Question

Prove that root 3 is irrational.

Class 10 th​

Answer 1

Answer:

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. ■

Answer 2

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suppose 3–√ is rational, then 3–√=ab for some (a,b) suppose we have a/b in simplest form.

3–√a2=ab=3b2

if b is even, then a is also even in which case a/b is not in simplest form.

if b is odd then a is also odd. Therefore:

ab(2n+1)24n2+4n+14n2+4n2n2+2n2(n2+n)=2n+1=2m+1=3(2m+1)2=12m2+12m+3=12m2+12m+2=6m2+6m+1=2(3m2+3m)+1

Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.