Try these, Recap Exercise, Student Activities, Think and answer

Solution: 

(i) −7/3 ÷ 5 = [−7/3] ÷ [5/1] = [−7/3] × [1/5] = −7/15

(ii) 5 ÷ (−7/3) = [5/1] × [3/−7] = 15/−7 = −2 (1/7)

(iii) −7/3 ÷ 35/62 = [−7/3] × [6/35] = −2/5

Try this

The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.

Solution:

Let 0 and 1/2 be two rational numbers 0 – 1/2 is a rational number

∴ Closure property for subtraction holds for rational numbers.

But consider the two rational number 5/2 and 0.

5/2  ÷ 0 = 5 / [2 × 0] = 5 / 0

Here denominator = 0 and it is not a rational number.

∴ Closure property is not true for division of rational numbers.

Solution:

Is 3/5 – 7/8 = 7/8 – 3/5 ?

LHS = 3/5 + 7/8 = [ (3 × 8) – (7 × 5) ] / 40 = [24 – 35] / 40 = −11/40

RHS = 7/8 − 3/5 = [ (7 × 5) – (3 × 8) ] / 40 = [35 – 24] / 40 = 11/40

LHS  ≠  RHS

∴ 3/5  ÷ 7/8  ≠ 7/8 – 3/5

∴ Subtraction of rational numbers is not commutative.

(ii) Is 3/5 ÷ 7/8 = 7/8 – 5/3 ?

LHS = 3/5  ÷  7/8 = 3/5  × 8/7 = 24/35

RHS = 7/8  ÷  5/3 = 7/8  × 3/5 = 21/40

∴ LHS  ≠  RHS

∴ 3/5  ÷ 7/8  ≠ 7/8 ÷ 5/3

∴ Commutative property not hold good for division of rational numbers.

Try this

 Check whether associative property holds for subtraction and division.

Solution:

Consider the rational numbers 2/3, 1/2  and 3/4

To verify (2/3 – 1/2) – 3/4 = 2/3 – (1/2 – 3/4)

LHS = (2/3 – 1/2) – 3/4 = ( [(2 × 2) – (1 × 3)] / 6 ) – (3/4)

= ([4 – 3]/6) – 3/4 = 1/6 – 3/4 = [(1 × 2) – (3 × 3)] / 12 = [2 – 9] / 12 = – 7 / 12

RHS = 2/3 – (1/2 – 3/4) = 2/3 – ( [2 – 3] / 4) =  ( 2/3 −  (−1/4) )

= 2/3 + 1/4  = [ (2 × 4) + ( 1 × 3) ] / 12 = [ 8 + 3] / 12 = 11 / 12

LHS  ≠  RHS

∴ ( 2/3 – 1/2) – 3/4  ≠ 2/3  − (1/2 − 3/4)

∴ Associative property not holds for subtraction of rational numbers

Also to verify (2/3 ÷ 1/2) ÷ 3/4 = 2/3 ÷ (1/2 ÷ 3/4)

LHS = (2/3  ÷  1/2) ÷  3/4 = (2/3 × 2/1) ÷ 3/4

= 4/3  ÷ 3/4 = 4/3 × 4/3 = 16 / 9

RHS = 2/3 ÷ (1/2 ÷ 3/4) = 2/3 ÷ ( 1/2 × 4/3) =  2/3 ÷  ( 2/3 )

= 2/3 × 3/2 = 1

LHS  ≠  RHS

ie. (2/3 ÷ 1/2) ÷ 3/4 ≠ 2/3 ÷ ( 1/2 ÷ 3/4)

∴ Associative property does not hold for division of rational numbers.

Think

Observe that, 

Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.

Solution:

1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/ 6.7 + 1/7.8 = 7 / 8

Think

1. Is the square of a prime number, prime?

Solution:

No, the square of a prime number ‘P’ has at least 3 divisors 1, P and P2. But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.

2. Will the sum of two perfect squares always be a perfect square? What about their difference and their product?

Solution:

The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. But the product of two perfect square is a perfect square.

Try these

• Which among 256, 576, 960, 1025, 4096 are perfect square numbers? (Hint: Try to extend the table of squares already seen).

Solution:

256 = 162

576 = 242

4096 = 642

∴ 256, 576 and 4096 are perfect squares

• One can judge just by look that each of the following numbers 82, 113, 1972, 2057, 8888, 24353 is not a perfect square. Explain why?

Solution:

Because the unit digit of a perfect square will be 0, 1, 4, 5, 6, 9. But the given numbers have unit digits 2, 3, 7, 8. So they are not perfect squares.

Think

Consider the claim: “Between the squares of the consecutive numbers n and (n+1), there are 2n non-square numbers”. Can it be true? How many non-square numbers are there between 2500 and 2601? Verify the claim.

Solution:

 If n = 50 ⇒ n2 = 502 = 2500

n + 1 = 51 ⇒ (n + 1)2 = 512 = 2601

Non−square numbers of 2500 and 2601 = 100 = 2 × 50 = 2 n

So it is true that between the classes of successive numbers n, n + 1, there are non−square numbers of 2n.

Think

In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?

