Number System: 9 Math
Solution of NCERT 1.5 (class ninth mathematics): 9th math
Question (1) Classify the following numbers as rational or irrational
(i) `2-sqrt5`
Answer : Irrational.
Explanation: Since, one part of the given expression i.e. `sqrt5` is an irrational number, thus given number is an irrational.
(ii) `(3+sqrt(23))-sqrt(23)`
Solution
Given, `(3+sqrt(23))-sqrt(23)`
`=3+\cancel(sqrt(23))- \cancel(sqrt(23))`
=3
Thus, Given number is rational. Answer
(iii) `(2sqrt7)/(2sqrt7)`
Solution
Given, `(2sqrt7)/(2sqrt7)`
= 0
Thus, given number is rational. Answer
(iv) `1/sqrt2`
Solution
Given, `1/sqrt2`
Since, `sqrt2` is an irrational number.
Thus, given number is irrational. Answer
(v) `2pi`
Solution
Given, `2pi`
Since `pi` is an irrational number.
Thus, given number is irrational. Answer
Question (2) Simplify each of the following expressions:
(i) `(3+sqrt3)(2+sqrt2)`
Solution
Given, `(3+sqrt3)(2+sqrt2)`
`=3xx2+3sqrt2+2sqrt3+sqrt3xxsqrt2`
`=6+3sqrt2+2sqrt3+sqrt6` Answer
(ii) `(3+sqrt3)(3-sqrt3)`
Solution
Given, `(3+sqrt3)(3-sqrt3)`
`=3^2-(sqrt3)^2`
[∵ (a+b)(a–b)=a2– b2]
=9 – 3
= 6 Answer
(iii) `(sqrt5+sqrt2)^2`
Solution
Given, `(sqrt5+sqrt2)^2`
`=(sqrt5)^2+(sqrt2)^2+2sqrt5\ sqrt2`
`=5+2+2sqrt10`
`=7+2sqrt10`Answer
(iv) `(sqrt5-sqrt2)(sqrt5+sqrt2)`
Solution
Given, `(sqrt5-sqrt2)(sqrt5+sqrt2)`
`=(sqrt5)^2-(sqrt2)^2`
= 5 – 2
= 3 Answer
Question (3) Recall, `pi` is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is `pi = c/d`. This seems to contradict the fact that `pi` is irrational. How will you resolve this contradiction?
Solution
When measuring with scale or a tape the findings of any rational numbers are only to an approximation. This is the cause that while measuring the circumference of a circle, i.e. `pi` only approximate value is taken. This is the cause that it appears that we take `pi` as rational but actually `pi` is irrational number.
Thus, there is no contradiction.
Question (4) Represent `sqrt(9.3)` on the number line.
Solution:
To represent `sqrt(9.3)` do the following steps
(a) Draw a horizontal line and cut a segment equal to 9.3 unit, say CA = 9.3 unit
(b) Mark AE equal to 1 unit
(c) Find the centre of line segment CE
(d) Now, draw a semicircle on line segment CE
(e) Draw a perpendicular at A, which cut the semicircle at point D
(f) Now, this AD will be equal to `sqrt(9.3)`
(g) Measure AD and cut equal to this length on number line, i.e. horizontal line taking from A which becomes AE
(h) Now, AD = AE `=sqrt(9.3)`
This AE is the representation of `sqrt(9.3)` on the number line
Proof:
It is clear from the figure drawn above that
CA = 9.3 unit
AB = 1 unit
∴ diameter of the circle = 9.3 + 1 = 10.3 unit
∴ radius, OB = OC = OD `=(10.3)/2` unit
Now, OA = OB – AB
`=>OA=(10.3)/2\-1`
`=>OA=(10.3-2)/2`
`=>OA=(8.3)/2`
Now, in triangle, ODA
∠ A = 900
∴ According to Pythagoras Theorem
OD2 = OA2 + AD2
⇒ AD2 = OD2 – OA2
`=>AD^2 =(( 10.3)/2)^2-((8.3)/2)^2`
`=>AD^2=(106.09)/4-(68.89)/4`
`=>AD^2=(106.09-68.89)/4`
`=>AD^2=(37.2)/4=9.3`
`:. AD= sqrt(9.3)`
Hence, AD = `sqrt(9.3)` proved
Question (5) Rationalize the denominators of the following
(i) `1/sqrt7`
Solution:
Given, `1/sqrt7`
After multiplying with `sqrt7/sqrt7` we get
`=1/sqrt7xxsqrt7/sqrt7`
`=sqrt7/7` Answer
(ii) `1/(sqrt7-sqrt6)`
Solution:
Given, `1/(sqrt7-sqrt6)`
After multiplying with `(sqrt7+sqrt6)/ (sqrt7+sqrt6)` we get
`=1/(sqrt7-sqrt6)xx(sqrt7+sqrt6)/ (sqrt7+sqrt6)`
`=(sqrt7+sqrt6)/ ((sqrt7-sqrt6) (sqrt7+sqrt6))`
Now, after applying (a + b) (a – b) = a2 – b2, we get
`(sqrt7+sqrt6)/((sqrt7)^2-(sqrt6)^2)`
`=(sqrt7+sqrt6)/(7-6)`
`=(sqrt7+sqrt6)/1`
`=(sqrt7+sqrt6)` Answer
(iii) `1/(sqrt5+sqrt2)`
Solution:
Given, `1/(sqrt5+sqrt2)`
After multiplying with `(sqrt5-sqrt2)/ (sqrt5-sqrt2)` we get
`1/(sqrt5+sqrt2)xx(sqrt5-sqrt2)/ (sqrt5-sqrt2)`
`=(sqrt5-sqrt2)/( (sqrt5+sqrt2) (sqrt5-sqrt2))`
Now, after applying (a + b) (a – b) = a2 – b2, we get
`=(sqrt5-sqrt2)/((sqrt5)^2-(sqrt2)^2)`
`=(sqrt5-sqrt2)/(5-2)`
`=(sqrt5-sqrt2)/3` Answer
(iv) `1/(sqrt7-2)`
Solution:
Given, `1/(sqrt7-2)`
After multiplying with `(sqrt7+2)/ (sqrt7+2)` we get
`1/(sqrt7-2)xx(sqrt7+2)/ (sqrt7+2)`
`=(sqrt7+2)/( (sqrt7-2) (sqrt7+2))`
Now, after applying (a + b) (a – b) = a2 – b2, we get
`=(sqrt7+2)/((sqrt7)^2-2^2)`
`=(sqrt7+2)/(7-4)`
`=(sqrt7+2)/3` Answer
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