]> Operations on real numbers

### Chapter 1: Differentiation; Section 1: Real Numbers; page 2

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# Operations

## Basic operations on real numbers

One can add, subtract, multiply and divide real numbers with the result also being a real number, except for the case of division by zero, which will be discussed later.

In case of addition, for  a  and  b  real numbers:

 $\text{if}\text{ }a+b=0\text{ }\text{then}\text{ }b=-a$ (1.1.2.1)

and we'll  call b  the inverse-sign of  a . ( The correct mathematical expression is "inverse-sign of  a  with respect to addition". ). In this way instead of subtraction one can use addition to the inverse-sign.

A closely related notion is the absolute value of a real number. If  a  is a real number, then his absolute value is:

 $|a|=\left\{\begin{array}{c}\text{\hspace{0.28em}}\text{ }a\text{ }for\text{\hspace{0.17em}}\text{\hspace{0.17em}}a\ge 0\\ -a\text{ }for\text{\hspace{0.28em}}a<0\end{array}\right\}$ (1.1.2.2)

In case of multiplication, for  a  and  b  real numbers:

 $\text{if}\text{ }ab=1\text{ }\text{then}\text{ }b=\frac{1}{a}$ (1.1.2.3)

and we'll  call b  the reciprocal of  a . ( The correct mathematical expression is "inverse of  a  with respect to multiplication". ) In this way instead of division one can use multiplication by the reciprocal.

However, there is a problem: what happens with the reciprocal of zero? This is surely not a regular real number. It is called infinity and its symbol is $\infty$ , it may also carry a sign depending on the circumstances. Here are some relations with non zero real numbers:

 $\text{for}\text{ }a\ne 0\text{\hspace{0.28em}}:\text{ }\text{\hspace{0.28em}}\frac{a}{0}=\infty \text{\hspace{0.28em}},\text{ }a\text{\hspace{0.17em}}\infty =\infty$ (1.1.2.4)

Other relations hold in case of any real numbers:

 $\text{for}\text{ }a=\text{any}\text{\hspace{0.28em}}:\text{ }\text{\hspace{0.28em}}\frac{a}{\infty }=0\text{\hspace{0.28em}},\text{ }\frac{\infty }{a}=\infty \text{\hspace{0.28em}},\text{ }a+\infty =\infty$ (1.1.2.5)

There is a more severe problem. What is zero divided by zero? In the case that  a  is any non-zero real number we know that:

If one wants to extend the value of  a  to zero, then any of these three values: 0, 1, $\infty$  can be a solution. As a matter of fact any real number can be a solution depending on the circumstances. The derivatives are such kind of ratios and we will have plenty of experience dealing with them.

It can be shown that there are more undefined expressions. Here is the list:

 $\frac{0}{0}\text{\hspace{0.28em}};\text{ }0\text{\hspace{0.17em}}\infty \text{\hspace{0.28em}};\text{ }\infty -\infty \text{\hspace{0.28em}};\text{ }\frac{\infty }{\infty }$ (1.1.2.6)

## Integer power

The expression:   ${r}^{n}$   is called the real number  r  raised to the power  n  .

If  n  is a natural number the definition is:

 ${r}^{n}=\underset{n\text{ }\text{times}}{\underbrace{r\text{\hspace{0.17em}}r\text{\hspace{0.17em}}r\text{\hspace{0.17em}}...\text{\hspace{0.17em}}r}}$ (1.1.2.7)

From the definition (1.1.2.7) it follows that:

 ${r}^{m+n}={r}^{m}{r}^{n}\text{ }\text{\hspace{0.17em}}\text{and}\text{ }\text{\hspace{0.17em}}{r}^{m-n}=\frac{{r}^{m}}{{r}^{n}}$ (1.1.2.8)

which allows one to extend the definition of a power to the rest of the integers. For negative power one has:

 ${r}^{-n}=\frac{1}{{r}^{n}}\text{ }\text{\hspace{0.28em}}\text{\hspace{0.28em}}\left(n>0\right)$ (1.1.2.9)

that is infinity for  r = 0 . For power zero one obtains:

 ${r}^{0}=1\text{ }\text{\hspace{0.17em}}\text{for}\text{\hspace{0.28em}}\text{\hspace{0.28em}}r\ne 0$ (1.1.2.10)

which is undefined for  r = 0 .

From the definition of natural powers (1.1.2.7) it also follows that:

 ${r}^{nm}={\left({r}^{m}\right)}^{n}={\left({r}^{n}\right)}^{m}$ (1.1.2.11)

which could also be extended to any integer powers.

## Rational power

In order to extend the definition of a power to rational numbers, we'll start by substituting a power with the reciprocal of an integer. By use of (1.1.2.11) and defining:

 $m=\frac{1}{n}$ (1.1.2.12)

one obtains:

 $r={\left({r}^{n}\right)}^{\frac{1}{n}}={\left({r}^{\frac{1}{n}}\right)}^{n}$ (1.1.2.13)

The expression of  r  raised to a power reciprocal of  n  is also called the  nth  root of  r  and is also denoted by:

 ${r}^{\frac{1}{n}}=\sqrt[n]{r}$ (1.1.2.14)

The case of  n = 2  is called square root, and the number  2  is usually omitted,.

The definition (1.1.2.14) poses some restrictions. Let's take the case of  n = 2 :

1. The square root (root of 2) is double valued: having both negative and positive sign. The two signs are important in some cases, e.g. when solving quadratic equations. We'll meet later cases, where only the positive sign will be considered.
2. The square root of a negative number is not a real number, and in order to define any root of a negative number we need to extend the realm of numbers beyond the real numbers, which we are not doing on this stage. For that reason we have to restrict the value of  r  to non-negative numbers.

