Math, asked by lohhamdunloh, 6 months ago

prove that
 \sqrt{ \:  \:  \:  \: 3}

is irrational no.​

Answers

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
4

To Prove

  • √3 is a irrational Number

Proof

We know that any rational number will be of the form p/q where p & q are Integers and q ≠ 0

Assume 3 to be Rational

→ √3 = p/q

→ √3q = p

[Squaring both sides]

→ (√3q)² = p²

→ 3q = p²

Since p² is divisible by 3 then p will also be divisible by p

Assume p to be 3a

→ 3p = (3a)²

→ 3p = 9a²

Dividing by 3

→ p = 3a²

So from this we see that p is also divisible by 3

Bit this is a contradiction as p & q will be co primes amd cannot be divisible by the same number and hence our assumption is wrong and √3 is a rational number

Hence Proved!!


BrainlyPopularman: Nice :)
Answered by ItzDinu
2

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Given:

√3

To prove:

√3 is an irrational number.

Proof:

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

We have two cases to consider now.

Case I

Suppose that r is even. Then r2 is even, and 3r2 is even which implies that q2 is even and so q is even, but this cannot happen. If both q and r are even then gcd(q,r)≥2 which is a contradiction.

Case  II

Now suppose that r is odd. Then r2 is odd and 3r2 is odd which implies that q2 is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N.

Therefore

q2=3r2

(2m−1)2=3(2n−1)2

4m2−4m+1=3(4n2−4n+1)

4m2−4m+1=12n2−12n+3

4m2−4m=12n2−12n+2

2m2−2m=6n2−6n+1

2(m2−m)=2(3n2−3n)+1

We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3.

Hence the root of 3 is an irrational number.

Hence Proved

√3 is an irrational number.

  • I Hope it's Helpful My Friend.
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