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Exercises on exponents and powers

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We offer free exercises on exponents and powers of numbers. The power of a number is obtained by multiplying this number by itself several times. The reader can first consult the chapter on exponents.

Mathematically, the powers of a number $a$ are defined by $a^n$ with $n$ integer. This is, $$a^n=a\times a\times\cdots\times a.$$

Selected exercises on exponents and powers of numbers

The following exercises use exponents and powers of real numbers. We assume that you are familiarised with the definitions.

Exercise: Write $1800$ as a product of powers:

Solution: Notice that $1800$ is an even number and $$1800=200\times 9= 8 \times 25\times 3^2=2^3\times 3^2\times 5^2.$$ Thus \begin{align*} 1800=2^3\times 3^2\times 5^2.\end{align*}

Exercise: Compute the number \begin{align*} \left(\frac{2}{3^2}\right)^{-3}\times \frac{2^2}{3^4}.
\end{align*}
Solution: Notice that \begin{align*}\left(\frac{2}{3^2}\right)^{-3}=\left(\frac{3^2}{2}\right)^{3}=\frac{(3^2)^3}{2^3}=\frac{3^6}{2^3}.
\end{align*}
Then \begin{align*} \left(\frac{2}{3^2}\right)^{-3}\times \frac{2^2}{3^4}=\frac{3^6}{2^3}\times \frac{2^2}{3^4}=\frac{3^2}{2}=\frac{9}{2}.
\end{align*}

Exercise: Let $a,b$ and $c$ be non zero numbers. Simplify the number\begin{align*}
A=\frac{ab^{-2}(bc^{-1})^{-2}b^3 c^2}{a^{-3}b^5 \left(\frac{1}{a^2}c^3\right)^2}.
\end{align*}Compute $A$ in the case: $a=10^3,\;b=10^2$ and $c=0.0001$.

Proof: First we calculate \begin{align*} & \left( bc^{-1}\right)^{-2}=b^{-2} (c^{-1})^{-2}=b^{-2}c^2,\cr & \left(\frac{1}{a^2}c^3\right)^2=\left( \frac{1}{a^2}\right)^2 (c^3)^2=\frac{1}{a^4}c^6=a^{-4}c^6.\end{align*} We replace these quantities in A, we find \begin{align*} A&=\frac{ab^{-2}b^{-2}c^2b^3c^2}{a^{-3}b^5a^{-4}c^6}=\frac{a b^{-1}c^4}{a^{-7}b^5c^6}\cr &=  a^8 b^{-1}c^{-2}.\end{align*} Now we take $a=10^3,$ $b=10^2$ and $c=0.0001=10^{-4}$. We then have \begin{align*} a^8= (10^3)^8=10^24,\quad b^{-1}= (10^2)^{-1}=10^{-2},\quad c^{-2}=(10^{-4})^{-2}=10^{-8}.\end{align*} We replace in $A,$ we get $A=10^{24}10^{-2}10^{-8}=10^{14}$.

Next, we give more advanced exercises on exponents and powers of numbers. Some remarkable identities are needed.

Exercise: Simplify the number \begin{align*}A=1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{32}.
\end{align*}
Proof: Observe that $32=2\times 4\times 4=2^5$. Then $A$ can be rewritten as
\begin{align*}
A&=1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^5}\cr &=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\cdots+\left(\frac{1}{2}\right)^5.
\end{align*} Now we will use a remarkable identity of order 6. Then \begin{align*}
1-\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\cdots+\left(\frac{1}{2}\right)^5\right)=\frac{1}{2} A
\end{align*} On the other hand, we compute\begin{align*} 1-\left(\frac{1}{2}\right)^6=1-\frac{1}{64}=\frac{63}{64}.\end{align*}Thus\begin{align*}\frac{1}{2} A=\frac{63}{64}.\end{align*}
That is $A=\frac{63}{32}$.

Exercise: Let $a$ and $b$ be two distinct numbers, and $n$ a relative number. Simplify the expression
\begin{align*} B=\frac{a^n b-a^{n+1}}{ab^n-b^{n+1}}\times \left(\frac{a}{b}\right)^{-n}. \end{align*}
Proof: We factor the numerator by $a^n$ and the denominator by $b^n$ in the fraction we obtain
\begin{align*} B&=\frac{a^n( b-a)}{b^n(a-b)}\times \left(\frac{a}{b}\right)^{-n}\cr &=\left(\frac{a}{b}\right)^{-n}\frac{( b-a)}{(a-b)}\times \left(\frac{a}{b}\right)^{-n}\cr &= – \left(\frac{a}{b}\right)^{n-n}=-1 \times 1=-1.\end{align*}

Nth roots and rational exponents

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We will shed some light on nth roots and rational exponents of numbers. We give a clear and rigorous definition of these root numbers. Examples and solved exercises are also given.

Generality on nth roots

We explain in a simple manner what is the nth roots of a number. We also deal with the rational exponents of numbers.

Square roots numbers and more roots

Although the square roots are standard and they are very restrictive. In fact, a number $a$ is the square root of a positive number $b$ if $a^2=b$. In this case, we write $a=\sqrt{b},$ sometimes we denote $a=b^{\frac{1}{2}}$. Thus square roots are solutions of equations of type $a^2=b$.

Naturel question: Does exists numbers $m$ positive integer number and a real number $x$ such that $x^m$ is not positive, i.e. negative? Yes, we can find examples. In fact, $(-3)^3=(-3)\times (-3)\times (-3)=-9$. Then $x=-3$ is the solution of the equation $x^3=-9$. We say that $-3$ is the cubic root of the number $-9$ and we write $-3=\sqrt[3]{-9}$. So unlike square roots, cubic roots can be negative.

More generally, let $k$ be a non-zero integer and $b$ a number, not necessarily an integer, and ask the question: Are their numbers $x$ such that $x^{2k}=b$. Nice that $x^{2k}=(x^k)^2$. Thus necessarily $b$ is positive and in this case $x$ is called $2k$-th root of $b,$ and we write $x=\sqrt[2k]{b}$ or $x=b^{\frac{1}{2k}}$. If $b$ is negative, then there is no solution. 

