prove that root 3 is irrational
divya7711:
hi
hello
Answers
Answered by
2
Heya Dear !!
------------------
Here Is Your Ans.----->
Let us assume that √3 is a rational number.
.
That is, we can find integersaandb(≠ 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 bare coprime.
√3b = a⇒ 3b2=a2(Squaring on both sides) → (1)Therefore, a2is divisible by 3Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integerc.Equation (1) becomes,3b2=(3c)2⇒ 3b2= 9c2∴ b2= 3c2
This means that b2is 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 andb are coprime.
This contradiction has arisen because ofour incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
Hence proved <=========
Hope It Helps ☺
------------------
Here Is Your Ans.----->
Let us assume that √3 is a rational number.
.
That is, we can find integersaandb(≠ 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 bare coprime.
√3b = a⇒ 3b2=a2(Squaring on both sides) → (1)Therefore, a2is divisible by 3Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integerc.Equation (1) becomes,3b2=(3c)2⇒ 3b2= 9c2∴ b2= 3c2
This means that b2is 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 andb are coprime.
This contradiction has arisen because ofour incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
Hence proved <=========
Hope It Helps ☺
Answered by
0
Answer:
177Step-by-step explanation:
Similar questions