice that every complex number $z = a+bi$ is described completely by specifying two real numbers, $a$ and $b$. In fancy mathematical language, way say that the complex numbers form an extension of the field of real numbers of degree $2$.

This allows us to plot complex numbers on the two dimensional cartesian plane, the complex number $z = a+bi$ corresponds to the point on the cartesian plane $(a, b)$.

We call such a graph an Argand diagram, named after the mathematician Argand, although he wasn't the first to describe it!

It now becomes clear that we can associate a length to every complex number - this is the length from the origin of the Argand diagram to the point corresponding to that complex number.

This length is a non-negative real number, and we call the length of a complex number the modulus of that number.

We denote the modulus of the complex number $z = a+bi$ by $|z| = \sqrt{a^2 + b^2}$.

We can also associate with every complex number a direction, this is the direction from the origin of the Argand diagram. We call this direction the argument of the complex number, and denote the argument of the complex number $z$ by $\arg(z)$. We usually measure the argument of a complex number in radians, and measure counterclockwise, starting from the $x$ axis.

We have to be careful though! If we rotate the complex number a full turn around the origin, we arrive back at where we started, thus we come back to the same number.

By convention, the argument of a complex number is usually chosen to lie in the interval $(-\pi, \pi]$.

We now introduce a result that pertains to the modulus.

For any complex numbers $z$ and $w$, we have $|zw| = |z||w|$. This means that we can either multiply $z$ and $w$ as complex numbers and then take the modulus, or we can take the modulus of $z$ and $w$ separately and then multiply them as real numbers, to get the same result.

We can also think about this in terms of a geometrical effect of multiplying the complex number $z$ by the complex number $w$ - the modulus of $z$ increases (or decreases, if $|w| < 1$) by a factor of $|w|$!

Can you prove this result?

We now introduce a result pertaining to the arguments of complex numbers.

For any complex numbers $z$ and $w$, we have $\arg(zw) = \arg(z) + \arg(w)$. Notice that we may have to add or subtract $\pi$ to get the result within the conventional range of $(-\pi, \pi]$.

Geometrically, this says that an effect of multiplying the complex number $z$ by the complex number $w$ is to add the argument of $w$ onto the original argument of $z$.

We now concentrate on complex numbers with modulus equal to $1$. These lie on the unit circle on the Argand diagram.

We use the symbol $\mathrm{e}^{i\theta}$ to denote the complex number with argument $\theta$ and unit modulus. Thus $|\mathrm{e}^{i\theta}| = 1$ and $\arg(\mathrm{e}^{i\theta}) = \theta$ (up to some multiple of $2\pi$).

By what we have learnt so far we know that the geometrical effect of multiplication by $\mathrm{e}^{i\theta}$ is to rotate by $\theta$ radians, while leaving the modulus the same.

In particular, $\mathrm{e}^{i\theta}\mathrm{e}^{i\phi} = \mathrm{e}^{i(\theta+\phi)}$. Think about this carefully! Perhaps draw a sketch, labelling all three angles.

Suppose $|z| = r$ and $\arg(z) = \theta$, then $z = r\mathrm{e}^{i\theta} = r(\cos(\theta)+i\sin(\theta))$.

Say we also have $|w| = \rho$ and $\arg(w) = \phi$, so that $w = \rho\mathrm{e}^{i\phi}$. Then $|zw| = |z||w| = r\rho$, and $\arg(zw) = \theta + \phi$. Thus $zw = r\rho\mathrm{e}^{i\theta}\mathrm{e}^{i\phi} = r\rho\mathrm{e}^{i(\theta+\phi)}$.

As $\mathbb{C}$ is a field extension of $\mathbb{R}$of degree $2$, $\mathbb{C}$ forms a vector space over $\mathbb{R}$ of dimension $2$. A standard basis for this vector space would be $\{1, i\}$.

Multiplication by an element $\alpha \in \mathbb{C}$ gives us a linear map $\mathbb{C} \to \mathbb{C}$. We denote the determinant of this map as $N(\alpha)$, the norm of $\alpha$.

By what we have said so far, in the standard basis mentioned above, the matrix representing non-zero $\alpha$ is a rotation matrix multiplied by $|\alpha|$.

As $|\mathbb{C}:\mathbb{R}| = 2$, there can be at most $2$ $\mathbb{R}$-homomorphisms in $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C})$. We have one nontrivial homomorphism - complex conjugation, that sends $a+bi \to a-bi$.

Polynomials

Definition

Division Algorithm

Remainder / Factor Theorem

** Rational Root Theorem**

** Identical Polynomials**

** More Exercises**

Factorizations

Definitions and Methods

Quadratics - Quadratic Formula (factoring linear out)

Cubics and Higher-degree Polynomials (factoring linear out )

Factor Theorem and its uses

Rational Root Theorem and its uses

Vieta's Formulas (related to factorization)

Factoring quadratics or higher out

Factoring Multivariable Polynomials

More Exercises