Thus for even numbers $n,$ of the form $n=2k$, the nth root must be positive. 

Now assume that we have an odd number $n=2k+1,$ and look for numbers $x$ satisfying the algebraic equation $x^n=b$. This is, equivalent to $x^{2k} x =b$. As $x^{2k}$ is positive, it follows that the numbers $x$ and $b$ have the same sign, either both positive or negative. In this case, the solution exists. Now if $x$ and $b$ have opposite signs, then there is no solution to the above equation.

In order to be more precise on the nth roots, let the following definition of the components of a radical expression

 

components-of-a-radical-expression


If we set $b=\sqrt[n]{x}$. Then we have the following cases
: If the index $n$ is even: then the radicant $x$ and $b$ must be positive. If index $n$ is an odd number: then either $x$ and $b$ are positive or $x$ and $b$ are negative.

Example: take am index $n=3,$ odd number, $b=-9$, the radicand $x=-3$ is negartive.

Rational exponents

A fraction exponent, also called fractional exponent, of a number $a$ is given by $$  \sqrt[n]{a^m}=\left( \sqrt[n]{a}\right)^m=a^{\frac{m}{n}}$$ where $n$ and $m$ are relative numbers. Simply the name rational means that the exponent of the number $a$ is a fraction of the forme $\frac{m}{n}$.

A selection of exercises on nth roots

Exercise: Simplify the following expressions \begin{align*} \left(\sqrt[4]{\left(\sqrt[3]{3} \right)^2} \right)^6,\quad \sqrt[n]{3^{2n+1}},\quad \frac{5^{4}{3}}{5^{-\frac{2}{3}}}.\end{align*}

Remarkable identities

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Remarkable identities are equations that simplify calculus in algebra and analysis. We give you the most famous remarkable identities, and we will provide you with beautiful proof of them.

Three famous Remarkable identities:

We recall here the most famous and used Remarkable identities.

  1. The square of an addition: let $a$ and $b$ be two real numbers. Then \begin{align*} (a+b)^2=a^2+2ab+b^2.\end{align*} Preuve: Using properties of exponents, we have \begin{align*} (a+b)^2&=(a+b)(a+b)=a(a+b)+b(a+b)\cr &= a^2+ab+ba+b^2\cr&=a^2+2ab+b^2.\end{align*}
  2. The square of a difference: \begin{align*}(a-b)^2=a^2-2ab+b^2.\end{align*}Preuve: We apply the above remarkable identity. In fact,\begin{align*}(a-b)^2&=(a+(-b))^2\cr &=a^2+2a(-b)+(-b)^2\cr &=a^2-2ab+b^2.\end{align*}
  3. Addition multiplied by subtraction:
    \begin{align*}a^2-b^2=(a+b)(a-b).\end{align*}Preuve: \begin{align*}(a+b)(a-b)&=a(a-b)+b(a-b)\cr &= a^2-ab+ba+b^2\cr&=a^2-b^2.\end{align*}

Exercise: Find $x$ such that $x^3-x=0$.

Solution: Factor by $ x $ and using the third remarkable identity, we get

\begin{align*} x^3-x&=x(x^2-1)=x(x^2-1^2)\cr &= x(x+1)(x-1).\end{align*}Thus $x^3-x=0$ is equivalent to $x(x+1)(x-1)=0$. Hence the set of solutions is ${-1,0,1}$.

Exercise: Factor the following expressions

\begin{align*}& 9x^2-6x+1\cr &x^2-6xy+9y^2\cr &a^2x^2+2acx+c^2.\end{align*}

Solution:  The idea is to rewrite these expressions as standard remarkable Identities and then use the formula above. For the first expression, we get \begin{align*}9x^2-6x+1&=(3x)^2-2(3x)\times 1+1^2\cr&=(3x-1)^2.\end{align*} For the second expression \begin{align*}x^2-6xy+9y^2&=x^2-2x(3y)+(3y)^2\cr &=(x-3y)^2.\end{align*}For the third expression\begin{align*}a^2x^2+2acx+c^2&=(ax)^2+2(ax)c+c^2\cr&=(ax+c)^2.\end{align*}

In the case of higher degrees exponents

In this paragraph, we show remarkable identities for exponents bigger than or equal to three. These formulas are important to determine the limits of some sequences, and also to simplify expressions.

Let $a$ be a real number such that $a\neq 1$ and $n$ be a non-zero integer. We have \begin{align*}a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1).\end{align*}  Let us give a simple proof of this identity: We set \begin{align*} A=a^{n-1}+a^{n-2}+\cdots+a+1.\end{align*}We have \begin{align*}(a-1)A&=a A-A\cr &=
(a^{n}+a^{n-1}+\cdots+a^2+a)-(a^{n-1}+a^{n-2}+\cdots+a+1)\cr &= a^n-1.\end{align*}

Let now $a$ and $b$ be two numbers such that $a\neq b$. Then
\begin{align*}a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}).\end{align*}

Preuve: We assume that $ a $ is non-zero. Then we factor by $a^n$, we get \begin{align*}a^n-b^n=a^n\left(1-\left(\frac{b}{a}\right)^n\right)\end{align*}By using the first higher remarkable identity, we have\begin{align*}
\left(1-\left(\frac{b}{a}\right)^{n}\right)=\left(1-\frac{b}{a}\right)\left( \left(\frac{b}{a}\right)^{n-1}+\left(\frac{b}{a}\right)^{n-2}+\cdots+\frac{b}{a}+1 \right)
\end{align*}On the other hand, we write $a^n=a a^{n-1}$. We obtain \begin{align*}a^n-b^n&=a\left(1-\frac{b}{a}\right)a^{n-1}\left( \left(\frac{b}{a}\right)^{n-1}+\left(\frac{b}{a}\right)^{n-2}+\cdots+\frac{b}{a}+1 \right)\cr & =(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}).
\end{align*}

Rules of Exponents

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Are you ready to unlock the secrets of exponents? These powerful mathematical tools, also known as powers or indices, are essential for mastering algebra and calculus. With a solid understanding of the rules of exponents, you’ll be able to simplify expressions, solve equations, and tackle even the most complex mathematical problems. In this article, we’ll dive into the key rules and properties of exponents, so you can take your math skills to the next level. Let’s get started!

