Finding a good math tutor can be essential to your success in school. Tutoring can do much more than just improving math grades. Here are 10 ways in which you can benefit from tutoring.

When working one-on-one outside of the classroom setting, you might find it easier to focus on the subject material and return to math class with increased confidence.

Supplementing your coursework with tutoring can help you understand the material more thoroughly so you can do better in school.

3. Work at a pace you are comfortable with

Your teacher might be moving through the material at a speed that is too fast for you, but when you are studying with a tutor they will structure their sessions at a pace that makes you feel comfortable. Alternatively, if you feel that your class is moving too slow, your tutor can work ahead with you to prepare for future material.

4. Build mathematical foundation

It is essential that you understand the basics before you can succeed in more difficult math. While your teacher must follow a strict curriculum,a tutor will work with you to review the material that you didn’t understand before. Building a strong mathematical foundation now will allow you to tackle the more complicated topics later.

5. Provide supplemental work

If you are confused about a certain topic, your tutor can find you extra material and work through it with you.

6. Tailor lessons to your needs

In a classroom setting, students work at different paces and understand things differently. In one-on-one tutoring, your lessons can be perfectly tailored to your own learning style, which can make learning math easier and more enjoyable.

7. Find a new challenge

Is your math class too easy for you? Tutoring might be the perfect way to push your own boundaries. A tutor can expand and enrich upon the material you’re learning in school, or work on interesting material that’s not even offered at your school. Your tutor can also help you find local math competitions to challenge you in a fun way.

8. Improve SAT/ACT scores

Tutoring is a great way to make sure you ace your college entrance exam. Tutors can help you learn test taking strategies, as well as helping you learn what kind of questions to expect.

9. Bring a fresh new perspective

If you’re feeling stuck in a mathematical rut, working with a tutor can help. Your tutor can discuss math from a new point of view that might make more sense to you than the way your teacher approaches the material.

10. Make math more interesting

Tutors are passionate about math and would love to share that passion with you. If there is a certain topic of interest to you, your tutor can build some lessons around that to keep you interested.

# Modular Arithmetic and Fermat’s Little Theorem

Modular arithmetic is a way of counting in which the numbers wrap around after reaching a certain value. The clock is often used as an analogy. While time always progresses forward, the 12-hour clock “resets” to 1 after passing 12 (13 o’clock is equivalent to 1 o’clock). If we replace 12 with 0, we have the set known as ℤ12 which is made up of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. If we consider midnight as t0=0, then 5 hours later is t5=5. 17 hours later will also be t17=5. 84 hours later will be t84 = 0. 85 hours later will be t85 = 1. There is a pattern here. The value a in the set ℤ12 corresponding to any other integer b is the remainder when b is divided by 12. We can write this mathematically as b ≡ a (mod 12), and we say “b is congruent to a modulo 12”

A clock represents modular arithmetic in ℤ12

More generally, we can write the set of integers for any modulus n as ℤn = {0, 1, 2, 3, … n-1} . Each element a in ℤn represents not only a itself, but all of the b’s for which b ≡ a (mod n). These elements are called congruence classes, and the set ℤn is a set made up of these classes. Sometimes, it’s more convenient to write the relationship as b-a ≡ 0 (mod n), i.e. each b is assigned to a congruence class depending on how much needs to be subtracted to be divisible by n. Addition and multiplication can be done using modular arithmetic if you are using the same modulus. If b ≡ a (mod n) and d ≡ c (mod n), then

1. (b+d) ≡ (a+c) (mod n)
2. (b*d) ≡ (a*c) (mod n)

Example 1. What is 84,872 (mod 5)? A simple way to approach this problem is to find the closest multiple of 5 to 84,872. That is 84,870. Therefore, we can write 84,872 (mod 5) ≡ 84,870 (mod 5) + 2 (mod 5) ≡ 0 (mod 5) + 2 (mod 5) ≡ 2 (mod 5). Example 2. In modulo 10, what is 19,374 · 3,172? One way to attempt this problem is to multiply out these numbers and then find the remainder when dividing by 10. However, since we want our answer in modulo 10, we can instead multiply just the representatives from their congruence classes. The remainder when 19,374 is divided by 10 is 4, so 19,374 ≡ 4 (mod 10) and 3,172 ≡ 2 (mod 10), so (19,374 · 3,172) ≡ (2·4) (mod 10) ≡ 8 (mod 10). Example 3. In modulo 4, what is 180,700,247 + 64,987,422? Again we first find the congruence class for each number modulo 4. We can simplify each by breaking it into the sum of smaller multiples of 4. One possible way is to write 180,700,247 ≡ 180,700,200 (mod 4) + 44 (mod 4) + 3 (mod 4) ≡ 0 (mod 4) + 0 (mod 4) + 3 (mod 4) ≡ 3 (mod 4) Similarly, 64,987,422 ≡ 2 (mod 4). Therefore, (180,700,247+ 64,987,422) ≡ (3 + 2) (mod 4) ≡ 5 (mod 4). However, 5 is not in the set ℤ4 = {0, 1, 2, 3}, so we must reduce 5 (mod 4) to something in the set. We can simplify by noticing that 5 (mod 4) ≡ 4 (mod 4) + 1 (mod 4) ≡ 0 (mod 4) + 1 (mod 4) ≡ 1 (mod 4).

## Fermat’s Little Theorem

One important application for modular arithmetic is Fermat’s Little Theorem which states that if p is a prime number and a is not divisible by p, then ap-1 ≡ 1 (mod p). This theorem is useful because allows you to find a remainder when dividing a really big number by a prime number. Example 4. What is the remainder when 2784 is divided by 11? 2784 is too big to put in most calculators and too time-consuming to multiply by hand. We can apply Fermat’s Little Theorem to make it easier to solve. From the theorem we know that 210 ≡ 1 (mod 11). And we also know that we can multiply congruence classes with the same modulus. Therefore, one strategy we can apply is to write 2784 as the product of as many 210’s as possible. We can simplify as follows: 2784= (210)78·(24) ≡ 178 · 24 (mod 11) ≡ 24 (mod 11) ≡16 (mod 11) ≡ 5 (mod 11). The remainder is 5.

# 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