Question

1. 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.

2. Show how √5 can be represented on the number line.

3. 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 OP1 of unit length. Draw a line segment P1P2 perpendicular to OP1 of unit length (see Fig. 1.9). Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in Fig. 1.9 :

Note : Refer the attachment for the figure

Continue of 3rd Question -
Constructing this manner, you can get the line segment Pn-1Pn by square root spiral drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …​

Answer 1

$\large\color{purple}\underline{\underline{{ \boxed {Solution \: 01 }}}}$

Statement (01): No, the square roots of all positive integers are not irrational.

Statement (02): Let us take a few examples whose square root of positive integers are rational numbers.

$\tt \: \sqrt{4} = 2 \\ \tt \: \sqrt{9} = 3$

Therefore by using statement (01) and statement (02) we can say that the square root of all positive integers includes both rational and irrational numbers.

$\large\color{purple}\underline{\underline{{ \boxed {Solution \: 02 }}}}$

1. Draw a number line and mark the center point as zero.
2. Mark right side of the zero as (1) and the left side as (-1).
3. We won’t be considering (-1) for our purpose.
4. With 2 units as length draw a line from (1) such that it is perpendicular to the line.
5. Now join the point (0) and the end of new line of 2 units length.
6. A right angled triangle is constructed.
7. Now let us name the triangle as ABC such that AB is the height (perpendicular), BC is the base of triangle and AC is the hypotenuse of the right angled triangle ABC.
8. Now length of hypotenuse, i.e., AC can be found by applying Pythagoras theorem to the triangle ABC.

$\tt \: AC {}^{2} = AB {}^{2} + BC {}^{2} \\ \tt \: ⟹ AC {}^{2} = 22+ 12 \\ \tt \: ⟹ AC {}^{2} = 4 + 1 \\ \tt \: ⟹ AC {}^{2} = 5 \\ \tt \: ⟹ AC = √5$

1. Now with AC as radius and C as the centre cut an arc on the same number line and name the point as D.
2. Since AC is the radius of the arc and hence, CD will also be the radius of the arc whose length is √5.
3. Hence, D is the representation of √5 on the number line.

$\large\color{purple}\underline{\underline{{ \boxed {Solution \: 03 }}}}$

$\tt \: Start \: with \: a \: point \: O \\ \tt \:draw \: a \: line \: segment \: OP1, of \: unit \: lengths \\ \tt \:Draw \: a \: line \: segment \: P_1, P_2 \: perpendicular \: to \: OP_1 \: of \: unit \: length\\ \tt \: (see \: attachment \: 02).\\ \tt \:Now, draw \: a \: line \: segment \: P_2P_3 \: perpendicular \: to \: OP_2. \\ \tt \:Draw \: a \: line \: segment \: P_3P_4 \: perpendicular \: to \: OP_3. \\ \tt \:Continuing \: in \: this \: manner, you \: can \: get \: the \: line \: segment \: Pn_1\\ \\ \tt \: Hence,you \: received \: the \: spiral.$

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$\sf \: @WildCat7083$