What are exponents in math?

Have you ever heard of exponents? They’re an important concept that can help you quickly calculate large numbers. Basically, an exponent tells you how many times a number is multiplied by itself. Let’s use the number 5 as an example. If the exponent of 5 is 3, then we get another number by multiplying 5 by itself three times. That means the number 5 is repeated three times! To write this in shorthand, we use the notation 5^3. In this case, 5 is called the base, 3 is the exponent, and 5^3 is the power. Pretty neat, huh?

exponent-power-numbers

The present discourse aims to provide a comprehensive expression for power numbers. Consider a real number denoted by $y$ and a natural number denoted by $n$. We may express the $n$-th power of $y$ as follows: \begin{align*} y^n=\underset{n \; {\rm times}}{\underbrace{y\times y\cdots\times y}}\end{align*}

Explore the Rules of Exponents

Product of Powers

When you multiply two powers with the same base, you can simplify the expression by adding their exponents: \begin{align*} y^{n+m}=y^n y^m\end{align*}For example $$ 3^4\times 3^2=3^{4+2}=3^6.$$

Quotient of Powers

When you divide two powers with the same base, subtract the exponent of the divisor from the exponent of the dividend: $$ \frac{a^n}{a^m}=a^{n-m}.$$For example $$ \frac{5^7}{5^3}=5^{7-3}=7^4.$$

Power of a Power

To raise a power to another exponent, multiply the exponents $$(a^n)^m=a^{nm}.$$ For example \begin{align*}(2^3)^4=2^{3\times 4}=3^{12}.\end{align*}

Negative Exponents

When you have a negative exponent, move the base to the denominator and change the exponent’s sign: $$ a^{-n}=\frac{1}{a^n}.$$ For example $$ 2^{-3}=\frac{1}{2^3}=\frac{1}{8}.$$

Zero Exponent

Any nonzero base raised to the power of zero is equal to 1: $$ a^0=1.$$ In fact $a^0=a^{1-1}=\frac{a}{a}=1$.

Product to a Power

When you have a product inside parentheses raised to an exponent, distribute the exponent to each term: $$ (a\times b)^n=a^n\times b^n.$$ For example $$ (4\cdot 2)^3=4^3\cdot 2^3=64\times 8=512.$$

  Exponents worksheets

Exercice 1:Compare the numbers $8^4$ and $2^{12}$.
We have \begin{align*}2^{12}= 2^{3\times 4}=(2^3)^4=8^4.\end{align*}
Exercice 2: Find a positive number $x$ such that $x^5=2^{15}$.
Let us first recall the following rule: $ a^n = b^n $ implies that $ a = b $. We will apply this property to answer the exercise. In fact, the idea is to rewrite $2^{15}$ as an exponent of $5$. Remark that $15=3\times 5$. Then\begin{align*}x^5=2^{15}=2^{3\times 5}=(2^3)^5=8^5.\end{align*}It follows that $x=8$.
Exercice 3: Simplify the expression\begin{align*}A(n)=\frac{2^n+1}{1+2^{-n}}.\end{align*}
We multiply $A(n)$ by $2^{-n}$, we get $$ 2^{-n}A(n)= 2^{-n} \frac{2^n+1}{1+2^{-n}}=\frac{1+2^{-n}}{1+2^{-n}}=1.$$ Thus $A(n)=2^n$.

You can also consult the page on the powers of numbers where you will find some exercises with detailed solutions.

Conclusion

In the world of math, the rules of exponents are like a superhero’s trusty sidekick – always there to simplify expressions and solve equations with ease. These rules are the building blocks for more complex algebraic and calculus concepts, making them an essential tool in any math whiz’s arsenal. By mastering these rules, you’ll be able to take on any mathematical challenge that comes your way with confidence and ease. So, gear up and get ready to conquer the world of math with the power of exponents!

Unlocking the Mystery of Square Roots

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Get ready to dive into the fascinating world of square roots! These mathematical marvels have captivated the minds of brilliant thinkers for centuries, and for good reason. Not only are they crucial in understanding the basics of geometry, but they also have countless applications in the scientific and engineering fields.

For centuries, humans have been fascinated by the power of square roots. From ancient civilizations to modern-day mathematicians, we’ve used square roots to unlock the secrets of geometry and beyond. Take, for example, a triangle with sides of equal length – let’s say 1 centimeter each. What’s the length of the third side, you ask? Well, that’s where the magic of square roots comes in. By applying the Pythagorean relation, we can determine that the length of the third side, let’s call it $a$, is equal to the square root of 2. But here’s where things get really interesting – mathematicians have shown that this number, $\sqrt{2}$, is not a rational number. Nope, it’s a real number, and in this section, we’re going to dive deep into the fascinating world of these types of numbers. So buckle up, folks – it’s going to be a wild ride!

In this article, we’ll take a comprehensive journey through the world of square roots. We’ll start by defining what they are and exploring their properties. Then, we’ll delve into the exciting ways they’re used in both mathematics and the real world.

What is a square number?

A square number, also known as a perfect square, is a number that can be expressed as the product of another number multiplied by itself. In mathematical notation, if ‘a’ is a number, then ‘$a^2$’ represents a square number. In fact, a square is just a number $a$ with exponent $2$.

For example, 9 is a square number because it can be expressed as $3^2$, 3 multiplied by 3. Square numbers have the property that when their square root is taken, the result is a whole number.

What is a square root of positive numbers?

So, what exactly is a square root? At its core, it’s a value that, when multiplied by itself, gives you the original number. We use the symbol $\sqrt{b}$ to represent the square root of a non-negative number ‘$b$’. Sometimes we also use the notation $b^{\frac{1}{2}}$ for the square root. It’s important to note that square roots of negative numbers are considered complex numbers and are beyond the scope of this article. Are you ready to explore the world of square roots? Let’s get started!

