# Introduction to Algebraic Rings

An algebraic ring is one of the most fundamental algebraic structures. It builds off of the idea of algebraic groups by adding a second operation  (For more information please review our article on groups). For rings we often use the notation of addition and multiplication because the integers are a good analogy for a ring, but the operations “+” and “⋅” do not necessarily represent addition and multiplication in the ring. A ring is defined as a set S combined with two operations (“+” and “⋅”) that has the following properties: I. The set S is a group under one operation (we will call this operation “+”): a. It is closed under this operation: For all elements a and b in the set S, a+b is also in S. b. It contains an additive identity element (which we will call “0”, although it does not necessarily represent the integer 0): There is some element 0 in the set S such that for every element a in S, a+0 = 0+a = a. c. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c). d. Inverses exist: For every element a in the set S, there is a (-a)  in S such that a+(-a)= (-a)+a = 0. II. The group operation is commutative: For all a and b in the set S, a + b = b + a. III. The second operation (written as “⋅”) follows these rules: a. It is closed under this operation. b. The set contains a multiplicative identity element (which we call “1”, although it does not necessarily represent the integer 1): such that for all elements a in the set S, a⋅1 = 1⋅a = a. c. It is associative: For all a, b, and c in the set S, (a⋅b)⋅c = a⋅(b⋅c). d. It is distributive over “+” on both the right and left side: For all elements a, b, and c in the set S, a⋅(b+c) = a⋅b + a⋅c and (b+c)⋅a = b⋅a + c⋅a. Note that inverses are not required for the “⋅” operation. If inverses do exist under “⋅”, we have a special kind of ring called a field.   We will now look at some examples of rings and sets that aren’t rings. Example 1. The integers under addition and multiplication. I. The integers are a group under addition (see our group article). I. Addition can be carried out in any order so it is commutative. III. Multiplication follows the additional rules laid out for rings. The integers are closed under multiplication. The identity element is 1. Multiplication is associative and it is also distributive. Therefore, the integers are a ring under addition and multiplication. ∎ This is the reason addition, multiplication, 0, and 1 show up in our notation. If a set “acts like” the integers by following the ring axioms, it is a ring. Example 2. The set of 2×2 matrices under addition and multiplication. I. The set of 2×2 matrices form a group under addition. a. For any 2×2 matrices A and B, A+B is also a 2×2 matrix. b. The additive identity “0” is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$. c. Matrix addition is associative. d. The additive inverse of a matrix A is just -A. Observe $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} + \begin{bmatrix} -a & -b \\ -c & -d \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$. II. Addition of 2×2 matrices is commutative, A+B = B+A. III. Multiplication of 2×2 matrices satisfies the following conditions: a. It is closed under the operation. The multiplication of two 2×2 matrices results in another 2×2 matrix. b. The identity matrix is I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ and for any 2×2 matrix A, A⋅I = I⋅A = A. c. Matrix multiplication is associative. d. Matrix multiplication distributes over addition on the left and right side (This can be easily proven, but to save space it will not be proven here). Therefore, the set of 2×2 matrices is a ring. ∎   Example 3. The set of 2×2 matrices with non-negative entries under addition and multiplication. This set satisfies all of the requirements for the multiplication operation, but it is not a group under addition. Without negative entries, most matrices in this set do not have additive inverses. Therefore, this set is not a ring. ∎ Like groups, rings do not have to be composed of real numbers or matrices. Some examples include sets of polynomials, the complex numbers, and integers modulo n. The word “ring” comes from the German word “Zahlring,” which means a collection of numbers that circle back on themselves. However, now that the field has been studied more extensively, we know that cyclic rings are just one type of rings. The symbol ℤ for the integers is also named after a German word for numbers “Zahlen.”

