Question 1:
Classify the following numbers as rational or irrational:\(\)
(i)\(\displaystyle {2-\sqrt 5}\)
(ii)\((3+\sqrt{23})-\sqrt{23}\)
(iii)\(\displaystyle{ 2\sqrt 7\over 7\sqrt 7}\)
(iv) \( \dfrac{1}{\sqrt 2}\)
(v) \(2\pi\)
Solution 1:
(i)\(2-\sqrt 5\)
\(\sqrt 5 = 2.236\dots\) ,is a non-terminating and non-repeating irrational number.
\(2 \sqrt 5 = 2- 2.236\dots\)2√5 ( By substituting the value of \(\sqrt 5\) )
\(= –0.236\dots \), which is also an irrational number.
Therefore, \(2-\sqrt 5\) is an irrational number.
Therefore, \(2-\sqrt 5\) is an irrational number.
(ii)\((3+\sqrt {23})-\sqrt {23}\)
\((3+\sqrt {23})-\sqrt {23}=3+√\sqrt {23}-\sqrt {23}=3\) which is in the form of \(\dfrac{p}{q}\) and it is a rational number.
Therefore, we conclude that \((3+\sqrt {23})-\sqrt {23}\) is a rational number.
(iii)\(\displaystyle{ 2\sqrt 7\over 7\sqrt 7}=\displaystyle{ 2\over 7}\)
Which is in the form of \(\dfrac{p}{q}\) and it is a rational number.
Therefore, we conclude that \(\displaystyle{ 2\sqrt 7\over 7\sqrt 7}\) is a rational number.(iv) \( \dfrac{1}{\sqrt 2}\)
\({\sqrt 2}=1.414\dots\) , is a non-terminating and non-repeating irrational number.
Therefore, we conclude that
\( \dfrac{1}{\sqrt 2}\) is an irrational number.
(v) \(2\pi=3.1415\dots\), which is an irrational number.
Observe, Rational × Irrational = Irrational
Therefore \(2\pi\) will also be an irrational number.
Simplify each of the following expressions:
(i)(3+√3)(2+√2)
(ii)(3+√3)(3-√3)
(iii)(√5+√2))2
(iv)(√5-√2)(√5+√2)
Solution 2:
(i)\((3+√3)(2+√2)\)
\((3+√3)(2+√2)=3(2+√2)+√3(2+√2)\)
\(=6+3√2+2√3+√6\)
(ii)\((3+√3)(3-√3)\)
Using the identity
\((a+b)(a-b)=a^2-b^2\)
\( \therefore (3+√3)(3-√3)=9-3=6\)
(iii)\((√5+√2)^2=√5^2+√2^2+2.√5.√2\)
\(=5+2√10+2\)
\(=7+2√10\)
(iv)\((√5-√2)(√5+√2)\)
Using the identity
(a+b)(a-b)=a2-b2
\(\therefore\) (√5-√2)(√5+√2) = 5 - 2 = 3
Question 4. Represent √9.3 on the number line.
Solution 4:
•Mark the distance 9.3 units from a fixed-point A on a given line to obtain a point B
such that AB = 9.3 units.
•From B mark a distance of 1 unit and mark the point as C.
•Find the mid-point of AC and mark the point as O.
•Draw a semi-circle with centre O and radius OC = 5.15 units
\((AC = AB + BC = 9.3 + 1=10.3\). Therefore \(OC = \dfrac{AC}{2} = \dfrac{10.3}{2}=5.15\).
•Draw a line perpendicular at B and draw an arc with centre \(B\) and let meet at semicircle AC at D
• BE = BD = \(\sqrt{9.3}\) (Radius of an arc DE).
Rationalize the denominators of the following:
i.\(\dfrac{1}{\sqrt{7}}\)
ii.\(\dfrac{1}{\sqrt{7}-\sqrt 6}\)
iii.\(\dfrac{1}{\sqrt{5}+√2}\)
iv.\(\dfrac{1}{\sqrt{7}-2}\)
Solution 5:
(i)\(\dfrac{1}{\sqrt 7}\)
Multiply and divide by \(\sqrt 7\)
\(\dfrac{1}{\sqrt 7}=\dfrac{1 \times \sqrt 7}{\sqrt 7 \times \sqrt 7}=\dfrac{\sqrt 7}{7}\)
(ii)\(\dfrac{1}{\sqrt 7+\sqrt 6}\)
Multiply and divide by \({\sqrt 7+\sqrt 6}\),We get
\(\dfrac{1×(\sqrt 7-\sqrt 6)}{(\sqrt 7+\sqrt 6) (\sqrt 7-\sqrt 6)}\)
=\(\dfrac{\sqrt 7-\sqrt 6}{7-6}\)
=\(\sqrt 7-\sqrt 6\)
iii.\(\dfrac{1}{\sqrt 5+\sqrt 2}\)
Multiply and divide by \({\sqrt 5-\sqrt 2}\),We get
\(1×\dfrac{\sqrt 5-\sqrt 2}{(\sqrt 5-\sqrt 2) ×(\sqrt 5+\sqrt 2)}\)
=\(\dfrac{\sqrt 5-\sqrt 2}{5-2}\)
=\(\dfrac{\sqrt 5-\sqrt 2}{3}\)
iv.\(\dfrac{1}{\sqrt 7-2}\)
Multiply and divide by \(\sqrt 7-2\), We get
\( \dfrac{1 \times \sqrt 7+2}{(\sqrt 7-2) ×(\sqrt 7+2)}\)
=\(\dfrac{\sqrt 7+2}{7-4}\)
=\(\dfrac{\sqrt 7+2}{3}\)
