prove √2, √3, √5and√7 are irrational no
Answers
- √2 :- let us suppose that √2 is irrational.
➡ √2 = p/q (q ≠ 0, p and q are coprime)
➡ √2q = p
squaring both sides,
➡ (√2q)² = (p)²
➡ 2q² = p² ----------(i)
therefore 2 divides p ---------(ii)
➡ 2 × r = p
again squaring both sides,
➡ (2 × r)² = (p)²
➡ 4 × r² = p²
➡ 4 × r² = 2q² (from equation (i)
➡ 2 × r² = q²
2 divides q², therefore 2 divides q ---------(iii)
from (ii) and (iii), we can say that p and q have common factor 2 which contradicts our assumption that p and q are coprime.
hence, our assumption is wrong and it's proved √2 is irrational.
- √3 :- let us suppose that √3 is irrational.
➡ √3 = p/q (q ≠ 0, p and q are coprime)
➡ √3q = p
squaring both sides,
➡ (√3q)² = (p)²
➡ 3q² = p² ----------(i)
therefore 3 divides p ---------(ii)
➡ 3 × r = p
again squaring both sides,
➡ (3 × r)² = (p)²
➡ 9 × r² = p²
➡ 9 × r² = 3q² (from equation (i)
➡ 3 × r² = q²
3 divides q², therefore 3 divides q ---------(iii)
from (ii) and (iii), we can say that p and q have common factor 3 which contradicts our assumption that p and q are coprime.
hence, our assumption is wrong and it's proved √3 is irrational.
- √5 :- let us suppose that √5 is irrational.
➡ √5 = p/q (q ≠ 0, p and q are coprime)
➡ √5q = p
squaring both sides,
➡ (√5q)² = (p)²
➡ 5q² = p² ----------(i)
therefore 5 divides p ---------(ii)
➡ 5 × r = p
again squaring both sides,
➡ (5 × r)² = (p)²
➡ 25 × r² = p²
➡ 25 × r² = 5q² (from equation (i)
➡ 5 × r² = q²
5 divides q², therefore 5 divides q ---------(iii)
from (ii) and (iii), we can say that p and q have common factor 5 which contradicts our assumption that p and q are coprime.
hence, our assumption is wrong and it's proved √5 is irrational.
- √7 :- let us suppose that √7 is irrational.
➡ √7 = p/q (q ≠ 0, p and q are coprime)
➡ √7q = p
squaring both sides,
➡ (√7q)² = (p)²
➡ 7q² = p² ----------(i)
therefore 5 divides p ---------(ii)
➡ 7 × r = p
again squaring both sides,
➡ (7 × r)² = (p)²
➡ 49 × r² = p²
➡ 49 × r² = 7q² (from equation (i)
➡ 7 × r² = q²
7 divides q², therefore 7 divides q ---------(iii)
from (ii) and (iii), we can say that p and q have common factor 7 which contradicts our assumption that p and q are coprime.
hence, our assumption is wrong and it's proved √7 is irrational.