AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.3

Question 1.

Take any three consecutive odd numbers and find their product, for example 1 × 3 × 5 = 15;

3 × 5 × 7 = 105: 5 × 7 × 9 = ……………

ii) Take any three consecutive even numbers and add them, say,

2 + 4 + 6 = 12; 4 + 6 + 8 = 18:

6 + 8 + 10 = 24; 8 + 10 + 12 = 30 ….

so on. Is there any pattern you can guess in these sums ? What can von conjecture about them ?

Solution:

i) 1 × 3 × 5 = 15

3 × 5× 7 = 105

5 × 7 × 9 = 315

7 × 9 × 11 = 693

- The product of any three consecutive odd numbers is odd.
- The product of any three consecutive odd numbers is divisible by ’3′.
- 2 + 4 + 6 = 12; 4 + 6 + 8= 18;

6 + 8 + 10 = 24; 8 + 10 + 12 = 30 - The sum of any three consecutive even numbers is even.
- The sum of any three consecutive even numbers is divisible by 6
- The sum of any three consecutive even numbers is a multiple of 6.

Question 2.

Go back to Pascal’s triangle.

Line-1: 1=11°

Line-2: 11 = 11^{1}

Line-3 : 121 = 11^{2}

Make a conjecture about line – 4 and line – 5.

Does your conjecture hold ? Does your conjecture hold for line – 6 too ?

Solution:

Line-4 : 1331 = 11^{3}

Line-5 : 14641 = 11^{4}

Line – 6 : 11^{5}

∴ Line – n = 11^{n-1}

Yes the conjecture holds good for line – 6 too.

Question 3.

Look at the following pattern.

i) 28 = 2^{2} × 7^{1};

Total number of factors

(2+ 1)(1 + 1) = 3 × 2 = 6

28 is divisible by 6 factors i.e.,

1, 2, 4, 7, 14, 28.

ii) 30 = 2^{1} × 3^{1} × 5^{1}, Total number of .

factors (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 = 8

30 Is divisible by 8 factors i.e., 1, 2, 3, 5, 6, 10, 15 and 30

Find the pattern.

[Hint : Product of every prime base exponent +1)

Solution:

24 = 2^{3} × 3^{1}

24 has (3+1) (1 + 1) = 4 × 2 = 8 factors

[1, 2, 3, 4, 6, 8, 12 and 24]

36 = 2^{2} × 3^{2}

Number of factors = (2 + 1) (2 + 1)

3 × 3 = 9 [ 1, 2, 3. 4, 6, 9, 12, 18 and 36]

If N = a^{p}. b^{q} . c^{r}…….. where N is a natural number.

a. b, c … are primes and p, q, r ……. are positive integers then, the number of factors of N =(p- 1)(q+ 1)(r + 1)

Question 4.

Look at the following pattern :

1^{2} = 1

11^{2} = 121

111^{2} = 12321;

1111^{2} = 1234321

11111^{2} = 123454321

Make a conjecture about each of the following

111111^{2} =

1111111^{2} =

Check if your conjecture is true.

Solution:

111111^{2} = 12345654321

1111111^{2} = 1234567654321

(111………. n-times)^{2} = (123 … (n- 1) n (n – 1) (n – 2) 1)

The conjecture is true.

Question 5.

List five axioms (postulates) used in text book.

Solution:

- Things which are equal to the same things are equal to one another.
- If equals are added to equals, the sums are equal.
- If equals are subtracted from equals, the differences are equal.
- When a pair of parallel lines are in-tersected by a transversal, the pairs of corresponding angles are equal.
- There passes infinitely many lines through a given point.

Question 6.

In a polynomial p(x) = x^{2} + x + 41 put different values of x and find p(x). Can you conclude after putting different values of x that p(x) is prime for all. Is ‘x’ an element of N ? Put x = 41 in p(x). Now what do you find ?

Solution:

p(x) = x^{2} + x + 41

p(0) = 0^{2} + 0 + 41 = 41 – is a prime

p(1) = 1^{2} + 1 + 41 = 43 – is a prime

p(2) = 2^{2} + 2 + 41 = 47 – is a prime

p(3) = 3^{2} + 3 + 41 = 53 – is a prime

p(41) = 41^{2} + 41 + 41

= 41(41 + 1 + 1)

= 41 x 43 is not a prime.

∴ p(x) = x^{2}+ x + 41 is not a prime for all x.

∴ The conjecture “p(x) = x^{2} + x + 41 is a prime” is false.