What is an Irrational Number?

There is no formal definition of irrational number, but the real number which not rational number is called irrational number.

In the article of rational number we have observed that every rational number can be represented by a point on the number line. A question now naturally arises: “Is the converse is true?” Or, in other words, “Is it possible to assign a rational number to every point on the number line?” A simple consideration, as explained below, will show that the answer to this question is negative.

Geometrical Representation of Irrational Number:

Let us construct a square with one of its sides as OA of unit length and take a point P on the number line such that OP is equal in length to the diagonal OB of this square OABC. It will now be shown that the point P cannot corresponds to a rational number i.e., the length of OP cannot have any rational number as its measure.

If possible let the measure of OP be a rational number p/q, so that

\[{{\left( \frac{p}{q} \right)}^{2}}=O{{P}^{2}}=O{{B}^{2}}=O{{A}^{2}}+A{{B}^{2}}={{1}^{2}}+{{1}^{2}}=2\]

\[\Rightarrow {{p}^{2}}=2{{q}^{2}}………..(1)\]

We assumed that the natural numbers p and q have no common factor, for such factors, if any, can be cancelled at the beginning.

We observe that if n is any natural number, 2n is an even number and (2n + 1) is an odd number. Also, we note that (2n)² = 4n², and (2n + 1)² = 4(n² + n) + 1 i.e., the square of an even number is even and the square of an odd number is an odd number.

Now, equation (1) shows that p² is an even number, so that p itself must be even.

Let then p be equal to 2m, where m is any integer. And we have p² = (2m)² = 4m² = 2q² ⇒ q² = 2m² ⇒ q² is even ⇒ q is also even.

Hence p and q have a common factor 2 and this conclusion contradicts the hypothesis that p and q have no common factor. Thus the measure √2 of OP is not a rational number.

There exists, therefore, a point on the number line that does not corresponds to any rational number.

Again, let us take a point L on the number line, such that the length OL is any rational multiple of the of OP, i.e., of p/q.

We observe that OL cannot have a rational measure. For if possible, let r/s is the measure of OL, where r and s are integers. Then

\[\frac{p}{q}\sqrt{2}=\frac{r}{s}\Rightarrow \sqrt{2}=\frac{rq}{ps}\]

Which means that √2 is a rational number, being equal to rq/ps. Thus we arrive at a contradiction. Hence the point L cannot correspond to any rational number.

Thus we see that there exist an unlimited large number of points on the number line which do not correspond any rational number.

Now, if we require that the set of numbers should be such that after the choice of unit length on the number line, every point on the line should correspond to a number (or, every line segment should be capable of measurement), we are obliged to extend the number system further beyond the set of rational numbers. This is done by introducing a set of new numbers, called Irrational numbers. With the introduction of irrational numbers, each point on the number line shall have its corresponding number, rational number or irrational number.

The total of rational and irrational numbers from the set of Real Numbers.

Prove that √3 + √2 is an irrational number.

Solution:

Let us assume the contrary, i.e., we assume that √3 + √2 is rational.

\[Since~~\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)=1\]

\[\Rightarrow \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}\]

is also a rational number, being the quotient of two rational number 1 and √3 + √2, which by assumption is real.

Now,

\[\sqrt{2}=\frac{1}{2}\left\{ \left( \sqrt{3}+\sqrt{2} \right)-\left( \sqrt{3}-\sqrt{2} \right) \right\}\]

Thus √2 , being the difference of two rational numbers, is rational. But √2 is not a rational number. So we arrive at a contradiction. Hence our initial assumption, that √3 + √2 is rational, is wrong.

Hence √3 + √2 is an irrational number.;