Prove √2 is a irrational number.

Question

Bytelearn · Accepted Answer

Assume Rational Number Definition: Assume that √2 is a rational number. By definition, a rational number can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1, and b is not zero.Express √2 as Fraction: Express √2 as a/b, where a and b are the smallest possible positive integers and have no common factors.Square Both Sides: Square both sides of the equation to eliminate the square root: (√2)^2 = (a/b)^2.Simplify Equation: This simplifies to 2 = a^2/b^2.Represent a as 2k: Multiply both sides by b^2 to clear the fraction: 2b^2 = a^2.Substitute a with 2k: This implies that a^2 is an even number since it is equal to 2 times another whole number (b^2).Solve for b^2: If a^2 is even, then a must also be even because the square of an odd number is odd. Let's represent a as 2k, where k is an integer.Contradiction Found: Substitute a with 2k in the equation 2b^2 = a^2: 2b^2 = (2k)^2.Final Conclusion: Simplify the equation: 2b^2 = 4k^2.Divide both sides by 2 to solve for b^2: b^2 = 2k^2.This implies that b^2 is also even, and therefore b must be even as well.We now have a contradiction because we assumed that a and b have no common factors other than 1. However, we have shown that both a and b must be even, which means they both are divisible by 2, and thus have a common factor of 2.Since our assumption that √2 is rational leads to a contradiction, our initial assumption must be false. Therefore, √2 cannot be expressed as a fraction of two integers with no common factors, which means √2 is irrational.

Prove $\sqrt{2}$ is a irrational number.

Full solution

More problems from Cube roots of positive and negative perfect cubes

Related topics

Prove 2\sqrt{2}2​ is a irrational number.

Prove $\sqrt{2}$ is a irrational number.