# Analysis — Chapter 1 Real Variables — part 7 — continued

Part 7. Irrational numbers (continued).

In the first two cases, we say that the section corresponds to a positive rational number a, which is l in the one case and r in the other. Conversely, it is clear that to any such number a corresponds a section which we shall denote by

$\alpha^{*}$. For we might take P and Q to be the properties expressed by

$x \leq a, x > a$

respectively, or by $x and $x \leq a$. In the first case, a would be the greatest number of L, and in the second case the least member of R. These are in fact just two sections corresponding to any positive rational number. In order to avoid ambiguity we select one of them; let us select that in which the number itself belongs to the upper class. In other words, let us agree that we will consider only sections in which the lower class L has no greatest number.

There being this correspondence between the positive rational numbers and the sections defined by means of them, it would be perfectly legitimate, for mathematical purposes, to replace the numbers by the sections, and to regard the symbols which occur in our formulae as standing for the sections instead of for the numbers. Thus, for example,

$\alpha > \alpha^{'}$ would mean the same as $a > a^{'}$. If $\alpha$ and $\alpha^{'}$ are

the sections which correspond to a and $a^{'}$.

But, when we have in this way substituted sections of rational numbers for the rational numbers themselves, we are almost forced to a generalization of our number system. For there are sections (such as that of blog on Chapter 1 — part 4) which do not correspond to any rational number. The aggregate of sections is a larger aggregate than that of the positive rational numbers; it includes sections corresponding to all these numbers, and more besides. It is this fact which we make the basis of our generalization of the idea of a number. We accordingly frame the following definitions, which will however be modified in the next blog, and must therefore be regarded as temporary and provisional.

A section of the positive rational numbers, in which both classes exist and the lower class has no greatest member, is called a positive real number.

A positive real number which does not correspond to a positive rational number is called a positive irrational

number.

More later,

Nalin Pithwa

# Analysis — Chapter I — Real Variables — Part 5 — Irrational numbers continued

We have thus divided the positive rational numbers into two classes, L and R, such that (i) every member of R is greater than every member of L, and (ii) we can find a member of L and a member of R, whose difference is as small as we please, (iii) L has no greatest and R has not least member. Our common-sense notion of the attributes of a straight line, the requirements of our elementary geometry and our elementary algebra, alike demand the existence of a number x greater than all the members of L and less than all the members of R, and of a corresponding point P on $\Lambda$ such that P divides the points which correspond to members of L from those which correspond to members of R.

Let us suppose for a moment that there is such a number x and that it may be operated upon in accordance with laws of algebra, so that, for example, $x^{2}$ has a definite meaning. Then $x^{2}$ cannot either be less than or greater than 2. For suppose, for example, that $x^{2}$ is less than 2. Then, it follows from what precedes that we can find a positive rational number $\xi$ such that $\xi^{2}$ lies between $x^{2}$ and 2. That is to say, we can find a member of L greater than x; and this contradicts the supposition that x divides the members of L from those of R. Thus, $x^{2}$ cannot be less than 2, and similarly, it cannot be greater than 2. We are therefore driven to the conclusion that $x^{2}=2$, and that x is the number which in algebra  we denote by $\sqrt{2}$. And, of course, this number $\sqrt{2}$ is not rational, for no rational number has its square equal to 2. It is the simplest example of what is called an irrational number.

But the preceding argument may be applied to equations other than $x^{2}=2$, almost word for word; for example, to

$x^{2}=N$, where N is an integer which is not a perfect square, or to

$latex$x^{3}=3\$ and $x^{2}=7$ and $x^{4}=23$,

or, as we shall see later on. to $x^{3}=3x+8$. We are thus led to believe for the existence of irrational numbers x and points P on $\Lambda$ such that x satisfies equations such as these, even when these lengths cannot (as $\sqrt{2}$ can) be constructed by means of elementary geometric methods.

The reader may now follow one or other of two alternative courses. He may, if he pleases, be content to assume that “irrational numbers” such as $\sqrt{2}$ and $\sqrt[5]{3}$ exist and are amenable to usual algebraic laws. If he does this, he will be able to avoid the more abstract discussions of the next few blogs.

If, on the other hand, he is not disposed to adopt so naive an attitude, he will be well advised to pay careful attention to the blogs which follow, in which these questions receive further consideration.

More later,

Nalin Pithwa

# Analysis — Chapter 1 — Real Variables — Part 4 Irrational numbers continued

Part 4. Irrational numbers (continued).

