Telangana SCERT Solution Class IX (9) Math Chapter 15 Proofs in Mathematics Exercise 15.4
1.)
(i) No, humans have more that only blue color of eyes
(ii) x+7 = 18, Yes
X = 18 -7
=11
X can only be 11 in this case.
(iii) No, today might be Sunday or 6 other days of the week.
(iv) Yes, it is true for all numbers.
(v) No, It be any time of the day or night.
2.)
(i) False, all side of square are equal but opposite sides of rectangular are equal.
(iii) For, n= 11, 2n2 +11= 253 which is not a prime number.
(iv) True.
(v) False , A rhombus has equal side but it is not a square.
3.)
Let,
2k be an even number
∴ 2k +1 is an odd number, where k is an integer.
Now,
Let,
2a +1 & 2b +1 are two odd numbers
a, b are integers
∴ (2a + 1) +( 2b +1) = 2a +2b +2
= 2 (a+b+1).
Now,
a+b+1 is in itself an integer since sum of two integers and constant gives an integers.
: (a+b+1) = k
: (2a+1) + (2b +1) = 2k
Now,
2k is an even number hence proved as shown above.
4.)
Let,
2k be an even number.
When k is an integer.
Let,
2a , 2b, be two even number
Where a, b an integers.
Now,
2a * 2b = 4 ab
= 4 (a*b)
=2* 2(a*b)
Now,
Multiplication of two integer is where one integer so we can write,
2a * 2b = 2 * 2k
When 2k is even number.
Now, twice of any even number is bound to be even.
Hence,
Product of two even numbers is even.
5.)
Let,
2k be an even number.
2k +1 be an odd number where k is any integer.
Let,
2a +1 be an odd numbers.
∴ (2a +1)2 = 4a2 + 2 * 2a * 1 + 1
= 4a2 +4a + 1
= 4a x a + 4a +1
∴ Now,
Product of two integer is always an integer.
∴ ( 2a +1)2 = 4a + 4a +1
= 8a +1
= 4 * 2a +1
= 4 * 2k +1
Here 2a = 2k
Sine a, k are both integers.
We know,
When an even number is multiplied with any number the result is even.
∴ 4 * 2k is an even number.
∴ ( 2a +1)2 = 4 *2k +1
= 2k +1 [∵ 2k is an even number]
∴ 2k +1 is an odd number.
∴ Square of any odd No. x is odd proved.
(ii)
Now,
7 x 11 x 13 = 1001
abc be a 3 digit number
∴ abc x 1001 = abc abc
∴ The six digit number abc abc is divisible by 7, 11, 213.