The following rule is important:

If $a$ is a postive number, then $$\sqrt{a^2}= a.$$ $\blacktriangleright$ Examples: +
The square root of 4 is calculated as: $\sqrt{4}=\sqrt{2^2}= 2$. Observe that $\sqrt{0}=\sqrt{0^2}= 0$ and $\sqrt{1}=\sqrt{1^2}= 1$.

Basic properties of square roots

If $a$ and $b$ are positive numbers, then square root of the product $ab$ is given by $$\sqrt{ab}=\sqrt{a}\times \sqrt{b}.$$ $\blacktriangleright$ Examples: +
$ \sqrt{12}=\sqrt{4\times 3}=\sqrt{4}\times \sqrt{3}=2\sqrt{3}$.
For a positive number $b$ the solutions of the quadratic equation $x^2=b$ are: \begin{align*}x=\sqrt{b}\quad\text{or}\quad x=-\sqrt{b},\end{align*} $\blacktriangleright$ Details: +

Are you ready to unlock the secrets of solving quadratic equations? Let’s dive in! First, let’s take a look at the equation $x^2=b$. We can rewrite this as $x^2-(\sqrt{b})^2=0$, which is equivalent to $(x-\sqrt{b})(x+\sqrt{b})=0$. This means that either $x-\sqrt{b}=0$ or $x+\sqrt{b}=0$, giving us the solutions $x=\sqrt{b}$ or $x=-\sqrt{b}$.

Now, let’s put this knowledge to the test and solve the equation $x^2=144$. It’s as easy as computing the square root of 144, which is 12. But wait, there’s more! We can simplify even further by breaking down $\sqrt{144}$ into $\sqrt{4\times 36}$. This gives us $\sqrt{4}\times \sqrt{36}$, which equals 2 times 6. Therefore, the solutions to the equation are $x=12$ or $x=-12$.

See how easy it is to solve quadratic equations? With a little bit of algebraic manipulation and some clever simplification, you can unlock the solutions to even the most complex equations. Keep practicing and soon you’ll be a quadratic equation-solving master!

Square Root of a Fraction $$ \sqrt{\frac{p}{q}}=\frac{\sqrt{p}}{\sqrt{q}}.$$ $\blacktriangleright$ Examples: +
Calculate $\sqrt{\frac{4}{9}}$. We have $$ \sqrt{\frac{4}{9}}= \frac{\sqrt{4}}{\sqrt{9}}=\frac{\sqrt{2^2}}{\sqrt{3^2}}=\frac{2}{3}.$$ Determine $\frac{\sqrt{12}}{\sqrt{3}}$. We have $$\frac{\sqrt{12}}{\sqrt{3}}=\sqrt{\frac{12}{3}}=\sqrt{4}=2.$$

Did you know that in the world of mathematics, it’s a well-known fact that square roots should never be left hanging out in the denominator of a number? Yup, it’s true! So if you want to impress your math teacher or your friends, make sure to always simplify those pesky square roots. Trust us, it’ll make your life a whole lot easier!

Simplify the following numbers $$ \frac{1}{\sqrt{2}},\quad \frac{2}{\sqrt{5}}.$$ $\blacktriangleright$ Details: +
Multiplying the denominator by the same square root number! It’s like casting a spell that transforms the equation into a simpler, more manageable form. In fact $$\frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{\sqrt{2}}{2}.$$ In addition $$ \frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{2\sqrt{5}}{5}.$$
For positive numbers $a$ and $b$, $$ (\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b.$$ $\blacktriangleright$ Details: +
We calculate \begin{align*} (\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})&=(\sqrt{a})^2-\sqrt{a}\times \sqrt{b}+\sqrt{a}\times \sqrt{b}-(\sqrt{b})^2\cr &= a-b.\end{align*}
Simplify the number $$ \frac{1}{\sqrt{3}+\sqrt{2}}.$$ $\blacktriangleright$ Details: +
We calculate \begin{align*} \frac{1}{\sqrt{3}+\sqrt{2}}&=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\cr &=\frac{\sqrt{3}-\sqrt{2}}{3-2}\cr &=\sqrt{3}-\sqrt{2}.\end{align*}

Applications of this class of numbers

Are you ready to discover the power of square roots? From geometry to computer graphics, these little symbols have a big impact on a variety of fields. In geometry, the Pythagorean theorem relies on square roots to solve problems involving right-angled triangles. Engineers and physicists use square roots to calculate everything from velocity to electrical voltage. In statistics and probability, square roots help calculate standard deviations and confidence intervals. Furthermore, in finance and investment, square roots are essential for calculating risk and volatility. Even computer graphics rely on square roots to calculate distances, angles, and transformations in virtual environments. So next time you see that little symbol, remember the big impact it can have in so many different areas.

Conclusion

Did you know that square roots are more than just boring math concepts? They’re actually super important tools that have practical applications in all sorts of fields! From geometry to finance, these little symbols help us understand the world around us and solve complex problems. So next time you come across the √ symbol, don’t just brush it off as another math thing. It’s actually the key to unlocking a whole world of possibilities and real-world applications. Square roots are the foundation of math and science, and they enrich our lives in ways we might not even realize. Pretty cool, huh?

Dividing Fractions: Principles and Techniques

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Dividing fractions may sound complex, but with the right approach, it becomes a straightforward mathematical operation. In this guide, we’ll break down the steps to divide fractions, making it accessible to everyone, from students to adults.

What is a fraction in mathematics?

Prior to delving into the division of fractions, it is imperative to possess a comprehensive understanding of the fundamental concepts. It is important to recollect that a fraction is comprised of two components, namely the numerator, which is the upper number, and the denominator, which is the lower number. The numerator signifies the proportion of the entirety under consideration, while the denominator denotes the total quantity of identical segments that constitute the entirety.

fraction-x-y

The number $x$ is called the numerator, and $y$ is called the denominator.

Maintaining a fraction in its simplified form is a more expedient approach. The simplification of a fraction renders it highly amenable for utilization in mathematical equations.