The result of our geometrical interpretation of the rational numbers is therefore to suggest the desirability of enlarging our conception of “number” by the introduction of further numbers of a new kind.

The same conclusion might have been reached without the use of geometrical language. One of the central problems of algebra is that of the solution of equations, such as

$x^{2}=1$, $x^{2}=2$.

The first equation has the two rational roots 1 and -1. But, if our conception of number is to be limited to the rational numbers, we can only say that the second equation has no roots; and the same is the case with such equations as $x^{3}=2$, $x^{4}=7$. These facts are plainly sufficient to make some generalization of our idea of number desirable, if it should prove to be possible.

Let us consider more closely the equation $x^{2}=2$.

We have already seen that there is no rational number x which satisfies this equation. The square of any rational number is either less than or greater than 2. We can therefore divide the rational numbers into two classes, one containing the numbers whose squares are less than 2, and the other those whose squares are greater than 2. We shall confine our attention to the positive rational numbers, and we shall call these two classes the class L, or the lower class, or the left-hand class, and the class R, or the upper class, or the right hand class. It is obvious that every member of R is greater than all the members of class R. Moreover, it is easy to convince ourselves that we can find a member of the class L whose square, though less than 2, differs from 2 by as little as possible, and a member of R whose square, though greater than 2, also differs from 2 by as little as we please. In fact, it we carry out the ordinary arithmetical process for the extraction of the square root of 2, we obtain a series of rational numbers, viz.,

1,1.4, 1.41, 1.414, 1.4142, $\ldots$

whose squares

1, 1.96, 1.9881, 1.999396, 1.99996164, $\ldots$

are all less than 2, but approach nearer and nearer to it, and by taking a sufficient number of the figures given by the process we can obtain as close an approximation as we want. And if we increase the last figure, in each of the approximations given above, by unity, we obtain a series of rational numbers

2, 1.5, 1.42, 1.415,1.413, $\ldots$

whose squares

4, 2.25, 2.0164, 2.002225, 2.00024449, $\ldots$

are all greater than 2, but approximate to 2 as closely as we please.

It follows also that there can be no largest member of L or smallest member of R. For if x is any member of L, then

$x^{2} < 2$. Suppose that $x^{2}=2-\delta$. Then we can find a member x, of L such that ${x_{1}}^{2}$ differs from 2 by less than $\delta$, and ${x_{1}}^{2}>x^{2}$ or $x_{1}>x$. Thus there are larger members of L than x; and, as x is any member of L, it follows that no member of L can be larger than all the rest. Hence, L has no largest member, and similarly, it has no smallest.

Note: A rigorous analysis of the above can be easily carried out. If you need help, please let me know and I will post it in the next blog.

More later,

Nalin Pithwa

# Analysis — Real Variables — Chapter 1 — Examples II

Examples II.

1) Show that no rational number can have its cube equal to 2.

Proof #1.

Let, if possible, $p/q, q \neq 0$, and p and q do not  have any common factor and are integers. Then, if $(p/q)^{3}=2$, we have

$p^{3}=2q^{3}$. So, p contains a factor of 2. So, let $p=2k$. So, q contains a factor of 2. Hence, both p and q have a common factor have a common factor 2, contradictory to out assumption. Hence, the proof.

2) Prove generally that a rational function $p/q$ in its lowest terms cannot be the cube of a rational number unless p and q are both perfect cubes.

Proof #2.

Let, if possible, $p/q = (m/n)^{3}$ where m,n,p,q are integers, with n and q non-zero and p and q are in lowest terms. This implies that m and n have no common factor.  Hence, $p=m^{3}, q=n^{3}$.

3) A more general proposition, which is due to Gauss and includes those which precede as particular cases, is the following: an algebraical equation

$z^{n}+p_{1}z^{n-1}+p_{2}z^{n-2}+ \ldots + p_{n}=0$ with integral coefficients, cannot have rational, but non-integral root.

Proof #3.

For suppose that, the equation has a root $a/b$, where a and b are integers without a common factor, and b is positive. Writing

$a/b$ for z, and multiplying by $b^{n-1}$, we obtain

$-(a^{n}/b)=p_{1}a^{n-1}+p_{2}a^{n-2}b+ \ldots + p_{n}b^{n-1}$,

a function in its lowest terms equal to an integer, which is absurd. Thus, $b=1$, and the root is a. It is evident that a must be a divisor of $p_{n}$.

