Number Systems (Chapter 1, Exercise 1.2) – Class 9 Maths Solutions) NCERT

Question 1: State whether the following statements are true or false. Justify your answers

i. Every Irrational Number is a Real number.

Answer: True.

Explanation:

We know that the irrational Numbers are the numbers that cannot be expressed in terms of \frac{p}{q}. However, the Real Numbers are the collection of all the numbers that can be represented in the number line. Therefore, the Real Numbers also includes the Irrational Numbers. Hence, it can be concluded that Every Irrational Number is a Real Number.

ii. Every point on the number line is of the form \sqrt {m}, where m is a Natural Number.

Answer: False

Explanation:

Let us consider two numbers as \frac{2}{3} and \frac{5}{11}. Both of the numbers are present in the number line.

If we express the above two numbers in terms of \sqrt {m}, then

\frac{2}{3}=\sqrt{\frac{4}{9}} and

\frac{5}{11}=\sqrt{\frac{25}{121}}

In the above expressions, the value of m is \frac{4}{9} and \frac{5}{121} respectively which are not Natural Numbers.

Therefore, it can be concluded that the given statement is false.

iii. Every Real Number is an Irrational Number.

Answer: False.

Explanation:

We know that the real numbers are the set of all the numbers that can be represented in the number line. It includes Natural Numbers, Whole Numbers, Integers, rational Numbers, Irrational Numbers etc. However, the Irrational Numbers are only those numbers that cannot be expressed in terms of \frac{p}{q}. Therefore, the real Numbers also contains the numbers that are not irrational.

Hence, Every Real Number is not an Irrational Number.

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.

Answer: No, the square root of all the positive integers are not irrational.

For example: \sqrt{16}=4

In the above example the square root of 16 is 4. 4 is a Rational Number as it can be expressed in term of \sqrt{{p}{q}}.

Therefore, square root of all positive integers are not irrational.

Question 3: Show how \sqrt{5} can be represented on the number line.

Answer:

Step 1: Draw two lines OA and AB of unit length such that AB\ bot OA as shown in the figure below and connect O and B.

Using Pythagoras Theorem,

(\overline {OB})^2=(\overline {OA})^2+(\overline {AB})^2

(\overline {OB})^2=1^2+1^2

(\overline {OB})^2=2

\overline {OB}=\sqrt{2}
show how root 5 can be represented on the number line

Step 2: Draw a line BC such that BC\bot OB and connect O and C.

(\overline {OC})^2=(\overline {OB})^2+(\overline {BC})^2

(\overline {OC})^2=(\sqrt 2)^2+1^2

(\overline {OC})^2=2+1=3

\overline {OC}=\sqrt{3}

Step 3: Similarly, draw a line CD such that CD\bot OC and connect O and D.

(\overline {OD})^2=(\overline {OC})^2+(\overline {CD})^2

(\overline {OD})^2=(\sqrt 3)^2+1^2

(\overline {OD})^2=3+1=4

\overline {OD}=\sqrt{4}

Step 4: Lastly, draw a line DE such that DE\bot OD and connect O and E.

(\overline {OE})^2=(\overline {OD})^2+(\overline {DE}^2

(\overline {OE})^2=(\sqrt 4)^2+1^2

(\overline {OE})^2=4+1=5

\overline {OE}=\sqrt{5}

Step 5: Draw an Arc whose centre is at O and radius is OE. The point in which the Arc cuts the Number Line is the point having value is \sqrt {5}.

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