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### Theory:

Till now, we learned two sets of numbers, namely rational and irrational.
The real numbers are numbers that include both rational and irrational. In short, the real number is the union of rational and irrational numbers. Therefore, the real number is either rational or irrational and is denoted by $$R$$.
Important!
Each real number can be represented by a unique point on the number line.  Conversely, each point on the number line represents a unique real number.
Procedure to represent irrational number $\sqrt{2}$ and $\sqrt{3}$ on the number line.

We use the concept of Pythagoras Theorem to form square root numbers.

Pythagoras theorem says that 'In a right triangle, the square of the hypotenuse is equal to the sum of the squares of its legs'.

Step 1: Draw a number line and label the centre point as zero.

Step 2: Mark right side of the zero as $$1, 2, 3,...$$ and the left side as $$-1, -2, -3,...$$. Step 3: Since we need the principal square root value, we won't consider the negative side for measuring.

Step 4: Measure the length between $$0$$ and $$1$$ and draw a line perpendicular to point $$1$$ such that the perpendicular line measures $$1$$ unit. Step 5: Join $$0$$ and end of the new line whose length is unity.

Step 6: Thus, we got a right-angled triangle. Step 7: Label the triangle and show base, height and the hypotenuse of it.

Step 8: Length of the hypotenuse can be determined by applying Pythagoras theorem. $\begin{array}{l}{\mathit{OB}}^{2}={\mathit{OA}}^{2}+{\mathit{AB}}^{2}\\ \\ \mathit{OB}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{{\mathit{OA}}^{2}+{\mathit{AB}}^{2}}\\ \\ \mathit{OB}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{{1}^{2}+{1}^{2}}\\ \\ \mathit{OB}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{2}\end{array}$

Step 9: Draw an arc on the number line by taking hypotenuse as radius and origin as the centre.

Step 10: Thus, the distance between origin and the new arc gives the representation of the square root of $$2$$ on the number line. $$OB = OP =$$ $\sqrt{2}$. Step 11:  Taking the hypotenuse side (length is\) $\sqrt{2}$ as a base and drawing an arc perpendicular to end of hypotenuse by taking a unit length, will result in another right triangle with legs measures $\sqrt{{\left(\sqrt{2}\right)}^{2}+{1}^{2}}=\sqrt{2+1}\phantom{\rule{0.147em}{0ex}}=\sqrt{3}$ Step 12: Now drawing an arc on the number line by taking the length of the resultant hypotenuse will represent $\sqrt{3}$ on the number line. Step 13: Proceeding in the same way, we can locate $\sqrt{n}$ for any positive integer $$n$$, after $\sqrt{n-1}$.