Dividing fractions: step-by-step

Are you ready to become a fraction master? Dividing fractions is a piece of cake once you’ve got the classical operators down pat. The key players here are the multiplication of two fractions and the inverse of a fraction. But wait, there’s more! You’ll also need to know how to add two functions. Don’t worry if you need a refresher on these operators, we’ve got you covered. Let’s dive in and become fraction superheroes!

  • Addition of two fractions: $$ \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.$$
  • Multiplication of two fractions: $$ \frac{a}{b}\times\frac{c}{d}=\frac{ac}{bd}.$$
  • The inverse of a fraction: $$ \frac{1}{\frac{a}{b}}=\frac{b}{a}.$$

The aforementioned operations on fractions provide a basis for the justification of the subsequent rule pertaining to the division of fractions.

Dividing fractions satisfy the following rule: $$ \frac{a}{b}\div\frac{c}{d}=\frac{ad}{bc}.$$
In Fact \begin{align*} \frac{a}{b}\div\frac{c}{d}&=\frac{a}{b}\frac{1}{\frac{c}{d}}\cr &=\frac{a}{b}\times\frac{d}{c}=\frac{ad}{bc}.\end{align*}

Fractions division worksheets

Are you ready to take your fraction skills to the next level? Look no further than our exciting fractions division worksheets! These worksheets will challenge you to divide fractions like a pro, with fun and engaging exercises that will keep you on your toes. Whether you’re a math whiz or just starting out, our worksheets are the perfect way to sharpen your skills and boost your confidence. So why wait? Dive into the world of fractions division today and see just how far you can go!

Exerise 1: Perform the division fraction division: $$ \frac{4}{5}\div (1+\frac{1}{2}).$$
Before applying the dividing fractions rule, we first calculate the sum $1+\frac{1}{2}$. Using the addition rule we have $$ 1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}.$$ Now replace $$\frac{4}{5}\div (1+\frac{1}{2})=\frac{4}{5}\div \frac{3}{2}=\frac{4}{5} \frac{2}{3} =\frac{8}{15}.$$
Dividing the fraction $$ \frac{n}{n+1}\div \frac{n^2}{1+ \frac{1}{n}},\quad n\ge 1.$$
First we calculate: $$ 1+\frac{1}{n}=\frac{n+1}{n}.$$ Second, $$ \frac{n^2}{\frac{n+1}{n}}= n^2 \times \frac{n}{n+1}=\frac{n^3}{n+1}.$$ Finally, \begin{align*} \frac{n}{n+1}\div \frac{n^2}{1+ \frac{1}{n}}&=\frac{n}{n+1}\div \frac{n^3}{n+1}\cr &= \frac{n}{n+1} \frac{n+1}{n^3} \cr &=\frac{1}{n^2}.\end{align*}

Conclusion

Dividing fractions may have seemed daunting at first, but the “Invert and Multiply” rule simplifies the process significantly. With practice, you’ll become proficient in dividing fractions, a skill that finds applications in various real-world scenarios, from recipes in the kitchen to complex mathematical calculations. Remember, mastering this skill is about understanding the concept and applying the straightforward steps outlined in this guide.

How to solve rational inequalities

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We offer top techniques on how to solve rational inequalities for high school students. It is an important part of algebra and calculus. Of course, there are a lot of methods to solve these inequalities. Some of them are based on advanced tools in analysis while others are mainly based on the simplification of the expression of the rational equation.

Basics of Rational Fractions

Before talking about rational inequalities, it is better to first master the notion of calculus for rational numbers and fractions.A rational number is a real number that can be written as $x=\frac{a}{b}$, where $a$ and $b$ are renumbers with $b$ different of zero, $b\neq 0$. The number $a$ is called the numerator and $b$ is the denominator. Then fraction is a numerator/denominator. For details on computation for this class of numbers, we refer to the fractions chapter

Sign of a fraction

Positive fraction: The fraction $\frac{a}{b}$ is positive if its numerator and denominator have the same sign, i.e. positive together or negative together.  This means that $a$ and $b$ are positive, or $a$ and $b$ are negative.
 
Negative fraction: The fraction $\frac{a}{b}$ is negative if the numerator and denominator have opposite signs. This means that $a$ is positive and $b$ is negative or $a$ is negative and $b$ is positive.

The first-order rational functions

In this paragraph, we will give you a unified and simple technique to solve rational inequalities of the for $$ \frac{f(x)}{g(x)}\ge 0,$$ where $f(x)$ and $g(x)$ are afine functions of the forme $ax+b$.
 
Example I: Solve the rational inequality 
$$\frac{x-2}{x+1}\ge 0,$$
Proof: As the fraction is positive then we have $x-2\ge 0$ and $x+1>0$ or  $x-2\le 0$ and $x+1<0$. This means that $$x\ge 0\quad\text{and}\quad x>-1,$$
or $$ x\le 2\quad\text{and}\quad x<-1.$$
It follows that $x\in [2,+\infty)$ or $x\in (-\infty,-1)$. Thus the set $\mathcal{S}$ of solutions of the rational inequality is $$\mathcal{S}=[2,+\infty)\cup (-\infty,1).$$
 
Example II: Determine the solution of the following rational inequality $$\frac{3x-2}{x+1}\ge 2.$$ This problem can also be reformulated as $$ \frac{3x-2}{x+1}-2\ge 0.$$
By calculating this difference, we only need to solve the following rational inequality $$\frac{x-3}{x+1}\ge 0.$$ Using the same arguments as in Example I, we deduce that the set of solution is $$ \mathcal{S}=[3,+\infty)\cup (-\infty,-1).$$

Rational inequalities with quadratic functions

Determine the set of solutions $\mathcal{S}$ of the following rational inequality $$ \frac{3x^2+5x+2}{x^2+2x}\ge 0.$$ The same principle can also be applied to these types of inequalities. As the solutions depend on the sign of the quadratic functions $3x^2+5x+2$ and $x^2+2x$, then the first work to do is to determine the signs of these functions.
 
