Saturday, February 23, 2013
X mathematics solutions chapter 1 exercise 1.3
EXERCISE 1.3
1. Prove that √5 is irrational.
Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b^2 = a^2 ……………(1)
Therefore, 5 divides a^2 and according to theorem of rational number, for any prime number p which is divides a^2 then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
and plug the value of a in equation (1) we get
5b^2 = (5c)^2
5b^2 = 25c^2
Divide by 25 we get
b^2/5 = c^2
again using same theorem we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it is contradicting .
Hence √5 is a non rational number
2. Prove that 3 + 2√5 is irrational.
Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2 we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradict the fact
Hence result is 3 + 2√5 is a irrational number
3. Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2
(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a√2)/b
Now multiply by b
b = a√2
divide by a we get
b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict
Hence result is 1/√2 is a irrational number
(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get
√5) =a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it is contradict
Hence result is 7√5 is a irrational number.
(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict
Hence result is 6 + √2 is a irrational number
real numbers class 10 cbse
X mathematics solutions chapter 1 exercise 1.4
1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal
expansion:
(i)13/3125 (ii)17/8 (iii)64/455 (iv)15/1600 (v)29/343 (vi)23/2^3*5^2 (vii)129/2^2* 5^7*7^5 (viii)6/15 (ix)35/50 (x)77/210
(i) 13/3125
Factorize the denominator we get
3125 =5 x 5 x 5 x 5 x 5 = 5^5
So denominator is in form of 5^m so 13/3125 is terminating .
(ii) 17/8
Factorize the denominator we get
8 =2 x 2 x 2 = 2^3
So denominator is in form of 2^n so 17/8 is terminating .
(iii)64/455
Factorize the denominator we get
455 =5 x 7 x 13
There are 7 and 13 also in denominator so denominator is not in form of 2^n*5^m . hence 64/455 is not terminating.
(iv)15/1600
Factorize the denominator we get
1600 =2 x 2 x 2 x2 x 2 x 2 x 5 x 5 = 2^6 x 5^2
so denominator is in form of 2^n x5^m
Hence 15/1600 is terminating.
(v) 29/343
Factorize the denominator we get
343 = 7 x 7 x 7 = 7^3
There are 7 also in denominator so denominator is not in form of 2^nx5^m
Hence it is none - terminating.
(vi)23/(2^3 x 5^2)
Denominator is in form of 2^n x 5^m
Hence it is terminating.
(vii) 129/(2^2 x 5^7 x 7^5 )
Denominator has 7 in denominator so denominator is not in form of 2^n x 5^n
Hence it is none terminating.
(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5^m so it is terminating.
(ix) 35/50 divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get
10=2 x 5
So denominator is in form of 2^n x5^m so it is terminating
(x) 77/210.
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get
30=2 x 3 x 5
Denominator has 3 also in denominator so denominator is not in form of 2^n x 5^n
Hence it is none terminating.
1. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
 |
| terminating and non terminating |
3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p , q you say about the prime factors of q?
(i) 43.123456789
it has certain number of digits so they can be represented in form of p/q . Hence they are rational number
As they have certain number of digit and the number which has certain number of digits is always terminating number and for terminating number denominator has prime factor 2 and 5 only .
(ii) 0.120120012000120000. . .
In this problem repetitions number are not same so it is not a irrational number
so prime factor of denominator Q will has a value which is not equal to 2 or 5. And irrational number is always none terminating
(iii) 43.123456789
In this number 0.123456789 repeating again and again so it is a rational number and it is none terminating so that the prime factor has a value which is not equal to 2 or 5
Subscribe to:
Posts (Atom)