Question

1. circle the irrational number in the list below.

$7.overline{27}$ $\frac{5}{9}$ $sqrt{15}$ $sqrt{196}$

Explanation:

Step1: Analyze \( 7.\overline{27} \)

A repeating decimal like \( 7.\overline{27} \) (where 27 repeats) can be expressed as a fraction. Let \( x = 7.2727\ldots \), then \( 100x = 727.2727\ldots \), subtracting \( x \) from \( 100x \) gives \( 99x = 720 \), so \( x=\frac{720}{99} \), which is a rational number.

Step2: Analyze \( \frac{5}{9} \)

A fraction \( \frac{5}{9} \) is by definition a rational number (ratio of two integers, denominator non - zero).

Step3: Analyze \( \sqrt{15} \)

To check if \( \sqrt{15} \) is rational, we assume \( \sqrt{15}=\frac{p}{q} \) where \( p \) and \( q \) are coprime integers (\( q
eq0 \)). Then \( 15=\frac{p^{2}}{q^{2}} \), or \( p^{2}=15q^{2} \). This means \( p^{2} \) is divisible by 15, so \( p \) is divisible by 15. Let \( p = 15k \), then \( (15k)^{2}=15q^{2}\Rightarrow225k^{2}=15q^{2}\Rightarrow15k^{2}=q^{2} \), so \( q^{2} \) is divisible by 15, and \( q \) is divisible by 15. But this contradicts the fact that \( p \) and \( q \) are coprime. So \( \sqrt{15} \) is irrational.

Step4: Analyze \( \sqrt{196} \)

We know that \( \sqrt{196}=\sqrt{14\times14} = 14 \), which is an integer and thus a rational number.

Answer:

The irrational number is \( \sqrt{15} \), so we circle \( \boldsymbol{\sqrt{15}} \).