Using the chapter of quadratic equations,  we can write $$ 3x^2+5x+2=(x+1)(x+\frac{2}{3})$$ and $$ x^2+2x=x(x+2).$$ We deduce that $3x^2+5x+2$ is positive if $$x\in (-\infty,-1]\cup [-\frac{2}{3},+\infty),$$ and it is negative if $$x\in [-1,-\frac{2}{3}].$$ Similarly, $x^2+2x$ is strictly positive if $$x\in (-\infty,-2)\cup (0,+\infty),$$ and it is negative if $x\in (-2,0)$. Now $x$ is a solution of the rational inequality if $$x\in (-\infty,-2)\cup [-1,-\frac{2}{3}].$$
 
Note: The same technique can also be applied for rational inequalities defined by the polynomial functions with higher degrees.

What is real analysis?

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In this article, we answer the question: what is real analysis about? Apparently the word real comes from the set of real numbers while analysis is what we study with the use of real numbers such as sequences, series, continuity, differentiability, and integrability of real functions.

The key to real analysis

Real numbers and sequences are the keys to real analysis. The topology of real numbers is heavily used to treat other disciplines in analysis. In fact, the set of real numbers enjoys a lot of nice properties such as the density of the set of rational numbers in $\mathbb{R}$. This means that any real number can be approximated by a sequence of rational numbers.

The Bolzano-Weierstrass theorem is a compacity theorem that gives information about the convergence of bounded sequences. In fact, usually, we do not that the convergence of the whole sequence, but only a part of this sequence, which we call subsequences. So this theorem says that any bounded sequence admits a convergent subsequence. From this, we can also deduce that the compact set of $\mathbb{R}$ are exactly the closed bounded sets.

Continuous functions 

The Heine theorem says that any continuous function on a compact set is uniformly continuous, bounded, and in particular reaches its upper and lower bounds. This is an important result in real analysis.

The dichotomy method is well-known in numerical analysis for its efficiency in approaching the roots of algebraic equations. This method is mainly based on the well-known theorem of real analysis, the intermediate value theorem. This theorem says that if a function $f$ is continuous on $[a,b]$, then for any $\lambda$ between $f(a)$ and $f(b)$, there exists $c\in [a,b]$ such that $f(c)=\lambda$. In particular, if $f(a)$ and $f(b)$ have an opposite sign then the function vanishes on a point in $[a,b]$. As an application, we have the following particular fixed by theorem: if $f:[a,b]\to [a,b]$ is continuous then there at least a fixed point $c$ of $f,$ i.e. $c\in [a,b]$ with $f(c)=0$. This number $c$ is unique if, in addition, $f$ is strictly monotone or a contraction function.

Simplifying fractions

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The process of simplifying fractions is a fundamental mathematical concept that involves reducing a given fraction to its lowest possible terms. This process is essential in various mathematical applications, including algebra, geometry, and calculus.

How to simplify fractions

Are you ready to simplify fractions like a pro? It’s all about reducing them to their simplest form, where the top and bottom numbers have no common factors except for 1. We call these simplified fractions “lowest terms” or “reduced fractions.” So let’s get started and make math a breeze!

Step 1: Identify the Greatest Common Divisor (GCD)
The first step in simplifying a fraction is to find the greatest common divisor (GCD) of the numerator and denominator. The GCD is the largest number that evenly divides both the numerator and the denominator.
Step 2: Divide by the GCD
Once you’ve determined the GCD, you’ll divide both the numerator and the denominator of the fraction by this common divisor. This step ensures that the fraction is reduced to its simplest form.
Step 3: Verify for Further Simplification
After dividing by the GCD, check if there are any common factors left between the numerator and the denominator. If there are, repeat steps 1 and 2 until no common factors remain other than 1.
Step 4: Express the Simplified Fraction
The result of these steps will be the simplified or reduced fraction, which is in its simplest form.

Simplifying fractions: Examples

Simplify the fractions: $$ \frac{12}{30},\quad \frac{33}{55},$$
For the first fraction, we remark that the factors of $12$ are $1,2,3,4,6,12,$ and the factors of $30$ are $1,2,3,5,6,10,15,30$. Therefore the GCD of $12$ and $30$ is $6$. Observe that $$ \frac{12}{30}=\frac{2\times 6}{5\times 6}=\frac{2}{5}.$$ For the seconde fraction, the factors of $33$ are $1,3,11,33$ and the factors for $55$ are $1,5,11,55$. Then the GCD of $33$ and $55$ is $11$. In addition, $33=3\times 11$ and $55=5\times 11$. Therefore $$\frac{33}{55}=\frac{3\times 11}{5\times 11}=\frac{3}{5}.$$

Conclusion

To simplify a fraction, one must divide both the numerator and denominator by their greatest common factor (GCF). The GCF is the largest number that divides both the numerator and denominator without leaving a remainder. By dividing both the numerator and denominator by the GCF, the fraction is reduced to its simplest form. This simplification process is crucial in solving mathematical problems that involve fractions, as it allows for easier manipulation and calculation. Therefore, mastering the skill of simplifying fractions is essential for any student or individual seeking to excel in mathematics.

Algebra for beginners at the college level

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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 formulas, 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. For example, the set that contains positive integer numbers is denoted by $\mathbb{N}:=\{0,1,2,\cdots\}$.

We say that $a$ belongs to a set $A$, and 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 $y=f(x)$.

Polar form of complex numbers

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The polar form of complex numbers is a mathematical representation that expresses a complex number in terms of its magnitude and argument. It is a useful alternative to the standard rectangular form, which represents a complex number as a sum of a real and imaginary part.

What is the polar form of complex numbers?

In the polar form, a complex number is written as $r(\cos\theta + i\sin\theta)$, where $r$ is the magnitude or modulus of the complex number, and $\theta$ is the argument or angle it makes with the positive real axis. This representation allows for a more intuitive understanding of complex numbers, as it relates them to the polar coordinate system.

Additionally, the polar form facilitates operations such as multiplication and division of complex numbers, as they can be performed by simply multiplying or dividing their magnitudes and adding or subtracting their arguments. Overall, the polar form of complex numbers provides a concise and geometrically meaningful representation that is widely used in various branches of mathematics and engineering.

