Solved on Jan 09, 2024

Find the rational numbers in the set $\left\{-\frac{15}{7},-\sqrt{2}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, \sqrt{3}, 3.25\right\}$ .

STEP 1

Assumptions
1. A rational number is a number that can be expressed as the quotient or fraction $\frac{p}{q}$ of two integers, with the denominator $q$ not equal to zero.
2. We are given the set of real numbers $\left\{-\frac{15}{7},-\sqrt{2}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, \sqrt{3}, 3.25\right\}$ .
3. We need to identify which numbers in the set are rational.

STEP 2

We will go through each number in the set and determine if it is rational or irrational.

STEP 3

The first number is $-\frac{15}{7}$ . This is a fraction of two integers, and the denominator is not zero. Therefore, $-\frac{15}{7}$ is a rational number.

STEP 4

The second number is $-\sqrt{2}$ . Since $\sqrt{2}$ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, $-\sqrt{2}$ is an irrational number.

STEP 5

The third number is $\sqrt{1}$ . Since $\sqrt{1} = 1$ , and $1$ can be expressed as $\frac{1}{1}$ , which is a fraction of two integers, $\sqrt{1}$ is a rational number.

STEP 6

The fourth number is $\sqrt{\frac{4}{9}}$ . Since both $4$ and $9$ are perfect squares, $\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$ , which is a fraction of two integers. Therefore, $\sqrt{\frac{4}{9}}$ is a rational number.

STEP 7

The fifth number is $1$ . Since $1$ can be expressed as $\frac{1}{1}$ , which is a fraction of two integers, $1$ is a rational number.

STEP 8

The sixth number is $\sqrt{3}$ . Since $3$ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, $\sqrt{3}$ is an irrational number.

STEP 9

The seventh number is $3.25$ . Since $3.25$ can be expressed as $\frac{13}{4}$ , which is a fraction of two integers, $3.25$ is a rational number.

STEP 10

Now that we have determined which numbers are rational, we can list them.

STEP 11

The rational numbers in the set are $-\frac{15}{7}$ , $\sqrt{1}$ , $\sqrt{\frac{4}{9}}$ , $1$ , and $3.25$ .

STEP 12

We can now express the set containing all the rational numbers from the given set.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, 3.25\right\}$

STEP 13

Simplify the set by replacing $\sqrt{1}$ and $\sqrt{\frac{4}{9}}$ with their simplified rational forms.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, 1, \frac{2}{3}, 1, 3.25\right\}$

STEP 14

Since $1$ is listed twice, we can remove the duplicate to present the final set of rational numbers.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, 1, \frac{2}{3}, 3.25\right\}$
This is the set containing all the rational numbers from the given set.

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Find the rational numbers in the set $\left\{-\frac{15}{7},-\sqrt{2}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, \sqrt{3}, 3.25\right\}$ .

STEP 1

STEP 2

We will go through each number in the set and determine if it is rational or irrational.

STEP 3

The first number is $-\frac{15}{7}$ . This is a fraction of two integers, and the denominator is not zero. Therefore, $-\frac{15}{7}$ is a rational number.

STEP 4

The second number is $-\sqrt{2}$ . Since $\sqrt{2}$ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, $-\sqrt{2}$ is an irrational number.

STEP 5

The third number is $\sqrt{1}$ . Since $\sqrt{1} = 1$ , and $1$ can be expressed as $\frac{1}{1}$ , which is a fraction of two integers, $\sqrt{1}$ is a rational number.

STEP 6

The fourth number is $\sqrt{\frac{4}{9}}$ . Since both $4$ and $9$ are perfect squares, $\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$ , which is a fraction of two integers. Therefore, $\sqrt{\frac{4}{9}}$ is a rational number.

STEP 7

The fifth number is $1$ . Since $1$ can be expressed as $\frac{1}{1}$ , which is a fraction of two integers, $1$ is a rational number.

STEP 8

The sixth number is $\sqrt{3}$ . Since $3$ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, $\sqrt{3}$ is an irrational number.

STEP 9

The seventh number is $3.25$ . Since $3.25$ can be expressed as $\frac{13}{4}$ , which is a fraction of two integers, $3.25$ is a rational number.

STEP 10

Now that we have determined which numbers are rational, we can list them.

STEP 11

The rational numbers in the set are $-\frac{15}{7}$ , $\sqrt{1}$ , $\sqrt{\frac{4}{9}}$ , $1$ , and $3.25$ .

STEP 12

We can now express the set containing all the rational numbers from the given set.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, 3.25\right\}$

STEP 13

Simplify the set by replacing $\sqrt{1}$ and $\sqrt{\frac{4}{9}}$ with their simplified rational forms.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, 1, \frac{2}{3}, 1, 3.25\right\}$

STEP 14

Since $1$ is listed twice, we can remove the duplicate to present the final set of rational numbers.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, 1, \frac{2}{3}, 3.25\right\}$
This is the set containing all the rational numbers from the given set.

