EXERCISE 1.3
QUE:-1 PROVE THAT √5 IS IRRATIONAL.
SOL:-
Let √5 is a rational number .
∴ we can find two integers a,b (b≠0) such that √5=a/b let a and b have common factore other then 1 .
then we can divide them common factor , and assume that a and b are co-prime
a=√5b
a²=5b²
∴ a² is divisible by 5 and it can be said that a is divisible by 5.
let a= 5k ,where k is an integer
(5k)²=5b²
5k²=b²
this means that b² is divisible by 5 and hence , b is divisible by 5.
this implise that a and b have 5 as a common factore .
and this a contadiction to the fact a and b are co-prime.
hence ,√5 cannot be expressed as p/q or it can be said that
√5 is irrational
QUE:-1PROVE THAT 3+2√5 IS IRRATIONAL.
SOL:-
Let 3+2√5 is rational.
∴ we can find two co-prime integers a,b(b≠0) such that
3+2√5=a/b
2√5=a/b-3
√5 =½(a/b-3)
since a and b are integers ,½(a/b-3) will also b rational nd therefore,
√5 is rational .
QUE :-2 PROVE THAT THE FOLLOWING ARE IRRATIONAL :
(i) 1/√2 (ii) 7√5 (iii) 6+√2
SOL :-
(i) 1/√2
let 1/√2 is rational .
therefore we can find two co-prime integer a,b (b≠0) such that
1/√2=a/b
√2 =b/a
b/a is rational as a and b integers .
therefore √2 is rational which contradicts to the fact that √2 is irrational .
hence , our assumption is false and 1/√2 is irrational.
(ii) 7√5
let 7√5 is rational .
therefore, we can find two co-prime integer a,b (b≠0) such that
7√5 =a/b
√5 =a/7b
a/7b is rationalas a and bare integers.
therfore , √5should be rational.
this contradict the fact that √5 is irrational
∴ our assumption is rational is false . hence ,7√5 is irrational.
(iii) 6+√2
let 6+√2 be rational .
∴ we can find two co-prime integer a,b (b≠0) such that
6+√2=a/b
√2=a/b-6
since a and b are integers , a/b-6 is also rational . therrfore , our assumption is false and hence , 6+√2 is irrational.
