show that √3 is an irrarional number
Answers
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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
Answer:
Let us assume on the contrary that
3 is a rational number.
Then, there exist positive integers a and b such that
√3 = a/b
where, a and b, are co-prime i.e. their HCF is 1.
Now,
√3 = a/b
3 = a²/b²
3b²=a²
3 divides a² [ 3 divides 3b² ]
3 divides a .........(i)
a = 3c for some integer c.
a² = 9c²
3b² = 9c² [ a²= 3b² ]
b² = 3c²
3 divides b² [ 3 divides 3c² ]
3 divides b .........(ii)
From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.
Hence, √3 is an irrational number.
Explanation:
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