Using Proof by Contradiction Prove that [latex]\sqrt{2}[/latex] is an irrational number.

Begin by assuming that [latex]\sqrt{2}[/latex] is not an irrational number. Then [latex]\sqrt{2}[/latex] is rational and can be written as the quotient of two integers a and b (b ≠ 0) that have no common factors. [latex] \begin{matrix} \sqrt{2} = \frac{a}{b} & & \quad \text{ Assume that } \sqrt{2}\text{ is a rational number.} \\ \\ 2b^{2} = a^{2} & & \text{ Square each side and multiply by } b^{2}. \end{matrix} [/latex] This implies that 2 is a factor of [latex]a^2[/latex]. So, 2 is also a factor of a. Let a = 2c. [latex]\begin{matrix} 2b^{2} = (2c)^{2} & & \text{ Substitute } 2c \text{ for } a. \quad\quad\quad\quad\quad\quad\quad\quad\\\\ b^{2} = 2c^{2} & & \text{ Simplify and divide each side by } 2. \quad\end{matrix}[/latex] This implies that 2 is a factor of [latex]b^2[/latex], and it is also a factor of b. So, 2 is a factor of both a and b. But this is impossible because a and b have no common factors. It must be impossible that [latex]\sqrt{2}[/latex] is a rational number. You can conclude that [latex]\sqrt{2}[/latex] is an irrational number.

Question Appendix.5: Using Proof by Contradiction Prove that √2 is an irrational ...

Elementary Linear Algebra

Using Proof by Contradiction

Prove that $\sqrt{2}$ is an irrational number.

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Begin by assuming that $\sqrt{2}$ is not an irrational number. Then $\sqrt{2}$ is rational and can be written as the quotient of two integers a and b (b ≠ 0) that have no common factors.

\begin{matrix} \sqrt{2} = \frac{a}{b} & & \quad \text{ Assume that } \sqrt{2}\text{ is a rational number.} \\ \\ 2b^{2} = a^{2} & & \text{ Square each side and multiply by } b^{2}. \end{matrix}

This implies that 2 is a factor of $a^2$ . So, 2 is also a factor of a. Let a = 2c.

\begin{matrix} 2b^{2} = (2c)^{2} & & \text{ Substitute } 2c \text{ for } a. \quad\quad\quad\quad\quad\quad\quad\quad\\\\ b^{2} = 2c^{2} & & \text{ Simplify and divide each side by } 2. \quad\end{matrix}

This implies that 2 is a factor of $b^2$ , and it is also a factor of b. So, 2 is a factor of both a and b. But this is impossible because a and b have no common factors. It must be impossible that $\sqrt{2}$ is a rational number. You can conclude that $\sqrt{2}$ is an irrational number.