Name: Surds: Introduction
Uploaded: 2019-07-19
Duration: 4 min 21 s
Description: We know that irrational numbers are written in the form of neither terminating nor recurring decimals. Let's learn to write another form of irrational number...

Question

Show that there exist points on the number line not representing rational numbers.

Accepted Answer

Consider a square $A B C D$ whose side is of one unit in length. The diagonal $AC$ is of length given by ${AC}^{2} = {AB}^{2} + {BC}^{2}$

$= 1^{2} + 1^{2} = 1 + 1 = 2$

$\Rightarrow A C = \sqrt{2}$ .

Let's assume $\sqrt{2}$ is a rational number.

As per the definition of rational numbers, we can express $\sqrt{2} = \frac{p}{q}$ , {where $p, q$ are co-primes and $q \neq 0$ }.

On squaring both sides, we get,

${(\sqrt{2})}^{2} = \frac{p^{2}}{q^{2}}$

$\Rightarrow 2 = \frac{p^{2}}{q^{2}}$

$\Rightarrow p^{2} = 2 q^{2} . . . . . . (i)$

Which means that, $p^{2}$ is a multiple of $2$ ,

$\Rightarrow 2$ divides $p$

Now, let $p = 2 m \Rightarrow p^{2} = 4 m^{2}$

Now putting this value of $p^{2}$ in equation $(i)$ ,

$4 m^{2} = 2 q^{2}$

$\Rightarrow 2 m^{2} = q^{2}$

$\Rightarrow 2$ divides $q^{2}$

$\Rightarrow 2$ divides $q$

This means that, $2$ is a common divisor of $p$ and $q$ , which contradicts the assumption done in the first step. { $p, q$ are co-primes}.

Therefore, $\sqrt{2}$ is an irrational number.

It is now seen that the number indicating the length of a diagonal of a square whose side is one unit in length, is not a rational number. Accordingly, the number represented on the number line by the point $P$ where $O P$ (the point $O$ represents the number zero) is equal to the length of a diagonal of a unit square is not a rational number. This indicates that there are points on the number line not representing rational numbers. [Proved]

Show that there exist points on the number line not representing rational numbers.

Important Questions on Number System

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