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prove that $\frac{1}{\sqrt{2}}$ is irrational

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Cube roots of positive perfect cubes

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Q. prove that

\frac{1}{\sqrt{2}}

is irrational

Assume Rational Number: Assume that $\frac{1}{\sqrt{2}}$ is a rational number. According to the definition of rational numbers, if a number 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. $\newline$ Let's express $\frac{1}{\sqrt{2}}$ as $\frac{a}{b}$ , where $a$ and $b$ are integers with no common factors. $\newline$ $\frac{a}{b}$ $0$
Express as Fraction: To remove the square root from the denominator, multiply both sides of the equation by $\sqrt{2}$ . $\frac{1}{\sqrt{2}} \times \sqrt{2} = \frac{a}{b} \times \sqrt{2}$ $\frac{\sqrt{2}}{2} = \frac{a\sqrt{2}}{b}$
Eliminate Square Root: Now, square both sides of the equation to eliminate the square root. $\newline$ $(\sqrt{2}/2)^2 = (a\sqrt{2}/b)^2$ $\newline$ $2/4 = 2a^2/b^2$ $\newline$ $1/2 = a^2/b^2$
Square Both Sides: Multiply both sides by $b^2$ to get rid of the fraction on the right side. $\newline$ $\frac{b^2}{2} = a^2$
Substitute $b=2k$ : This implies that $b^2$ is even since it is equal to $2$ times some integer ( $a^2$ ). $\newline$ If $b^2$ is even, then $b$ must also be even because the square of an odd number is odd. $\newline$ Let's say $b = 2k$ , where $k$ is an integer.
Find a Contradiction: Substitute $b = 2k$ back into the equation $\frac{b^2}{2} = a^2$ .
$\frac{(2k)^2}{2} = a^2$
$\frac{4k^2}{2} = a^2$
$2k^2 = a^2$
Find a Contradiction: Substitute $b = 2k$ back into the equation $\frac{b^2}{2} = a^2$ .
$\frac{(2k)^2}{2} = a^2$
$\frac{4k^2}{2} = a^2$
$2k^2 = a^2$ This implies that $a^2$ is even, and therefore $a$ must also be even.
Let's say $a = 2m$ , where $m$ is an integer.
Find a Contradiction: Substitute $b = 2k$ back into the equation $\frac{b^2}{2} = a^2$ . $\frac{(2k)^2}{2} = a^2$ $\frac{4k^2}{2} = a^2$ $2k^2 = a^2$ This implies that $a^2$ is even, and therefore $a$ must also be even.Let's say $a = 2m$ , where $m$ is an integer.We now have a contradiction because we assumed that $a$ and $\frac{b^2}{2} = a^2$ $0$ have no common factors other than $\frac{b^2}{2} = a^2$ $1$ , but we have shown that both $a$ and $\frac{b^2}{2} = a^2$ $0$ are even, which means they both have at least the common factor $\frac{b^2}{2} = a^2$ $4$ .This contradiction means our initial assumption that $\frac{b^2}{2} = a^2$ $5$ is rational is incorrect.Therefore, $\frac{b^2}{2} = a^2$ $5$ must be irrational.

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