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Find the rational numbers in the set {−157,−2,1,49,1,3,3.25}\left\{-\frac{15}{7},-\sqrt{2}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, \sqrt{3}, 3.25\right\}{−715​,−2​,1​,94​​,1,3​,3.25}.

STEP 1

STEP 2

We will go through each number in the set and determine if it is rational or irrational.

STEP 3

The first number is −157-\frac{15}{7}−715​. This is a fraction of two integers, and the denominator is not zero. Therefore, −157-\frac{15}{7}−715​ is a rational number.

STEP 4

The second number is −2-\sqrt{2}−2​. Since 2\sqrt{2}2​ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, −2-\sqrt{2}−2​ is an irrational number.

STEP 5

The third number is 1\sqrt{1}1​. Since 1=1\sqrt{1} = 11​=1, and 111 can be expressed as 11\frac{1}{1}11​, which is a fraction of two integers, 1\sqrt{1}1​ is a rational number.

STEP 6

The fourth number is 49\sqrt{\frac{4}{9}}94​​. Since both 444 and 999 are perfect squares, 49=49=23\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}94​​=9​4​​=32​, which is a fraction of two integers. Therefore, 49\sqrt{\frac{4}{9}}94​​ is a rational number.

STEP 7

The fifth number is 111. Since 111 can be expressed as 11\frac{1}{1}11​, which is a fraction of two integers, 111 is a rational number.

STEP 8

The sixth number is 3\sqrt{3}3​. Since 333 is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, 3\sqrt{3}3​ is an irrational number.

STEP 9

The seventh number is 3.253.253.25. Since 3.253.253.25 can be expressed as 134\frac{13}{4}413​, which is a fraction of two integers, 3.253.253.25 is a rational number.

STEP 10

Now that we have determined which numbers are rational, we can list them.

STEP 11

The rational numbers in the set are −157-\frac{15}{7}−715​, 1\sqrt{1}1​, 49\sqrt{\frac{4}{9}}94​​, 111, and 3.253.253.25.

STEP 12

We can now express the set containing all the rational numbers from the given set.Rational numbers set={−157,1,49,1,3.25} \text{Rational numbers set} = \left\{-\frac{15}{7}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, 3.25\right\} Rational numbers set={−715​,1​,94​​,1,3.25}

STEP 13

Simplify the set by replacing 1\sqrt{1}1​ and 49\sqrt{\frac{4}{9}}94​​ with their simplified rational forms.Rational numbers set={−157,1,23,1,3.25} \text{Rational numbers set} = \left\{-\frac{15}{7}, 1, \frac{2}{3}, 1, 3.25\right\} Rational numbers set={−715​,1,32​,1,3.25}

STEP 14

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Find the rational numbers in the set $\left\{-\frac{15}{7},-\sqrt{2}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, \sqrt{3}, 3.25\right\}$ .

The first number is $-\frac{15}{7}$ . This is a fraction of two integers, and the denominator is not zero. Therefore, $-\frac{15}{7}$ is a rational number.

The second number is $-\sqrt{2}$ . Since $\sqrt{2}$ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, $-\sqrt{2}$ is an irrational number.

The third number is $\sqrt{1}$ . Since $\sqrt{1} = 1$ , and $1$ can be expressed as $\frac{1}{1}$ , which is a fraction of two integers, $\sqrt{1}$ is a rational number.

The fourth number is $\sqrt{\frac{4}{9}}$ . Since both $4$ and $9$ are perfect squares, $\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$ , which is a fraction of two integers. Therefore, $\sqrt{\frac{4}{9}}$ is a rational number.

The fifth number is $1$ . Since $1$ can be expressed as $\frac{1}{1}$ , which is a fraction of two integers, $1$ is a rational number.

The sixth number is $\sqrt{3}$ . Since $3$ is not a perfect square, it cannot be expressed as a fraction of two integers. Therefore, $\sqrt{3}$ is an irrational number.

The seventh number is $3.25$ . Since $3.25$ can be expressed as $\frac{13}{4}$ , which is a fraction of two integers, $3.25$ is a rational number.

The rational numbers in the set are $-\frac{15}{7}$ , $\sqrt{1}$ , $\sqrt{\frac{4}{9}}$ , $1$ , and $3.25$ .

We can now express the set containing all the rational numbers from the given set.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, \sqrt{1}, \sqrt{\frac{4}{9}}, 1, 3.25\right\}$

Simplify the set by replacing $\sqrt{1}$ and $\sqrt{\frac{4}{9}}$ with their simplified rational forms.
$\text{Rational numbers set} = \left\{-\frac{15}{7}, 1, \frac{2}{3}, 1, 3.25\right\}$