Solution:

108  = 2 × 2 × 3 × 3 = 22 × 32 × 3

If we multiply the factors by 2, then we get

22 × 32 × 3 × 3 ⇒ 22 × 32 × 32 = (2 × 3 × 3)2

Which is perfect square.

∴  Again if we divide by 3 then we get 22 × 32 ⇒ (2 × 3)2, a perfect square.

∴  We have to multiply or divide 108 by 3 to get a perfect square.

Try these

Find the square root by long division method.

1. 400 2. 1764 3. 9801

Solution:

Try these

Without calculating the square root, guess the number of digits in the square root of the following numbers:

1. 14400 2. 390625 3. 100000000

Solution:

(1) √14400 = √[144×100] = √l44 × √100

= 12 × 10 = 120.

(2) √390625 = √ [25 × 25 × 25 × 25]

= √[25×25] × √[25×25] = 25 × 25 = 625

(3) √100000000 = √[10000×10000]

= √10000 × √10000

= 100 × 100= 10,000

Try these

Find the s quare root of

1. 5.4756, 2. 19.36, 3. 116.64

Activity

Attempt to prepare a table of square root problems as in the above case to show that if a and b are two perfect square numbers, then √[a/b] = √a/√b (b ≠ 0). We can use this idea to compute certain square-root problems easily.

Try these

Using this method, find the square root of the numbers 1.2321 and 11.9025.

 Solution:

(i) √1.2321 = √[12321/10000] = 111/100 = 1.11

(ii) √11.9025 = √119025 / √10000 = 345/100 = 3.45

Try these

Write the numbers in ascending order.

1. 4, √14 , 5

2. 7, √65 , 8

Solution:

(i) 4, √14, 5

Squaring all the numbers we get 42, (√l4)2, 52 ⇒ 16, 14, 25

∴ Ascending order : 14, 16, 25

Ascending order : √14, 4, 5

(ii) 7, √65, 8

Squaring 7, √65 and 8 we get 72, (√65)2 , 82 ⇒ 49, 65, 64

Ascending order : 49, 64, 65

Ascending order : 7, 8, √65

Try these

Find the ones digit in the cubes of each of the following numbers.

1. 12

2. 27

3. 38

4. 53

5. 71

6. 84

Solution:

(i) 12

12 ends with 2, so its cube ends with 8 i.e, ones digit in 123 is 8.

(2) 27

27 ends with 7, so its cube end with 3. i.e., ones digit in 273 is 3.

(3) 38

38 ends with 8, so its cube ends with 2 i.e, ones digit in 383 is 2.

(4) 53

53 ends with 3, so its cube ends with 7. i.e, ones digit in 533 is 7.

(5) 71

71 ends with 1, so its cube ends with 1. i.e, ones digit in 713 is 1

(6) 84

84 ends with 4, so its cube ends with 4. i.e, ones digit in 843 is 4.

Think

In this question, if the word ‘multiplied’ is replaced by the word ‘divided’, how will the solution vary?

Try these

Expand the following numbers using exponents:

1.8120 2. 20305 3. 3652.01 4. 9426.521

Solution:

(i) 8120 = (8 × 1000) + (1 × 100) + (2 × 10) + 0 × l

 = (8 × 103) + (1 × 102) + (2 × 101)

(2) 20305 = (2 × 10000) + (0 × 1000) + (3 × 100) + ( 0 × 10) + (5 × 1)

 = (2 × 104) + (3 × 102) + 5 

(3) 3652.01 = 3000 + 600 + 50 + 2 + 0/10 + 1/100

= (3 × 1000) + (6 × 100) + (5 × 10) + (2 × 1) + 1 × 1/100)

 = (3 × 103) + (6 × 102) + (5 × 101) + 2 + (1 × 10−2)

(4) 9426.521 = (9 × 1000) + (4 × 100) + (2 × 10) + (6 × 1) + (5/10) + (2/100) + (1/1000)

= (9 × 103) + (4 × 102) + (2 × 101) + 6 + (5 × 10−1) + (2 × 10−2) + (1 × 10−3)

Try these

Verify the following rules (as we did above). Here, a,b are non-zero integers and m, n are any integers.

1. Product of same powers to power of product rule: a m × b m = (ab)m

2. Quotient of same powers to power of quotient rule: an/bm = (a/b) m

3. Zero exponent rule: a0 = 1.

Verification:

Let a = 2; b = 3; m = 2

1. am × bm = 22 × 32 = 4 × 9 = 36 = (2 × 3)2

2. am / bm = 22 / 32 = 4 / 9 = (2 / 3)2

3. a0 = 20 = 1.

Try these

1. Write in standard form: Mass of planet Uranus is 8.68 × 1025 kg.

Solution:

Mass of Planet Uranus = 86800000000000000000000000 kg

2. Write in scientific notation: (i) 0.000012005 (ii) 4312.345 (iii) 0.10524 (iv)The distance between the Sun and the planet Saturn 1.4335×1012 miles.

Solution:

(i)  0.000012005 = 1.2005 × 10−5

(ii)  4312.345 = 4.312345 × 103

(iii)  0.l 524 = 1.0524 × 10−1

(iv) The distance between Sun and the planet Saturn is 1.4335 × 1012 miles