On the other hand for the case of  n = 3 , the situation is different:

1. The root of  3  is single valued.
2. The root is defined for positive as well as for negative numbers.

In order to summarize: we'll keep the restriction of  r  to non-negative numbers as a general rule, but we'll bear in mind that in some cases we could also permit negative numbers.

The next step will be to extend the definition of a power to the rational numbers, which is done by expressing the power  p  as a ratio of the integers  n  over  m  and by the use of (1.1.2.11) and (1.1.2.14). One obtains:

 ${r}^{p}={r}^{\frac{n}{m}}={\left({r}^{\frac{1}{m}}\right)}^{n}={\left({r}^{n}\right)}^{\frac{1}{m}}\text{ }\text{\hspace{0.17em}}\text{for}\text{ }\text{\hspace{0.17em}}r\ge 0$ (1.1.2.15)

However, the order of the operations (first the numerator and then the denominator of the power or in reverse order) could result in an ambiguous sign, e.g. in the case of  m = 2  of (1.1.2.15).

Zero raised by the power  p , by definition should be:

 ${0}^{p}=\text{\hspace{0.28em}}\left\{\begin{array}{l}0\text{ }\text{for}\text{ }p>0\\ \infty \text{\hspace{0.28em}}\text{\hspace{0.28em}}\text{\hspace{0.17em}}\text{for}\text{ }p<0\end{array}$ (1.1.2.16)

therefore  p = 0  leads to an undefined value.

A real number raised by the power  p = $\infty$   by definition should be:

 ${r}^{\infty }=\left\{\begin{array}{l}0\text{ }\text{for}\text{ }01\end{array}$ (1.1.2.17)

which leaves   ${1}^{\infty }$  undefined.

Here is the list of undefined expressions involving powers:

 ${0}^{0}\text{\hspace{0.28em}};\text{ }{1}^{\infty }\text{\hspace{0.28em}};\text{ }{\infty }^{0}$ (1.1.2.18)

The definition of a power will be extended to any real number on the next page ( p.3 Sequences ).

## Binomial expansion

Factorial of a non-negative integer n is written as n! and its definition is

 $\begin{array}{l}0!=1!=1\text{ }2!=2\text{ }\text{and for}\text{ }n>2\text{ }:\\ \text{or}\text{\hspace{0.17em}}\left\{\begin{array}{l}n!=\left(1\right)\left(2\right)\left(3\right).........\left(n-2\right)\left(n-1\right)n\\ n!=\left(n-1\right)!\text{\hspace{0.28em}}n\end{array}\end{array}\right\}$ (1.1.2.19)

which means for  n > 1  - the product of all the natural numbers starting from 1 and ending by  n . As an example  n!  is the number of permutations for ordering  n  different items.

A binomial expresses the sum of two numbers, raised to the power  p, meaning: ${\left(a+b\right)}^{p}$ . We are only going to deal now with natural powers p (Newton's binomial). The expansion of such binomial is

 $\left\{\begin{array}{l}{\left(a+b\right)}^{n}=\sum _{k=0}^{n}\left(\begin{array}{l}n\\ k\end{array}\right){a}^{n-k}{b}^{k}\\ \left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}\end{array}\right\}$ (1.1.2.20)

where the symbol  $\sum _{k=0}^{n}$  means summation of the term that follows with  k  taking the values from  0  to  n . For example

The proof of (1.1.2.20) can be done by induction: First one has to prove it for the lowest  n  (n = 1  in this case - which is trivial), then by assuming that it is correct for a given  n , one has to prove the relation for  n+1 . This way all possible values of  n  are covered. In this particular case the proof is long and cumbersome and it will not be presented here, but we will give another example of proof by induction in order to introduce the method.

Example of induction

One has to prove the following relation by induction:

 $\sum _{k=1}^{n}{k}^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ (1.1.2.21)

It is trivial to see that the relation holds for  n = 1 . By use of (1.1.2.21) one has to prove that

 $\sum _{k=1}^{n+1}{k}^{2}=\frac{\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{6}$ (1.1.2.22)

therefore one has to obtain (1.1.2.22) by use of (1.1.2.21)

The right-hand side of the last equation should be equal to the right-hand side of (1.1.2.22). This is easily achieved by basic operations, and it is left as a simple exercise for the user (hint:  n+1  is a common factor).

## Exercises

Exercise 1. Use the definition of rational numbers for the following questions:

• Is it a correct statement that always
irrational number − rational number = irrational number ?
• As consequence of the previous, is it correct to state that always:
irrational number − irrational number = rational number ?
Give examples.

Exercise 2. Prove by contradiction that the

multiplication of a rational number by an irrational number gives an irrational number.
Is there any exception?

Exercise 3. Calculate the two expressions and find out what the difference in the results is:

$\frac{{\left({6}^{\frac{3}{2}}\right)}^{4}}{{3}^{5}{2}^{3}}\text{ }\text{\hspace{0.17em}}\text{and}\text{ }\text{\hspace{0.17em}}\frac{{\left({6}^{12}\right)}^{\frac{1}{2}}}{{6}^{3}9}$

Exercise 4. Show that ${\infty }^{0}$ is undefined.

Exercise 5. Prove by induction that : $\sum _{k=1}^{n}\frac{1}{\left(2k-1\right)\left(2k+1\right)}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{n}{2n+1}$

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