De Moivre’s Theorem

Step into the captivating world of complex numbers, where De Moivre’s Theorem reigns supreme as one of the most vital and practical theorems. This remarkable theorem serves as a bridge between the enigmatic realm of complex numbers and the captivating realm of trigonometry. But its usefulness doesn’t stop there! De Moivre’s Theorem also unveils the intricate relationships between trigonometric functions of multiple angles, making it an indispensable tool in the mathematician’s arsenal.

De Moivre’s Theorem: For all $n$ and $\theta$, $$ (\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).$$

Known by various names, such as “De Moivre’s Identity” and “De Moivre’s Formula,” this theorem pays homage to the brilliant mind of the esteemed mathematician, De Moivre. His contributions to the field of mathematics were nothing short of extraordinary, particularly in the realms of probability theory and algebra. So, let us delve into the depths of De Moivre’s Theorem and unlock the secrets it holds, as we follow in the footsteps of this mathematical luminary.

A selection of exercises about the polar form

Exercise: Rewrite in the polar and the exponential polar form: \begin{align*} z_1=-1 – i \sqrt{3},\quad z_2= 3 + 3i,\qquad z_3=-\frac{4}{3}i \end{align*}
The modulus of $z_1$ is \begin{align*} |z_1|=\sqrt{(-1)^2+(-\sqrt{3})^2}=\sqrt{4}=2. \end{align*} Let $\theta_1$ be the argument of $z_1$. We then have \begin{align*} \cos(\theta_1)&=\frac{{\rm Re}(z_1)}{|z_1|}=\frac{-1}{2},\cr \sin(\theta_1)&=\frac{{\rm Im}(z_1)}{|z_1|}=\frac{-\sqrt{3}}{2}. \end{align*} This shows that \begin{align*} \theta_1=\frac{4\pi}{3}+2k\pi,\quad k\in\mathbb{Z}. \end{align*} Thus \begin{align*} z_1=|z_1|e^{i\theta}=2 e^{i\frac{4\pi}{3}}. \end{align*} We need to give a representation of $z_1$ with a principal argument, i.e. belonging to $(-\pi,\pi]$. We can write $\frac{4\pi}{3}=\frac{-2\pi}{3}+2\pi$. Then \begin{align*} z_1=2 e^{i\frac{-2\pi}{3}}=2\left( \cos(\frac{-2\pi}{3})+i\sin(\frac{-2\pi}{3})\right). \end{align*} We have $z_2=3(1+i)$. Then $|z_2|=3\sqrt{1^2+1^2}=3\sqrt{2}$. Let $\theta_2$ be an argument of $z_2$. We have \begin{align*} \cos(\theta_2)=\frac{3}{3\sqrt{2}}=\frac{\sqrt{2}}{2}. \end{align*} Similarly, \begin{align*} \sin(\theta_2)=\frac{\sqrt{2}}{2}. \end{align*} We then obtain \begin{align*} \theta_2=\frac{\pi}{4}+2k\pi,\quad k\in\mathbb{Z}. \end{align*} Hence \begin{align*} z_2&=|z_2|e^{i\theta_2}=3\sqrt{2} e^{i\frac{\pi}{4}}\cr &= 3\sqrt{2}\left(\cos(\frac{\pi}{4})+\sin(\frac{\pi}{4})\right). \end{align*}Observe that $i=e^{i\frac{\pi}{2}}$. Then \begin{align*} z_3=-\frac{4}{3}i=-\frac{4}{3}e^{i\frac{\pi}{2}}. \end{align*} But this is not the good form as $-\frac{4}{3} < 0$. To overcome this obstacle we write $(-1)=e^{i\pi}$, so that \begin{align*} z_3&=\frac{4}{3}e^{i\pi} e^{i\frac{\pi}{2}}\cr &= \frac{4}{3} e^{i\frac{3\pi}{2}} \end{align*} We need a representation with a principal argument. We then write $e^{i\frac{3\pi}{2}}=e^{i\frac{4\pi-\pi}{2}}$. Thus \begin{align*} z_3=\frac{4}{3} e^{i\frac{-\pi}{2}}=\frac{4}{3} \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right). \end{align*}
Exercise: Solve the equation $z^2=1+i$.

We recall that if $\alpha= r e^{i\theta},$ here $r>0$ and $\theta$ is a principal argument of $\alpha$, then the complex equation $z^2=\alpha$ has two roots: \begin{align*}z_1&=\sqrt{r}e^{i\frac{\theta}{2}}\;\text{and,}\cr z_2&=\sqrt{r}e^{i\left(\frac{\theta}{2}+\pi\right)}\end{align*} On the other hand, if we write $z$ and $\alpha$ is the algebraic form $z=x+iy$ and $\alpha=a+i b$ then using the fact that $z^2=x^2-y^2+i (2xy),$ we obtain \begin{align*} x^2-y^2=a,\quad 2xy=b. \end{align*} It suffices then to solve the above system to obtain $z$. After a small effort we obtain \begin{align*} |x|=\sqrt{\frac{\sqrt{a^2+b^2}+a}{2}},\quad |y|=\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}. \end{align*} To determine $x$ and $y$ we shall remark that the sign of the product $xy$ is the same as the sign of $b$, this is very important!!!.

Let us now solve the equation $z^2=1+i$ here we have $\alpha=1+i$ and then $a=1$ and $b=1$. Moreover, $\alpha=\sqrt{2}e^{i\frac{\pi}{4}}$, so that the principal argument of $\alpha$ is $\theta=\frac{\pi}{4}$. We have the two roots \begin{align*} z_1=\sqrt{\sqrt{2}}e^{i\frac{\pi}{8}},\quad z_2=\sqrt{\sqrt{2}}e^{i\frac{9\pi}{8}}=-z_1. \end{align*}

Si $z=x+iy$ then we have \begin{align*} \begin{cases} x^2-y^2=1\cr x^2+y^2=\sqrt{2}\cr 2xy=1 \end{cases} \end{align*} Thus \begin{align*} x^2=\frac{1+\sqrt{2}}{2},\quad y^2=\frac{-1+\sqrt{2}}{2}. \end{align*} As $x$ and $y$ have the same sign, we then have \begin{align*} z_1= \sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}},\quad z_2=-z_1. \end{align*}

