Mathematical Introduction for Physics and Engineering

by Samuel Dagan (Copyright © 2007-2020)

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**Rational numbers** are defined as a **ratio of integers**:

$r=\frac{n}{m}$ | (1.1.1.1) |
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where ** n** and

$m\ne 0$ | (1.1.1.2) |
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**excludes the entities which do not represent numbers** (discussed later).

A **reduced form of a rational number** is the case when ** n** and

The following link discusses the representation of the rational numbers on an infinite axis:
**Fig. Rational numbers**.

The **natural numbers** form a set of infinite items, which **are countable**, meaning that we can order them in a row so that each one is associated to an ordinal number (1,2,3,....). **All the integers are also countable**, since we can arrange them in an ordered row and associate to each one an ordinal number. One way of such arrangement is:

$0,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}-2,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}-3,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}-4,\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}-5,\text{\hspace{0.17em}}6,\text{\hspace{0.17em}}.\text{\hspace{0.17em}}.\text{\hspace{0.17em}}.\text{\hspace{0.17em}}.$ | (1.1.1.3) |
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Notice that after each positive number, the corresponding negative number was added. As an example, the nineth ordinal number belongs to the integer −4 .

Although the integers appear to be only a tiny part of all the rational numbers, it can be shown that the **rational numbers are countable** as well, as seen from **Fig. Countability**.

It was shown in **Fig. Rational numbers** that all the rational numbers have their location on an infinite axis. But **not all the points of the axis can be represented as rational numbers**. For instance the well known number ** π** has a location on the axis but is not rational.

All the numbers that are located on this axis and are not rational are called irrational. The **rational and the irrational** numbers together **are called real numbers** and the axis is called the axis of the real numbers. It can be shown that **between two different real numbers**, no matter how close they are, there **are at least one rational and one irrational number**.

As an example of an irrational number, we are going to prove that ** p** , the square-root of 2
($\sqrt{2}$) , is irrational:

$\text{if}\text{\hspace{1em}}{p}^{2}=p\text{\hspace{0.17em}}p=2\text{\hspace{1em}}\text{\hspace{1em}}\text{then}\text{\hspace{1em}}p=\text{irrational}$ | (1.1.1.4) |
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Let's **assume that p** is a rational number and

$p=\frac{n}{m}\text{\hspace{1em}}\text{\hspace{1em}}\text{and}\text{\hspace{1em}}\text{\hspace{1em}}\frac{{n}^{2}}{{m}^{2}}=2$ | (1.1.1.5) |
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From (1.1.1.5) it follows that:

${n}^{2}=2{m}^{2}$ | (1.1.1.6) |
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If 2 is a factor of ** n** , it means that

$n=2k$ | (1.1.1.7) |
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where ** k** is another integer, then (1.1.1.6) is true.
On the other hand, if 2 is not a factor of

$n=2k+1\text{\hspace{1em}}\text{\hspace{1em}}\text{and}\text{\hspace{1em}}\text{\hspace{1em}}{n}^{2}=4k(k+1)+1$ | (1.1.1.8) |
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therefore ** n²** does not satisfy (1.1.1.6). From (1.1.1.6) and (1.1.1.7) one obtains:

${m}^{2}=2{k}^{2}$ | (1.1.1.9) |
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and therefore 2 is a factor of ** m** , which is inconsistent with the assumption that

It can be shown that **the set of irrational numbers** (and therefore - of the real numbers) **is not countable**. In a loose way one can say that the axis of real numbers is more densely populated with irrational, than with rational numbers.

**Exercise 1.** Use the definition of rational numbers {(1.1.1.1) and (1.1.1.2)} to prove that the results of addition (sum), multiplication (product) and division (ratio) of rational numbers are also rational.

**Exercise 2.** Prove that all the rational numbers ** r** in the interval:

**Exercise 3.** Prove that ** p**, the square-root of 3
($\sqrt{3}$), is irrational ( hint: see the example above for square-root of 2 ).

**Exercise 4.** By following the method of the proofs that
**$\sqrt{2}$
** and
**$\sqrt{3}$** are irrational, would you expect to be able to prove that **$\sqrt{4}$** is also irrational.
If you could, then find your mistake.

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