4) Show that if $p_{n}=1$ and neither of

$1+p_{1}+p_{2}+p_{3}+\ldots$ and $1-p_{1}+p_{2}-p_{3}+\ldots$

is zero, then the equation cannot have a rational root.

Proof #4. Please try this and send me a solution.. I do not have a solution yet 🙂

5) Find the rational roots, if any, of $x^{4}-4x^{3}-8x^{2}+13x+10=0$.

Solution #5.

Use problem #3.

The roots can only be integral, and so find the roots by trial and error. It is clear that we can in this way determine the rational roots of any such equation.

More later,

Nalin Pithwa

# Analysis — Chapter I — part 3 — Real Variables — Irrational numbers

Part 3. Irrational numbers.

If the reader will mark off on the line all the points corresponding to the rational numbers whose denominators are 1,2,3, …in succession, he will readily  convince himself that he can cover the line with rational points, as closely as he likes. We can state this more precisely as follows: If we take any segment BC on A, we can find as many rational points on it as we please on BC.

Suppose, for example, that BC falls within the segment $A_{1}A_{2}$. it is evident that if we choose a positive integer k such that

$k.BC>1$ Equation I

(The assumption that this is possible is equivalent to the assumption of what is known as the Axiom of Archimedes.)

and divide $A_{1}A_{2}$ into k equal parts, then at least one of the points of division (say P) must fall inside BC, without coinciding with either B or C. For if this were not so, BC would be entirely included in one of the k parts into which $A_{1}A_{2}$ has been divided, which contradicts the supposition I. But P obviously corresponds to a rational number whose denominator is k. Thus at least one rational point P lies between B and C. But, then we can find another such point Q between B and P, another between B and Q, and so on indefinitely; that is, as we asserted above, we can find as many as we please. We may express this by saying that BC includes infinitely many

rational points. (We will investigate the meaning of infinite more closely later).

From these considerations, the reader might be tempted to infer that an adequate view of the nature of the line could be obtained by imagining it to be formed simply by the rational points which lie on it. And, it is certainly the case that if we imagine the line to be made up of  solely of the rational points, and all other points (if there are any such) to be eliminated, the figure would possess most of the properties which common sense attributes to the straight line, and would, to put the matter roughly, look and behave very much like a line.

A little further consideration, however, shows that this view would involve us in serious difficulties.

Let us look at the matter for a moment with the eye of common sense, and consider some of the properties which we may reasonably expect a straight line to possess if it is to satisfy the idea which we have formed of it in elementary geometry.

The straight line must be composed of points, and any segment of it by all the points which lie between its end points.  With any such segment must be associated a certain entity called its length, which must be a quantity capable of numerical measurement in terms of any standard or unit length, and these lengths must be capable of combination with another, according to the ordinary rules of algebra, by means of addition or multiplication. Again, it must be possible to construct a line whose length is the sum or product of any two given lengths. If the length PQ along a given line is a, and the length QR, along the same straight line, is b, the length PR must be $a+b$.

Moreover, if the lengths OP and OQ, along one straight line, are 1 and a, and the length OR along another straight line is b, and if we determine the length OS by Euclid’s construction for a fourth proportional to the lines OP, OQ, OR, this length must be ab, the algebraic fourth proportional to 1, a and b. And, it is hardly necessary to remark that the sums and products thus defined must obey the ordinary laws of algebra; viz.,

$a+b=b+a$

$a+(b+c)=(a+b)+c$

$ab=ba$

$a(bc)=(ab)c$

$a(b+c)=ab+ac$

The lengths of our lines must also obey a number of obvious laws concerning inequalities as well as equalities: thus, if A, B, C are three points lying along A from left to right, we must have $AB, and so on. Moreover, it might be possible, on our fundamental line $\Lambda$ to find a point P such that $A_{0}P$ is equal to any segment whatever taken along $\Lambda$ or along any other straight line. All these properties of a line, and more, are involved in the presuppositions of our elementary geometry.

Now, it is very easy to see that the idea of a straight line as composed of a series of points, each corresponding to a rational number, cannot possibly satisfy all these requirements. There are various elementary geometrical constructions, for example, which purport to construct a length x such that $x^{2}=2$. For instance, we may construct an isosceles right angled triangle ABC such that $AB=AC=1$.. Then, if $BC=x$, $x^{2}=2$. Or we may determine the length x by means of Euclid’s construction for a mean proportional to a and 2, as indicated in the figure. Our requirements therefore involve the existence of a length measured by a number x, and a point P on $\Lambda$ such that $A_{0}P=x$, $x^{2}=2$.

