Real Numbers Ex 1.4 Solutions
Question - 1 : - Show that the cube of a positive integer of the form 6q + r, q is aninteger and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.
Answer - 1 : -
Solution:
6q + r is a positive integer, where q is aninteger and r = 0, 1, 2, 3, 4, 5
Then, the positive integers are ofform 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.
Taking cube on L.H.S and R.H.S,
For 6q,
(6q)³ = 216 q³ = 6(36q)³ + 0
= 6m + 0, (where m is an integer =(36q)³)
For 6q+1,
(6q+1)³ = 216q³ + 108q2 + 18q+ 1
= 6(36q³ + 18q2 + 3q) + 1
= 6m + 1, (where m is an integer = 36q³ +18q2 + 3q)
For 6q+2,
(6q+2)³ = 216q³ + 216q2 + 72q+ 8
= 6(36q³ + 36q2 + 12q + 1) +2
= 6m + 2, (where m is an integer = 36q³ +36q2 + 12q + 1)
For 6q+3,
(6q+3)³ = 216q³ + 324q2 + 162q+ 27
= 6(36q³ + 54q2 + 27q + 4) + 3
= 6m + 3, (where m is an integer = 36q³ +54q2 + 27q + 4)
For 6q+4,
(6q+4)³ = 216q³ + 432q2 + 288q+ 64
= 6(36q³ + 72q2 + 48q + 10) +4
= 6m + 4, (where m is an integer = 36q³ +72q2 + 48q + 10)
For 6q+5,
(6q+5)³ = 216q³ + 540q2 + 450q+ 125
= 6(36q³ + 90q2 + 75q + 20) +5
= 6m + 5, (where m is an integer = 36q³ +90q2 + 75q + 20)
Hence, the cube of apositive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5is also of the form 6m + r.
Question - 2 : - Prove that one and only one out of n, n + 2and n + 4 is divisible by 3, where n is any positive integer.
Answer - 2 : -
Solution:
According to Euclid’s division Lemma,
Let the positive integer = n
And b=3
n =3q+r, where q is the quotient and r is theremainder
0
Therefore, n may be in the form of 3q, 3q+1,3q+2
When n=3q
n+2=3q+2
n+4=3q+4
Here n is only divisible by 3
When n = 3q+1
n+2=3q=3
n+4=3q+5
Here only n+2 is divisible by 3
When n=3q+2
n+2=3q+4
n+4=3q+2+4=3q+6
Here only n+4 is divisible by 3
So, we can conclude that one and only one outof n, n + 2 and n + 4 is divisible by 3.
Question - 3 : - Without actually performing the long division, state whether thefollowing rational numbers will have a terminating decimal expansion or anon-terminating repeating decimal expansion:
Answer - 3 : -
(i) 13/3125 (ii)17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2352) (vii) 129/(225775) (viii) 6/15 (ix) 35/50 (x)77/210
Question - 4 : - Write down the decimal expansions of those rational numbers in Question1 above which have terminating decimal expansions.
Answer - 4 : - 
Question - 5 : - The following real numbers have decimalexpansions as given below. In each case, decide whether they are rational ornot. If they are rational, and of the form, p q what can you say about theprime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000. . .
Answer - 5 : -
(i) 43.123456789
Since it has a terminating decimal expansion,it is a rational number in the form of p/q and q has factors of 2 and 5 only.
(ii) 0.120120012000120000. . .
Since, it has non-terminating and non-repeating decimal expansion, it is an irrational number.
Since it has non-terminating but repeatingdecimal expansion, it is a rational number in the form of p/q and q has factorsother than 2 and 5.