# Introduction to Algebraic Groups

One of the most fundamental algebraic structures in mathematics is the group. A group is a set of elements paired with an operation that satisfies the following four conditions: I. It is closed under an operation (represented here by “+”, although it does not necessarily mean addition): For all elements a and b in the set S, a+b is also in S. II. It contains an identity element (often written as “e”): There is some element e in the set S such that for every element a in S, a+e = e+a = a. III. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c). IV. Inverses exist: For every element a in the set S, there is an a-1 in S such that a+a-1 = a-1+a = e We will look at some examples of groups and sets that aren’t groups. Example 1. Integers under addition, (ℤ, +). I. The group operation is addition. For any two integers n and m, n+m is also an integer. II. The identity element is 0 because for any integer n, n+0 = 0+n = n. III. Addition is associative. IV. Inverses exist. The inverse of any integer n is -n, and n + (-n) = (-n) + n = 0. Therefore, (ℤ, +) is a group. ∎ Example 2. Integers under multiplication, (ℤ, ⋅) I. The group is closed under multiplication, because for any two integers n and m, n⋅m is also an integer. II. The identity element is 1. For any integer n, n⋅1 = 1⋅n = n. III. Multiplication is associative. IV. Inverses do NOT exist. For any integer n, 1/n is not an integer except when n=1. Since the the set does not meet all four criteria, (ℤ, ⋅) is not a group. ∎ Example 3. The set of rational numbers not including zero, under multiplication (ℚ – 0, ) I. The group is closed under multiplication. II. 1 is the identity element. III. Multiplication is associative. IV. Inverses exist in this group. If q is a rational number, 1/q is also a rational number. And q⋅ (1/q) = (1/q) ⋅ q = 1. Notice that if 0 were in this set, it would not be a group because 0 has no inverse. Therefore, ℚ – 0 is a group. ∎ Example 4. The set {0, 1, 2} under addition I. This set is not closed under addition because 1 + 2 = 3, and 3 is not part of the set. Therefore, the set cannot be a group. ∎ Example 5. The trivial group {e} This group consists only of the identity element. We don’t need to specify the operation here because it works for both multiplication and addition. For addition, e=0. 0+0=0 so the group is closed under addition, has an identity element, and is closed under inverses (and addition is associative). For multiplication, e=1 and similarly, 1⋅1 = 1. ∎ The examples here involved only numbers, but there are many different types of groups. For example, the ways to transform a triangle is called a Dihedral Group, with the group operation being the act of rotating or reflecting the shape around an axis. The possible permutations of a Rubik’s Cube are also a group, with the operation being a sequence of moves. It’s powerful to know that a set is a group because it gives you an understanding of how the elements will always behave. Groups have many other properties that are useful to mathematicians, and there is a whole field of study built off of this knowledge called Group Theory.

The set of permutations of a Rubik’s Cube is considered an algebraic group.

## L’Hospital’s Rule

L’Hospital’s Rule is a useful way to evaluate tricky limits. It is most often used for limits of indeterminate form. The rule is as follows:

If $f(x)$ and $g(x)$ are differentiable on some interval around the number $a$ (or if $a=∞$, $f(x)$ and $g(x)$ are differentiable for all $x>ε$ for some $ε$), and ${\lim\limits_{x \to a} g(x)} ≠ 0$, then

$$\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$

We can take the derivatives of the numerator and the denominator and then take the limit and it will be the same as the limit of the original ratio. Indeterminate form includes limits of the forms $\frac{0}{0}$, $\frac{∞}{∞}$, $0\cdot∞$, $∞-∞$, $0^0$, $1^∞$ and $∞^0$.

Example 1. $\lim\limits_{x \to 0} \frac{\sin x}{x}$

In this example, we see that taking the limit of the numerator and denominator results in $\frac{0}{0}$, which is an indeterminate form. And since there is not an obvious way to simplify the fraction in order to make the limit easier to compute, it is a good candidate for L’Hospital’s Rule. We take the limit of the numerator and denominator as follows:

$$\lim\limits_{x \to 0} \frac{\sin x}{x} = \lim\limits_{x \to 0} \frac{\frac d{dx} (\sin x)}{\frac d{dx} (x)} = \lim\limits_{x \to 0} \frac{\cos x}{1} = cos(0) = 1$$

Example 2. $\lim\limits_{x \to ∞} \frac{e^x}{x^2+x+4}$

Here we see that attempting to take the limit as is results in $\frac{∞}{∞}$. So again we use L’Hospital’s Rule:

$$\lim\limits_{x \to ∞} \frac{e^x}{x^2+x+4} = \lim\limits_{x \to ∞} \frac{\frac d{dx} (e^x)}{\frac d{dx} (x^2+x+4)} = \lim\limits_{x \to ∞} \frac{e^x}{2x+1}$$

Taking the limit at this point still results in $\frac{∞}{∞}$, so we apply L’Hospital’s Rule again.

$$\lim\limits_{x \to ∞} \frac{e^x}{2x+1} = \lim\limits_{x \to ∞} \frac{\frac d{dx} (e^x)}{\frac d{dx} (2x+1)} = \lim\limits_{x \to ∞} \frac{e^x}{2} = ∞$$

$∞$ is not an indeterminate form and so that is our final answer.

