Full solution

Q. Is

\sqrt{2}

an irrational number?

\newline

Choices:

\newline

(A) yes

\newline

(B) no

Understand irrational numbers: Step $1$ : Understand the concept of an irrational number. An irrational number is a number that cannot be expressed as a simple fraction. It's decimal representation is non-terminating and non-repeating.
Consider $\sqrt{2}$ : Step $2$ : Consider the number $\sqrt{2}$ . $\newline$ We need to determine if $\sqrt{2}$ can be expressed as a fraction.
Attempt fraction expression: Step $3$ : Attempt to express $\sqrt{2}$ as a fraction. $\newline$ Assume $\sqrt{2}$ can be written as $\frac{a}{b}$ , where $a$ and $b$ are integers with no common factors other than $1$ , and $b$ is not zero.
Square to eliminate root: Step $4$ : Square both sides to eliminate the square root. $\newline$ $(\sqrt{2})^2 = (\frac{a}{b})^2$ $\newline$ $2 = \frac{a^2}{b^2}$ $\newline$ $2b^2 = a^2$
Analyze equation $2b^2 = a^2$ : Step $5$ : Analyze the equation $2b^2 = a^2$ . This equation implies that $a^2$ is an even number (since it's $2$ times another integer). Therefore, $a$ must also be even (since only even numbers squared give even results).
Substitute $a = 2k$ : Step $6$ : If $a$ is even, there exists an integer $k$ such that $a = 2k$ . Substitute $a = 2k$ into the equation: $2b^2 = (2k)^2$ $2b^2 = 4k^2$ $b^2 = 2k^2$
Follow $b^2 = 2k^2$ : Step $7$ : From $b^2 = 2k^2$ , it follows that $b^2$ is also even, and hence $b$ is even. This contradicts our initial assumption that $a$ and $b$ have no common factors other than $1$ , as both are divisible by $2$ .
Conclude $\sqrt{2}$ is irrational: Step $8$ : Conclude that $\sqrt{2}$ cannot be expressed as a fraction of two integers. Since our assumption led to a contradiction, $\sqrt{2}$ must be irrational.