From the exponential representation of the roots, we immediately deduce that \begin{align*} \cos(\frac{\pi}{8})&=\frac{\sqrt{1+\sqrt{2}}}{\sqrt{2\sqrt{2}}}\cr \sin(\frac{\pi}{8})&=\frac{\sqrt{-1+\sqrt{2}}}{\sqrt{2\sqrt{2}}} \end{align*}

Q&A about the polar form of complex numbers

Q1: Can you convert a complex number from standard to polar form and vice versa?
A1: Yes, you can convert between standard and polar forms using trigonometric functions. For example, to convert $a+ib$ to polar form, calculate $$ r=\sqrt{a^2+b^2},\quad \theta=\arctan\left(\frac{a}{b}\right).$$
Q2: Are there any restrictions on the values of $r$ and $\theta$ in the polar form?
A2: $r$ must be non-negative, and $\theta$ can take any real value, although it is often expressed within the range $[0,2\pi)$ radians.
Q3: How does the polar form simplify complex number operations?
A3: In polar form, multiplying complex numbers involves multiplying their magnitudes and adding their arguments. Dividing complex numbers requires dividing their magnitudes and subtracting their arguments.

Conclusion

The polar form of complex numbers offers a geometric perspective that enhances our understanding and simplifies complex arithmetic. Its elegance and versatility make it an indispensable tool in various mathematical and scientific disciplines. As an educator, sharing the insights and applications of the polar form with your students can deepen their appreciation for the beauty of mathematics and its practical relevance in the real world. Together, we embrace the power of this mathematical representation, unlocking new horizons of understanding in the realm of complex numbers.

Arithmetic of integers worksheets

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We offer a selection of arithmetic integers worksheets with step-by-step solutions. In this course, one can learn about the division of integers and the properties of prime numbers. In fact, some of the exercises are in the pre-algebra category, while others are in the algebra category. Integers are a subclass of real numbers.

Arithmetic integers worksheets, a selection of exercises

Exercise: The greatest common divisor, ”gcd” of two integers $a$ and $b$ will be denoted by ${\rm gcd}(a,b)$, which is the largest positive integer that divides each of the integers. On the other hand, we say that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, and one writes ${\displaystyle a\mid b}$ if there exists an integer $k$ such that $b=ka$.

  • Let $a,b,a’,b’$ and $d$ be non nulls integers such that \begin{align*} d={rm gcd}(a,b),\quad a=da’\quad\text{and}\; b=db’. \end{align*} Prove that ${\rm gcd}(a’,b’)=1$
  • Let $a$ and $b$ be non-nulls integers. Prove that \begin{align*} {\rm gcd}(a,b)=|a|\;\Longleftrightarrow\; {\displaystyle a\mid b}. \end{align*}
  • Let $a,b$ and $c$ be non-nulls integers. Prove that \begin{align*} {\rm gcd}(ab,ac)=|a|{\rm gcd}(b,c). \end{align*}
  • Let $a,b$ and $n$ be non nulls integers such that ${displaystyle a\mid n}$ and ${\displaystyle a\mid n}$ . Prove that \begin{align*} {\rm gcd}(a,b)=1\;\Longleftrightarrow\; {\displaystyle ab\mid n}. \end{align*}

Solution: 1) As $d={\rm gcd}(a,b),$ there exists $(p,q)\in\mathbb{Z}^2$ such that \begin{align*} d=pq+qb. \end{align*} But $a=da’$ and $b=db’$. Then $d=(pa’+qb’)d$. As $d\neq 0,$ then one can simplify by $d$ and get $1=pa’+qb’$. Now Bezout Theorem implies that ${\rm gcd}(a’,b’)=1$

2) The direct implication: Let ${\rm gcd}(a,b)=|a|$. Then by definition ${\displaystyle |a|\mid b}$, which means that there exists $k\in\mathbb{Z}$ such that $b=k|a|$.Then $b=ka$ or $b=(-k)a$. In both cases, we have ${\displaystyle a\mid b}$.

Then converse implication: If ${\displaystyle a\mid b}$ then there exists $q\in\mathbb{Z}$ such that $b=qa$. Then $b\mathbb{Z}\subset a\mathbb{Z}$. This implies that \begin{align*} a\mathbb{Z}+b\mathbb{Z}\subset a\mathbb{Z}=|a|\mathbb{Z}. \end{align*} Evidently, we have $a\mathbb{Z}\subset a\mathbb{Z}+b\mathbb{Z}$. This yields \begin{align*} a\mathbb{Z}+b\mathbb{Z}=|a|\mathbb{Z}, \end{align*} which means that ${\rm gcd}(a,b)=|a|$.

3) We first observe that for any $(m,n)\in \mathbb{N}^\ast\times \mathbb{N}^\ast,$ $n\mathbb{Z}=m\mathbb{Z}$ is equivalent to $m=n$. In fact, ${\displaystyle n\mid m}$ and ${\displaystyle m\mid n}$ implies that $n=m$.

We have \begin{align*} \left({\rm gcd}(ab,ac)\right)\mathbb{Z}&=ab\mathbb{Z}+ac\mathbb{Z}\cr &= |a|\left(b\mathbb{Z}+c\mathbb{Z}\right)\cr &=|a|{\rm gcd}(b,c)\mathbb{Z}. \end{align*} With our fist observation, we conclude that \begin{align*} {\rm gcd}(ab,ac)=|a|{\rm gcd}(b,c). \end{align*}

4) As ${\displaystyle a\mid n}$ and ${\displaystyle b\mid n}$, then there exist $q,q’\in \mathbb{Z}$ such that $n=qa$ and $n=q’b$. On the other hand, since ${\rm gcd}(a,b)=1$, by Bezout theorem, there exists $(u,v)\in\mathbb{Z}^2$ such that $au+bv=1$. This implies that \begin{align*} n&=nau+nbv\cr &= q’ba u+qab v\cr &= (q’u+qv)ab. \end{align*} This means that ${\displaystyle ab\mid n}$.