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Prove that
sqrt5 is irrational.

Prove that $\sqrt{5}$ is irrational.

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Positive and negative square roots

Full solution

Q. Prove that

\sqrt{5}

is irrational.

Assume Rational Number: Assume that $\sqrt{5}$ is a rational number. According to the definition of rational numbers, if $\sqrt{5}$ is rational, 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.
Square of $\sqrt{5}$ : Since $\sqrt{5} = \frac{a}{b}$ , then $5 = \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$ . This implies that $5b^2 = a^2$ .
Express $a$ as $5k$ : The equation $5b^2 = a^2$ indicates that $a^2$ is a multiple of $5$ . Therefore, $a$ must also be a multiple of $5$ , because the square of a non-multiple of $5$ cannot be a multiple of $5$ . Let's write $a$ as $5k$ , where $5k$ $1$ is an integer.
Substitute $a$ in equation: Substitute $a$ with $5k$ in the equation $5b^2 = a^2$ to get $5b^2 = (5k)^2$ . This simplifies to $5b^2 = 25k^2$ , and further to $b^2 = 5k^2$ .
b is a multiple of $5$ : The equation $b^2 = 5k^2$ shows that $b^2$ is also a multiple of $5$ , and hence $b$ must be a multiple of $5$ as well.
Contradiction Found: Since both $a$ and $b$ are multiples of $5$ , this contradicts our initial assumption that $a$ and $b$ have no common factors other than $1$ . Therefore, our initial assumption that $\sqrt{5}$ is rational is incorrect.
Final Conclusion: We have reached a contradiction, which means our initial assumption is false. Hence, $\sqrt{5}$ cannot be expressed as a fraction of two integers with no common factors other than $1$ . Therefore, $\sqrt{5}$ is irrational.

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