Example 3. $\lim\limits_{x \to 0} (x \cdot \ln x)$

If we plug zero into the limit, we get the indeterminant form $0 \cdot (-∞)$. We can’t apply L’Hospital’s Rule right away because we do not have a ratio. Instead, we must manipulate the function so that it becomes a ratio. We can move $x$ to the denominator by rewriting as $\frac {1}{x}$.

$$\lim\limits_{x \to 0} (x \cdot \ln x) = \lim\limits_{x \to 0} \frac{\ln x}{\frac {1}{x}}$$

Now we have a ratio and the indeterminate form $\frac {-∞}{∞}$. Applying L’Hospital’s Rule,

$$\lim\limits_{x \to 0} \frac{\ln x}{\frac {1}{x}} = \lim\limits_{x \to 0} \frac{\frac d{dx} (\ln x)}{\frac d{dx} (\frac {1}{x})} = \lim\limits_{x \to 0} \frac{ \frac {1}{x}}{- \frac {1}{x^2}}$$

We still have an indeterminate form $\frac {∞}{-∞}$, but before applying L’Hospital’s Rule again, we should notice that we can cancel the $\frac {1}{x}$ in the top and the bottom. It is always best to simplify after applying L’Hospital’s Rule if possible.

$$\require{cancel} \lim\limits_{x \to 0} \frac{ \frac {1}{\cancel {x}}}{- \frac {1}{x^\cancel{2}}} = \lim\limits_{x \to 0} \frac {1}{- \frac{1}{x}} = \lim\limits_{x \to 0} (-x) = 0$$

If we had tried to use L’Hospital’s Rule before canceling, our ratio would become more complicated with each application of the rule.

Example 4. $\lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}}$

Although L’Hospital’s Rule can always be applied to functions that meet the criteria of being differentiable around an interval, it does not always yield a useful answer, even if we see indeterminate forms. Here we will try to use L’Hospital’s Rule.

$$\lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}} = \lim\limits_{x \to ∞} \frac {\frac d{dx} (e^x – e^{-x})}{\frac d{dx}(e^x + e^{-x})} = \lim\limits_{x \to ∞} \frac {e^x + e^{-x}}{e^x – e^{-x}}$$

Noticing that this ratio isn’t any simpler, we can try to apply L’Hospital’s Rule again.

$$\lim\limits_{x \to ∞} \frac {e^x + e^{-x}}{e^x – e^{-x}} = \lim\limits_{x \to ∞} \frac {\frac d{dx} (e^x + e^{-x})}{\frac d{dx}(e^x – e^{-x})} = \lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}}$$

This is the original limit we started with, so we are not getting anywhere using L’Hospital’s Rule. Instead, we should try to simplify our problem by dividing everything by the element with the highest power, in this case $e^x$.

$$\lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}} = \lim\limits_{x \to ∞} \frac {\frac {e^x}{e^x} – \frac {e^{-x}}{e^x}}{\frac {e^x}{e^x} + \frac{e^{-x}}{e^x}} = \lim\limits_{x \to ∞} \frac {1 – e^{-2x}}{1 + e^{-2x}}$$

Noticing that $e^{-2x}$ approaches $0$ as $x$ approaches infinity, we can now take the limit

$$\lim\limits_{x \to ∞} \frac {1 – e^{-2x}}{1 + e^{-2x}} = \frac {1-0}{1+0} = 1$$

L’Hospital’s Rule is often useful to evaluate limits of indeterminate form. It is most useful when it makes the numerator and denominator simpler than they were before. Remember to always simplify after applying the rule. L’Hospital’s Rule is applicable to all limits of ratios of differentiable functions, but is not guaranteed to make the limit easier to find.

