Question

$prove \: that \: \sqrt{3 \: \: \: is \: an \: irrational \: \: number}$


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Answer 1

Answer:

Step-by-step explanation:

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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Answer 2

$\huge{Answer:}$



Let the √3 is rational so it can be written in the form of p/q where p and q are co prime

√3=a/b
squaring on both sides we get

3=a²/b²
3b²=a²


hence a²divide 3b² so it also divide 3b

now let a =3c


3b²=(3c)²

3b²= 9c²

hence 3b² divide 9c² and hence 9c


thus a and b are not co prime



hence √3 is irrational