Question

Show that √3 is not a rational number?

plzz do fast !!

Answer 1

$\huge \mathfrak{Ello ♡}$

$\bold{Here \: is \: ur \: answer !!}$

$\mathfrak{{by \: contradiction \: method.}}$

suppose√3 is a rational number. then,

√3 = p/q, p and q are integers and have no common factor.
also, q ≉ 0

squaring both sides, 3 = p²/q² ⇒ p² = 3q² .......(1)

⇒ p² is a multiple of 3 ⇒p is a multiple of 3.

let p = 3m ⇒p² = 9m²

putting this value in (1), we get

9m² = 3q² ⇒q² = 3m²

⇒q² is a multiple of 3 ⇒ q is a multiple of 3

(2) and (3) ⇒ p and q both are multiples of 3

⇒ 3 is a common factor of p and q.

this contradicts our supposition that p and q have no common factor.

Hence, √3 is not a rational number.

ᴛʜᴀɴᴋ ʏᴏᴜ,
# ʙᴇ ʙʀᴀɪɴʟʏ

Answer 2

Hlo mate :-

Solution :-
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☆ 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. 

Now

●√2 = 1.4142... 
●√3 = 1.7321... 


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