1. State whether the following statements are true or
false. Justify your answers.
(i) 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.
(iii) Every real number is an
irrational number.
solution :
(i) Every irrational number is a real number.
TRUE
Irrational numbers : A number is said to be irrational, if it cannot be written in the ` \p/q\ `, where p and q are integers and q ≠ 0.
Example : Irrational numbers= ` \sqrt{6}\ `, ` \pi\ `, ` \sqrt{7}\ `, ` \7+\sqrt{11}\ `, 9.123456745367..., 0.505005005000................
Real numbers : The collection of both rational and irrational numbers are known as real numbers.
Example : Real numbers= ` \sqrt{8}\ `,` \sqrt{7}\ `, 0.102, 1, ` \sqrt{5}\ `, 5, 200, ......................
Or, we can say that every irrational numbers is real numbers.
Hence, 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.
FALSE
Hence, No negative number can be the square root of any natural number.
(iii) Every real number is an irrational number.
FALSE
Real numbers : The collection of both rational and irrational numbers are known as real numbers.
Example : Real numbers = ` \sqrt{8}\ `, ` \sqrt{7}\ `, 0.102, 1, ` \sqrt{5}\ `, 5, 200, ......................
Irrational numbers : A number is said to be irrational, if it cannot be written in the ` \frac{p} {q}\ `, where p and q are integers and q ≠ 0.
Example : Irrational numbers = ` \sqrt{6}\ `, ` \pi\ `, ` \sqrt{7}\ `, ` \7+\sqrt{11}\ `, 9.123456745367..., 0.505005005000................
Or, we can say that every real numbers is not irrational numbers.
Hence, Every real number is not irrational number.
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.
solution :
No, the square roots of all positive integers are not irrational.
Example : ` \sqrt{9}= 3\ ` is a rational
number.
`
\sqrt{16}=4\ ` is a rational number.
Hence, the square roots of positive integers are not irrational.
3. Show how ` \sqrt{5}\ ` can be represented on the number line.
solution :
Step 1: Let draw a line of 5 unit.
Step 2: Let line AB be of 2 unit on a number line.
Step 3: At B, draw a perpendicular line BC of length 1 unit, and join AC.
Step 4: Now, ABC is a right angle triangle.
Applying Pythagoras theorem,
` \AB^2+BC^2=AC^2\ `
` \2^2+1^2=AC^2\ `
` \4+1=AC^2\ `
` \AC^2=5\ `
` \AC=\sqrt{5}\ `
Thus, AC is a line of length `\sqrt{5}\ ` unit.
Step 5: Taking AC as a radius and A as a center draw an arc touching the number line.
Thus, ` \sqrt{5}\ ` is represented on the number line as shown in the figure.
4. 
Classroom activity (Constructing the ‘square root spiral’): Take a large sheet of paper and construct the ‘square root spiral’ in
the following fashion. Start with a point ` \O\ ` and draw a line segment `
\OP_1\ ` of unit length. Draw a line segment ` \P_1P_2\ ` perpendicular to
` \OP_1\ ` of unit length (see Fig. 1.9). Now draw a line segment `
\P_2P_3\ ` perpendicular to ` \OP_2\ `. Then draw a line segment `
\P_3P_4\ ` perpendicular to ` \OP_3\ `. Continuing in this manner, you can
get the line segment ` \P_{n-1}P_n\ ` by drawing a line segment of unit
length perpendicular to ` \OP_{n-1}\ `. In this manner, you will have
created the points ` \P_2, P_3,...., P_n,....,\ ` and joined them to
create a beautiful spiral depicting ` \sqrt{2}, \sqrt{3}, \sqrt{4}, .....\
`
solution :
Step 1: Mark a point O on the paper.
Step 2: O will be the center of the square root spiral.
Step 3: From O, draw a straight line, OA, of 1cm.
Step 4: From A, draw a perpendicular line, AB, of 1 cm and join OB.
Step 5: OB will be of ` \sqrt{2}\ `.
Step 6: Now, from B, draw a perpendicular line of 1 cm and mark the end point C and join OC.
Step 7: OC will be of ` \sqrt{3}\ `.
Step 8: Repeat the steps to draw ` \sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}, \sqrt{10},........\ `
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