But, it is easy to see that there is no rational number such that its square is 2. In fact, we may go further and say that there is no rational number whose square is $m/n$, where $m/n$ is say positive fraction in its lowest terms, unless m and n are both perfect squares.

For suppose, if possible, that $\frac {p^{2}}{q^{2}}=m/n$.

p having no common factor with q, and m no common factor with n. Thus, $np^{2}=mq^{2}$. Every factor of $q^{2}$ must divide $np^{2}$, and as p and q have no common factor, every factor of $q^{2}$ must divide n. Hence,

$n={\lambda}q^{2}$, where $\lambda$ is an integer. But, this involves $m={\lambda}p^{2}$: and as m and n have common factor, $\lambda$ must be unity. Thus, $m=p^{2}$ and $n=q^{2}$, as was to be proved. In particular, it follows by taking $n=1$, that an integer cannot be the square of a rational number, unless that rational number is itself integral.

it appears that our requirements involve the existence of a number x and a point P, not one of the rational points already constructed, such that $A_{0}P=x$ and $x^{2}=2$; and, (as the reader will remember from elementary algebra) we write $x = \sqrt {2}$.

Alternate proof.

The following alternate proof that no rational number can have its square equal to 2 is interesting.

Suppose, if possible, that $p/q$ is a positive fraction, in its lowest terms such that $(p/q)^{2}=2$. It is easy to see that this involves $(2q-p)^{2}=2(p-q)^{2}$, and so $\frac {2q-p}{p-q}$ is also another fraction having the same property. But, clearly,

$q and so $p-q. Hence, there is another fraction equal to $p/q$ and having a smaller denomination, which contradicts the assumption that $p/q$ is in its lowest terms.

In the next blog, we shall look at examples,

More later,

Nalin Pithwa

# Analysis — Chapter I Part I — Real Variables

Real Variables.

1Rational Numbers. A fraction $r=p/q$, where p and q are positive or negative integers, is called a rational number. We can assume (i) that p and q have no common factors, as if they have a common factor we can divide each of them by it, and (ii) that q is positive, since

$p/(-q)=(-p)/q$ and $(-p)/(-q)=p/q$.

To the rational numbers thus defined we may add the “rational number 0” obtained by taking $p=0$.

We assume that you are familiar with the ordinary arithmetic rules for the manipulation of rational numbers. The examples which follow demand no knowledge beyond this.

Example I. 1. If r and s are rational numbers, then $r+s$, $r-s$, $rs$ and $r/s$ are rational numbers, unless in the last case $s=0$ (when $r/s$ is meaningless, of course).

2. If $\lambda$, m, and n are positive rational numbers, and $m>n$, then

$\lambda (m^{2}-n^{2})$, $2\lambda mn$, and $\lambda (m^{2}+n^{2})$ are positive rational numbers. Hence, show how to determine any number of right angled triangles the lengths of all of whose sides are rational.

Proof: Let the hypotenuse be $\lambda (m^{2}+n^{2})$ and the two arms of the right angled triangle be

$2\lambda mn$ and $\lambda (m^{2}-n^{2})$. Then, the Pythagoras’s theorem holds. But, the sides and the hypotenuse are all rational.

3. Any terminated decimal represents a rational  number whose denominator contains no  factors other than 2 or 5. Conversely, any such rational number can be expressed, and in one way only, as a terminated decimal.

(We will look into this matter a bit deeper, a little later).

4. The positive rational numbers may be arranged in the form of a simple series as follows:

$1/1,2/1,1/2, 3/1,2/2, 1/3,4/1,3/2,2/3,1/4, \ldots$.

Show that $p/q$ is the $\{ (1/2)(p+q-1)(p+q-2) +q \}$th term of the series.

(In this series, every rational number is repeated indefinitely. Thus 1 occurs as $1/1, 2/2, 3/3, \ldots$ We can of course avoid this by omitting every number which has already occurred in a simple form, but then the problem of determining the precise position of $p/q$ becomes more complicated.) Check this for yourself! If you do not get the answer, just write back in the comment section and I will help clarify the matter.

More later…

Nalin Pithwa