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Is $\sqrt{5}$ an irrational number? $\newline$ Choices: $\newline$ (A) yes $\newline$ (B) no

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Identify rational and irrational square roots

Full solution

Q. Is

\sqrt{5}

an irrational number?

\newline

Choices:

\newline

(A) yes

\newline

(B) no

Understand Square Root of $5$ : Step $1$ : Understand the square root of $5$ . $\newline$ To find if $\sqrt{5}$ is irrational, we first need to know what $\sqrt{5}$ means. It's the number that, when multiplied by itself, gives $5$ .
Check Fraction Expression: Step $2$ : Check if $\sqrt{5}$ can be expressed as a fraction. $\newline$ An irrational number can't be written as a simple fraction ( $\frac{a}{b}$ , where $a$ and $b$ are integers, and $b$ is not zero). We need to check if there's any fraction that squares to $5$ .
Attempt Fraction Expression: Step $3$ : Attempt to express $\sqrt{5}$ as a fraction. Assume $\sqrt{5} = \frac{a}{b}$ , where $a$ and $b$ are integers with no common factors, and $b \neq 0$ . Then $5 = \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$ . This implies $5b^2 = a^2$ .
Analyze Integer Solutions: Step $4$ : Analyze the equation for integer solutions. $\newline$ From $5b^2 = a^2$ , both $a^2$ and $b^2$ must be multiples of $5$ . If $b^2$ is a multiple of $5$ , then $b$ must also be a multiple of $5$ . Substituting back, we get that $a$ must also be a multiple of $5$ . This leads to an infinite loop of multiplying by $5$ , which means we can't find integers $a$ and $b$ that fit our initial assumption.
Conclude Nature of $\sqrt{5}$ : Step $5$ : Conclude the nature of $\sqrt{5}$ . $\newline$ Since we can't express $\sqrt{5}$ as a fraction of two integers, it must be an irrational number.

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