Although the rule was named for Guillaume de l’Hôpital, it was actually discovered by his employee Johann Bernoulli

# Absolute Value and Logarithms

Absolute values often turn up unexpectedly in problems involving logarithms. That’s because you can’t take the log of a negative number. Let’s first review the definition of the logarithm function: Logx = y ⇔ by = x (The double arrow is a bi-conditional, which means that one side is true if and only if the other side is true). There are several restrictions on this definition when using real numbers (those not involving i = √-1 ).
1. b cannot equal 1
2. b cannot be negative
3. x cannot be less than or equal to zero.
If any of these conditions are not met, we risk introducing imaginary solutions or an infinite number of solutions. To prevent this from happening, we must follow these rules when working with real numbers, and to do so we often need to introduce the absolute value.   Example 1. Expand the function f(x,y) = log(xy). We must always have a positive number inside the log function, and since x and y are being multiplied together, they must then always have the same sign. Therefore, the domain of our function is xy>0 which means that f(x,y) lies in Quadrants I and III on the x-y plane. To expand, we must follow the above rules, with emphasis in this case on Rule 3. To make sure that the values inside the log are positive, we introduce the absolute value. log(xy) = log|x| + log|y| If we had not included the absolute values, log(x) + log(y) would only be defined when x and y are both positive (Quadrant I). Since the domain is different than that of log(xy), they cannot be equivalent functions. Example 2. Simplify the function g(x,y) = log|x| + log|y| It is important to notice that the above solution is not true in the reverse direction. g(x,y) = log|x| +  log|y| is defined for all real, nonzero values of x and y. Therefore, to simplify this function, we must make sure the domain remains all real nonzero numbers. To do so, we use the absolute value. Simplifying, log|x| + log|y| = log|xy| Example 3. Evaluate the indefinite integral ∫ 1⁄x dx This common integral can be found in many integral tables. It is equal to ln|x| + C. The absolute value is important because this is an indefinite integral, which means x might range through the entire real number line (There is a singularity at x=0, but log(0) is undefined too). We introduce the absolute value into the log to ensure that the antiderivative is defined everywhere the integral is. When working in the real numbers with log functions, remember to always follow the three rules listed above. For many expressions involving logs, it is often necessary to introduce absolute values to ensure that these rules are met.

# Absolute Value and Square Roots

Absolute values often show up in problems involving square roots. That’s because you can’t take the square root of a negative number without introducing imaginary numbers (those involving i = √-1 ).

Example 1: Simplify √x².

This problem looks deceptively simple. Many students would say the answer is x and move on. However, that is only true for positive values of x.

Try x = 2

√x² =  x

Left-hand side: √2² =  √4 =  2

Right-hand-side:  2

The equality holds for x = 2 (and actually for all values of x that are ≥  0).

Try x = -5.

√x² =  x

Left-hand side: √(-5)² = √25 = 5

Right-hand-side: -5

Since 5 ≠ -5, this equation does not hold for x = -5. Notice that when x < 0, the left hand side is actually equal to -x. Therefore, for all x < 0, √x² = -x

To summarize, for all x ≥ 0, √x² = x, and for all x < 0, √x² = -x.

We can write this as a piecewise function as follows:

$\sqrt{x^2} = \begin{cases} x, & \text{if$x≥0$} \\ -x, & \text{if$x \lt 0$} \end{cases}$

Our assignment was to simplify, and a piecewise function is surely not simpler than √x². However, this piece-wise function is actually the exact definition of absolute value. Therefore, we can write √x² = |x|, and our simplification is complete. This holds true, for any expression in the square root that has a even exponent. Even though x² is positive, it can be product of two negatives ex: (-5) (-5) or two positives (5)(5). Therefore, the absolute value ensures that both cases are addressed in the solution.

Example 2: Solve for x:  x²=z+9

x²=z+9

To solve for x, we first take the square root of both sides.

√x²=√z+9

As we discussed earlier, √x² = |x|, so

|x| = √z+9

The equation isn’t quite solved for x yet. To remove the absolute value, we write:

x = ±√z+9, and our work is done.

When working on problems involving square roots, remember to always check the positive and negative cases and be careful that you don’t miss the absolute value.

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