-- searchonmath.com $\newline$ ( $3$ ) $\newline$ Prove that $\sqrt{3}$ is irrational

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Q. -- searchonmath.com

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3

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Prove that

\sqrt{3}

is irrational

Assume $\sqrt{3}$ rational: Assume $\sqrt{3}$ is rational, meaning it can be expressed as a fraction $\frac{a}{b}$ , where $a$ and $b$ are integers with no common factors other than $1$ , and $b$ is not zero.
Equation derivation: Then $\sqrt{3} = \frac{a}{b}$ implies that $3 = \frac{a^2}{b^2}$ .
Substitute $a=3k$ : Multiplying both sides by $b^2$ gives us $3b^2 = a^2$ .
Simplify equation: This means that $a^2$ is divisible by $3$ , so $a$ must also be divisible by $3$ .
Contradiction reached: Let $a = 3k$ , where $k$ is an integer. Substituting back into the equation gives us $3b^2 = (3k)^2$ .
Contradiction reached: Let $a = 3k$ , where $k$ is an integer. Substituting back into the equation gives us $3b^2 = (3k)^2$ . Simplifying the right side, we get $3b^2 = 9k^2$ .
Contradiction reached: Let $a = 3k$ , where $k$ is an integer. Substituting back into the equation gives us $3b^2 = (3k)^2$ . Simplifying the right side, we get $3b^2 = 9k^2$ . Dividing both sides by $3$ gives us $b^2 = 3k^2$ .
Contradiction reached: Let $a = 3k$ , where $k$ is an integer. Substituting back into the equation gives us $3b^2 = (3k)^2$ . Simplifying the right side, we get $3b^2 = 9k^2$ . Dividing both sides by $3$ gives us $b^2 = 3k^2$ . This implies that $b^2$ is also divisible by $3$ , and hence $b$ is divisible by $3$ .
Contradiction reached: Let $a = 3k$ , where $k$ is an integer. Substituting back into the equation gives us $3b^2 = (3k)^2$ . Simplifying the right side, we get $3b^2 = 9k^2$ . Dividing both sides by $3$ gives us $b^2 = 3k^2$ . This implies that $b^2$ is also divisible by $3$ , and hence $b$ is divisible by $3$ . But if both $k$ $0$ and $b$ are divisible by $3$ , this contradicts our initial assumption that $k$ $3$ is in its lowest terms, since $3$ is a common factor.
Contradiction reached: Let $a = 3k$ , where $k$ is an integer. Substituting back into the equation gives us $3b^2 = (3k)^2$ . Simplifying the right side, we get $3b^2 = 9k^2$ . Dividing both sides by $3$ gives us $b^2 = 3k^2$ . This implies that $b^2$ is also divisible by $3$ , and hence $b$ is divisible by $3$ . But if both $k$ $0$ and $b$ are divisible by $3$ , this contradicts our initial assumption that $k$ $3$ is in its lowest terms, since $3$ is a common factor. Therefore, our initial assumption that $k$ $5$ is rational is incorrect, and $k$ $5$ must be irrational.