In article <346CC27B.email@example.com>, Emanuel Binnun <firstname.lastname@example.org> wrote: >Hi > >Please help me to proove that Square root of 3 is an irrational number >
Adding a few extra lines to the standard proof, we can get a "constructive" proof of the irrationality of sqrt(3): for every (positive) rational number x, we can find another positive rational number r such that abs(x - sqrt(3)) > r.
Consider x = p/q where p, q are positive integers, which we can choose without a common factor (greater than 1). We first show tat
(*) abs(p^2 - 3 * q^2) >= 1.
If to the contrary, abs(p^2 - 3 * q^2) < 1 then p^2 = 3 * q^2.
So, discussing the cases p=3n, p=3n+1, p=3n+2 separately, we find that p must be divisible by 3: p=3n. Then we cancel 3's and obtain q^2 = 3 * n^2, and q would be divisible by 3. We would see that p and q are both divisible by 3, contrary to our assumption (no common factor).
Here the standard proof concludes that p^2/q^2 is never 3. We go further: Rationalize at the right moment and use sqrt(3) < 2 to get