Home Algebra Algebra for beginners at the college level

Algebra for beginners at the college level

algebra-for-beginners

We give a concise summary of math that helps in learning algebra for beginners at the college level. What you need to know is set theory and in particular the set of real numbers. This will help you to understand some important classes of sets such as groups, rings, polynomial rings, and fields. Also, you need to know some fundamental equalities, like remarkable identities, binomial expansion formula, etc.

The basics of algebra for beginners at the college level

We collect the important part of algebra to help beginners to well understand their first lectures on math at the University.

The basics of Sets

A set is just a collection of objects. For example, a set can be a collection of sports cars, or a set can be a collection of trigonometric functions. As example the set that contain positive integer numbers is denoted by $\mathbb{N}:=\{0,1,2,\cdots\}$.

We say that $a$ belongs to a set $A$, we write $a\in A,$ if $a$ is one of the collections of objects in $a$. In this case,  $a$ is called an element of $A$.

We say that another set $B$ is a subset of the set $A$, and we write $A\subset B$ if all elements of $B$ are also elements of $A$. By the way, two sets $E$ and $F$ are equal if and only if $E\subset F$ and $F\subset E$.

Example: We denote by $\mathbb{R}$ the set of real numbers and by $\mathbb{Q}$ the set of rational numbers. Then $\mathbb{Q}$ is not equal to $\mathbb{Q}$. This is because $\mathbb{Q}\subset \mathbb{R}$ and $\sqrt{2}\in \mathbb{R}$, but $\sqrt{2}\notin \mathbb{Q}$. Let us give you a more advanced property of $\mathbb{R}$. We assume that the reader is familiarized with the convergence of sequences. Let $x$ be an arbitrary real number. We select the following sequence $x_n=\frac{[nx]}{n^2}$ for $n\in\{1,2,\cdots\}$, where $[nx]$ is the integer part of the real number $nx$. It verifies $nx-1<[nx]\le nx$. Thus $x-\frac{1}{n}<x_n\le x$. Hence the sequence $(x_n)_n$ converges to $x$ when $n$ goes to $+\infty$. Remark that $x_n\in\mathbb{Q}$ for any $n$. Thus any real number is a limit of a sequence of rational numbers.

The empty set denoted by $\emptyset$ is a set that contains no elements. For any set $A,$ we have $\emptyset A$. On the other hand, the subsets of $A$ form another set denoted by $\mathscr{P}(A)$. So that $$\mathscr{P}(A)=\{ B: B\subset A\}.$$ Remark that $\emptyset$ and $A$ are both elements of $\mathscr{P}(A)$.

If we have two sets $A$ and $B$ then we can define other sets like the union of $A$ and $B$ denoted by $A\cup B$, and the intersection of $A$ and $B$ denoted by $A\cap B$. The product of $A$ and $B$ defined by $$ A\times B=\{(a,b):a\in A,\;b\in B\}.$$

Map between two sets

A map $f$ between two sets $A$ and $B$ is any subset of $A\times B$. It will be denoted by $f: A\to B$ $x\mapsto f(x)$. We can define a map between $\mathbb{R}$ and the set $\{0,1\}$ bt $f(x)=1$ is $x\in [0,+\infty)$ and $f(x)=0$ if $x\in (-\infty,0)$. A map is also called a function.

A function $f:A\to B$ is called in injective if $f(x)=f(y)$ implies $x=y$. As example the function $f:\mathbb{N}\to\mathbb{}N$ defined by $f(n)=2^n$ is injective. In fact, if $f(n)=f(m)$ then $2^n=2^{m}$. Now we apply the logarithmic function, we get $n\log(2)=m\log(2)$, which implies that $n=m$.

The map $f$ is said to be surjective if for every element of $B,$ there is at least one element $x$ in $A$ such that $y=f(x)$. The function $f(n)=2^n$ is not surjective. If not there will be $n\in \mathbb{N}$ with $3=2^n$. Thus $n=\frac{\log(3)}{2}\notin \mathbb{N},$ absurd. let $\mathbb{C}$ be the set of complex numbers. The maps $f:\mathbb{R}\times \mathbb{R}\to\mathbb{C}$ defined by $f(a,b)=a+ib$ with $i$ is the complex number satisfying $i^2=-1$. Then $f$ is surjective. In fact, we know that for any $z\in\mathbb{C}$ one can find real numbers $a\in\mathbb{R}$ and $b\in \mathbb{R}$ such that $z=a+ib=f(a,b)$.

We say that $f$ is bijective, or one to one, if it is injective and surjective. This means that for any $y\in B,$ there is only one element $x\in A$ such that $y=f(x)$.

Previous articleComplex numbers exercises
Next articleSimplify expressions of real numbers and fractions