Question

prove that √3 is a irrational number​

Answer 1

Proof: Once again we will prove this by contradiction. Suppose that there exists a rational number r=ab such that r2=3. Let r be in lowest terms, that is the greatest common divisor of a and b is 1, or rather gcd(a,b)=1. And so:

(1)

r2=3a2b2=3 a2=3b2

We have two cases to consider now. Suppose that b is even. Then b2 is even, and 3b2 is even which implies that a2 is even and so a is even, but this cannot happen. If both a and b are even then gcd(a,b)≥2 which is a contradiction.

Now suppose that b is odd. Then b2 is odd and 3b2 is odd which implies that a2 is odd and so a is odd. Since both a and b are odd, we can write a=2m−1 and b=2n−1 for some m,n∈N. Therefore:

(2)

a2=3b2(2m−1)2=3(2n−1)24m2−4m+1=3(4n2−4n+1)4m2−4m+1=12n2−12n+34m2−4m=12n2−12n+22m2−2m=6n2−6n+12(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. ■

hope it helps you

Answer 2

Answer:

let us assume that √3 is a rational number

√3 = a/B

b√3 =. a

square on both sides

(b√3)2 = (a)2

3b2 = a2......... (1)

therefore, 3 divides a2 then it will divide a also

let a = 3c

putting value of a in equation 1

3b2 = (3c)2

3b2 = 9c2

divide 9

b2/3 = c2

similarly we get that b will divide 3

so we already get that a divides 3

therefore this contracts the fact that √3 is irrational