A common place where you will see irrational numbers is when you are working with circles or spheres. Since pi is related to a circle, you will need to work with irrational numbers when solving problems involving circles. Take a look at this dilemma.

Henrietta knew that to find the circumference of a circle, she needed to multiply the diameter by \begin{align*}\pi\end{align*}. If the diameter of Henrietta’s circle is 6 inches, what is the approximate circumference of the circle?

**Pay attention and you will learn how to successfully solve this problem.**

### Guidance

There are many different ways to classify or name numbers.

**All numbers are considered** *real numbers.*

When you were in the lower grades, you worked with *whole numbers.***Whole numbers are counting numbers. We consider whole numbers as the set of numbers \begin{align*}\{0, 1, 2, 3, 4 \ldots \}\end{align*}**.

In middle school, you may also have learned about ** integers**.

**The set of integers includes whole numbers, but also includes their opposites.**Therefore, we can say that whole positive and negative numbers are part of the set of integers \begin{align*}\{ \ldots -2, -1, 0, 1, 2, 3 \ldots \}\end{align*}.

We can’t stop classify numbers with whole numbers and integers because sometimes we can measure a part of a whole or a whole with parts. These numbers are called *rational numbers.***A rational number is any number that can be written as a fraction where the numerator or the denominator is not equal to zero.**

Take a look at these definitions and then write them in your notebook.

Sometimes, you will need to find estimates of irrational numbers to solve an equation. The easiest way to do this is to find a decimal value on your calculator that is close to the irrational number. Remember that the more decimal points you include, the more accurate your answer will be. For these purposes, it is usually okay to round an irrational number to the nearest hundredth or thousandth. Once you have found the decimal approximate, solve the equation normally. It is crucial to use words or signage to show that your answer is approximate, not exact. The symbol \begin{align*}\approx\end{align*} means approximately equal to, and is more appropriate than an equals sign in these situations.

Take a look at this situation.

**Solve for \begin{align*}a\end{align*}: \begin{align*}a=4 \pi\end{align*}.**

First find a decimal approximation for \begin{align*}\pi\end{align*} using your calculator. The value of \begin{align*}\pi\end{align*} is 3.1415927... This can be rounded to 3.14 for these purposes.

**To solve the equation, multiply 3.14 by 4. This will be the approximate value of \begin{align*}a\end{align*}.**

\begin{align*}3.14 \times 4=12.56\end{align*}

**So the value of \begin{align*}a\end{align*} is approximately 12.56. \begin{align*}a \approx 12.56\end{align*}**

Evaluate each solution.

#### Example A

\begin{align*}a\end{align*}: \begin{align*}a=3 \pi\end{align*}

**Solution: \begin{align*}9.42\end{align*}**

#### Example B

\begin{align*}x\end{align*}: \begin{align*}x=7 \pi\end{align*}

**Solution: \begin{align*}21.98\end{align*}**

#### Example C

\begin{align*}y\end{align*}: \begin{align*}y=5 \pi\end{align*}

**Solution: \begin{align*}15.7\end{align*}**

Now let's go back to the dilemma from the beginning of the Concept.

First, translate the information in the problem into an equation. Let \begin{align*}C\end{align*} equal the circumference.

\begin{align*}C=\pi \times 6\end{align*}

The value of \begin{align*}\pi\end{align*} is 3.1415927... Rounded to the nearest hundredth, this value is 3.14. You can substitute this value back into the equation to find the value of \begin{align*}C\end{align*}. Remember to use the approximately equals sign after you make this estimation.

\begin{align*}C&=\pi \times 6\\ C &\approx 3.14 \times 6\\ C &\approx 18.84\end{align*}

**The circumference of Henrietta’s circle is approximately 18.84 inches.**

### Vocabulary

- Whole Numbers
- the set of positive counting numbers.

- Integers
- the set of whole numbers and their opposites.

- Rational Numbers
- any number that can be written in fraction form including terminating and repeating decimals.

- Irrational Numbers
- numbers that do not have an end point when written in decimal form – the decimal values continue indefinitely. These numbers do not fit into the set of rational numbers.

- Pi
- \begin{align*}\pi\end{align*}, the ratio of the diameter to the circumference of a circle. We use 3.14 to represent this irrational number.

- Real Numbers
- the set of rational and irrational numbers make up the set of real numbers.

### Guided Practice

Here is one for you to try on your own.

Solve for \begin{align*}y\end{align*}: \begin{align*}12-\sqrt{7}=y\end{align*}.

**Solution**

First find a decimal approximation for \begin{align*}\sqrt{7}\end{align*} using your calculator. The value of \begin{align*}\sqrt{7}\end{align*} is 2.64575... This can be rounded to 2.65 for these purposes.

To solve the equation, subtract 2.65 from 12. This will be the approximate value of \begin{align*}y\end{align*}

\begin{align*}12-2.65=9.35\end{align*}

**So, the value of \begin{align*}y\end{align*} is approximately 9.35.** \begin{align*}y \approx 9.35\end{align*}

### Video Review

### Practice

Directions: Approximate the solution for each equation given the irrational numbers.

- \begin{align*}a\end{align*}: \begin{align*}a=3 \pi\end{align*}
- \begin{align*}x\end{align*}: \begin{align*}x=8 \pi\end{align*}
- \begin{align*}b\end{align*}: \begin{align*}b=9 \pi\end{align*}
- \begin{align*}c\end{align*}: \begin{align*}c=12 \pi\end{align*}
- \begin{align*}a\end{align*}: \begin{align*}a=2 \pi\end{align*}
- \begin{align*}y\end{align*}: \begin{align*}y=6 \pi\end{align*}
- \begin{align*}b\end{align*}: \begin{align*}b=7 \pi\end{align*}
- \begin{align*}d\end{align*}: \begin{align*}d=12 \pi-6\end{align*}
- \begin{align*}a\end{align*}: \begin{align*}a=14 \pi-9\end{align*}
- \begin{align*}x\end{align*}: \begin{align*}x=11 \pi-5\end{align*}
- \begin{align*}\sqrt{2}+5=x\end{align*}
- \begin{align*}8 = \sqrt{2} + x\end{align*}
- \begin{align*}t=\pi-5.3\end{align*}
- \begin{align*}\sqrt{h}=\sqrt{6}-\frac{3}{4}\end{align*}
- Mrs. DeFazio wrote the following equation on the board. \begin{align*}w=\sqrt{11}-2^2\end{align*} What is the value of \begin{align*}w\end{align*} in Mrs. DeFazio’s equation?