prove that root 3 is an irrational number
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Raghunath answered 3 year(s) ago
Prove that root 3 is an irrational number.
Prove that 3 is an irrational number.
Find two irrational numbers lying between 2 and 3
Class-IX Maths
person
Asked by Prashanth
Dec 1
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Raghunath , SubjectMatterExpert
Member since Apr 11 2014
Sol:
Let us assume that √3 is a rational number.
That is, we can find integers a and b (≠ 0) such that √3 = (a/b)
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
√3b = a
⇒ 3b2=a2 (Squaring on both sides) → (1)
Therefore, a2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b2 =(3c)2
⇒ 3b2 = 9c2
∴ b2 = 3c2
This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
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Raghunath answered 3 year(s) ago
Prove that root 3 is an irrational number.
Prove that 3 is an irrational number.
Find two irrational numbers lying between 2 and 3
Class-IX Maths
person
Asked by Prashanth
Dec 1
0 Like
2930 views
editAnswer
Like
Follow
1 Answers
Top Recommend
|
Recent
person
Raghunath , SubjectMatterExpert
Member since Apr 11 2014
Sol:
Let us assume that √3 is a rational number.
That is, we can find integers a and b (≠ 0) such that √3 = (a/b)
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
√3b = a
⇒ 3b2=a2 (Squaring on both sides) → (1)
Therefore, a2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b2 =(3c)2
⇒ 3b2 = 9c2
∴ b2 = 3c2
This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
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