Exercise 1.2 Number Systems NCERT Solutions Class 9

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Chapter 1 Ex.1.2 Question 1

State whether the following statements are true or false. Justify your Answers.


Video Solution

(i) Every irrational number is a real number.


True, because the set of real numbers consists of rational numbers and irrational numbers.

(ii) Every point on the number line in of the form \(\begin{align}\,\sqrt m ,\end{align}\) where \(m\) is a natural number.


False, for example \(\begin{align}\sqrt {\frac{2}{3}} \end{align}\) is a real number on the number line but \(\begin{align}\frac{2}{3}\end{align}\) is not a natural number (Note that \(\begin{align}\frac{2}{3}\end{align}\) is a rational number).

(iii) Every real number is an irrational number.


False. Note that the set of real numbers contains both rational and irrational numbers. As an example, \(\begin{align}\frac{1}{2}\end{align}\) is a rational number and hence it is real. But it is not an irrational number.

Chapter 1 Ex.1.2 Question 2

Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.



Video Solution


The square roots of positive integers are not irrationals.

Example: \(\begin{align}\,\sqrt 4 = 2\end{align}\) and \(2\) in a rational number \(\begin{align} \left(\because 2=\frac{2}{1}\right) \end{align}\)

Chapter 1 Ex.1.2 Question 3

Show how \(\begin{align}\sqrt 5 \end{align}\) represented on the number line.



Video Solution

What is known?

Integer \(5\).

What is unknown?

Point representing \(\begin{align}\sqrt 5 \end{align}\) on the number line.


We shall write \(5\) on the sum of two squares in the form \(\begin{align}5=2^{2}+1^{2}=4+1=5 \end{align}\). This shows we need to construct a right triangle with sides \(2\) and \(1\) units. So the hypotenuse becomes \(\begin{align} \sqrt{5} \end{align}\) units on the number line. We shall proceed as follows.


On the number line take \(2\) units from \(O\) and represent the point as \(A\). At \(A\) draw the perpendicular and mark \(B\) such that \(\begin{align} AB =1\end{align}\) unit with \(O\) as center and \(\begin{align} OB\end{align}\) as radius. Draw an arc to cut the number line at \(C\). Now \(C\) represents \(\begin{align}\sqrt 5 \end{align}\).

In \(\Delta OAB,\)

\[\begin{align}   O{{B}^{2}}&=O{{A}^{2}}+A{{B}^{2}} \\  &={{2}^{2}}+{{1}^{2}} \\  &=5 \\  \therefore OB &=\sqrt{5}=OC \\ \end{align}\]