# 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

This site uses Akismet to reduce spam. Learn how your comment data is processed.