prove root 2 irrational

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Assume for contradiction: Assume, for the sake of contradiction, that √2 is rational. This means it can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1 (i.e., the fraction is in its simplest form).Express as coprime integers: Express √2 as a/b, where a and b are coprime integers (their greatest common divisor is 1). √2 = a/bSquare both sides: Square both sides of the equation to eliminate the square root. (√2)^2 = (a/b)^2 2 = a^2/b^2Multiply by b^2: Multiply both sides of the equation by b^2 to clear the denominator. 2b^2 = a^2Imply a is even: This implies that a^2 is an even number since it is equal to 2 times another number (b^2).Represent a as 2k: 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. a = 2kSubstitute a with 2k: Substitute a with 2k in the equation 2b^2 = a^2. 2b^2 = (2k)^2 2b^2 = 4k^2Divide by 2: Divide both sides of the equation by 2 to simplify. b^2 = 2k^2Imply b^2 is even: This implies that b^2 is also an even number since it is equal to 2 times another number (k^2).Imply b is even: If b^2 is even, then b must also be even (because the square of an odd number is odd).Contradiction in common factor: However, if both a and b are even, then they have a common factor of 2, which contradicts our initial assumption that a and b are coprime (have no common factors other than 1).Conclusion of irrationality: Since our assumption that √2 is rational leads to a contradiction, we must conclude that our assumption is false. Therefore, √2 cannot be expressed as a fraction of two integers and is irrational.

prove $\sqrt{2}$ irrational

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