How to prove that square root of 3 is not a rational number?

Answer 1

We can prove a general result. If p and n are positive integers and p^(1/n) is not an integer - that is, there's no positive integer that raised to power n gives p - then p^(1/n) is irrational.

An easy way to prove this is to use the rational roots theorem. Consider the polynomial P(x) = x^n - p. Then, p^(1/n) is a real root of P. According to the rational roots theorem, if the rational r = q1/q2, q1 and q2<>0 integers with GCD = 1, is a root of P, then q1 divides the independent term -p - and so divides p - and q2 divides the coefficient of the leading term, 1. So, q2 = 1 or q2 = -1, which implies r must be an integer satisfying r^n = p. But since, by assumption, there's no such integer, it follows P has no rational roots.

Since p^(1/n) is a real root of P, it follows p^(1/n) is irrational.

If we put p = 3 and n =2, we conclude sqrt(3) is irrational.

Answer 2

pointy has the right idea, but you still have to prove that \sqrt{3} is irrational. You can prove that the square root of any prime number is irrational in a similar way that you prove it for the square root of 2. Suppose that (m/n)^2 = p , where p is a prime. (You will use p=3). Further, suppose that the fraction (m/n) is fully reduced : GCD(m,n)=1 . Then: m^2 = p*n^2 .Thus, p divides m^2. Since p is prime, this means that p divides m. Say m=k*p. Now we have the equation: (k^2)(p^2)/(n^2) = p . Which gives us: p*k^2 = n^2. For the same reason as before, we get that p divides n. Say j*p=n . Our original fraction: (m)/(n) = (k*p)/(j*p) . Our original assumption that the fraction was in most simplified form has been contradicted. Thus, the square root of p must be irrational.

Answer 3

If a number is rational then it can be written as the quotient of two integers p/q and without loss of generality, p and q have no common factors, i.e. they're relatively prime.
Assume sqrt3 = p/q where p and q are relatively prime
Square both sides: 3 = p^2/q^2
or 3q^2 =p^2.
So p^2 has 3 as a factor and since 3 is a prime p must have a factor of 3. Therefore 3^2 is a factor of p^2. Say p^2 = 9P
Therefore 3q^2 = p^2 = 9P or q^2 = 3P
But this says q^2 has a factor or 3 and as above a factor of q has a factor of 3.
Therefore both p and q have factors of 3 contradicting their being relatively prime. Therefore sqrt3 not rational. Note: this argument is identical for any prime no.

Answer 4

every rational number can be written as a/b where a and b are integers and relatively primes,
if square root of 3 is rational number then there exist a and b which are relatively primes and sqrt(3)=a/b ===> a^2=3b^2 so 3 divides a^2 and so a. if a=3k then
3k^2=b^2 again 3 divides b
a,and b can be divided by 3 which is a contradiction (we already assumed a and b are relatively primes)
this contradiction shows square root of 3 is not rational.

Answer 5

There is a standard proof from contradiction.
Let's assume that sqrt(3)=p/q where p and q are relatively prime, i.e. GCD(p,q)=1.
Let's square this expression, then p^2=3q^2, it means that p has factor 3 so can be written as p=3m, then p^2=9m^2 and 9m^2=3q^2, so q^2=3m^2. In a similar way q should have a factor of 3. But in that case both p and q have 3 as a factor, that contradicts GCD(p,q)=1. So it is not possible to find such relatively prime p and q so that sqrt(3)=p/q, so sqrt(3) is irrational.

Answer 6

I would advice the following link
http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm