An irrational number isn't as scary as it sounds; it's just a number that can't be expressed as a simple fraction or, to put it another way, an irrational number is a never-ending decimal that continues an infinite number of places past the decimal point. You can perform most operations on irrational numbers just as you'd do with rational numbers, but when it comes to taking square roots, you're going to have to learn to approximate the value.

## What's an Irrational Number?

So what is an irrational number, anyway? You may already be familiar with two very famous irrational numbers: π or "pi," which is almost always abbreviated as 3.14 but in fact continues infinitely to the right of the decimal point; and "e," a.k.a. Euler's number, which is usually abbreviated as 2.71828 but also continues infinitely to the right of the decimal point.

But there are a lot more irrational numbers out there, and here's an easy way to spot some of them: If the number beneath a square root sign is not a perfect square, then that square root is an irrational number.

That's an awfully big mouthful, so here's an example to make it clear. It also helps to remember that a perfect square is a number whose square root is an integer:

**Is √8 an irrational number?** If you've memorized your perfect squares or take the time to look them up, you'll know that

Since √8 is in between those two numbers, but there's no integer between 2 and 3 to be its root, √8 is irrational.

## Taking the Square Root of an Irrational Number

When it comes to calculating the square root of an irrational number, you have two choices. Either put the irrational number into a calculator or an online square root calculator (see Resources), in which case the calculator will return an approximate value for you – or you can use a four-step process to estimate the value yourself.

**Example 1:** Estimate the value of the irrational number √8.

## Find a Starting Value

## Divide by Your Estimate

## Compute the Average

## Repeat Steps 2 and 3 as Needed

Find the perfect squares that would be to either side of √8 on the number-line. In this case, √4 = 2 and √9 = 3. Choose the one that's closest to your target number. Since 8 is much closer to 9 than to 4, choose

Next, divide the number whose root you want – 8 – by your estimate. Continuing the example, you have:

Now, find the average of the result from Step 2 with the divisor from Step 2. Here, that means averaging 3 and 2.67. First add the two numbers together, and then divide by two:

(This is actually the repeating decimal 5.6666666666, but it has been rounded to four decimal places for the sake of brevity.)

The result from Step 3 still isn't exact, but it's getting closer. Repeat Steps 2 and 3 as needed, using the result from Step 3 as the new divisor in Step 2 every time.

To continue the example, you would divide 8 by the result from Step 3 (2.83335), which gives you:

(Again, rounding to four decimal places for the sake of brevity.)

You would then average the result of your division with the divisor, which gives you:

You can continue this process, repeating Steps 2 and 3 as needed, until the answer is as exact as you need it to be.

## What About Irrational Square Roots?

Sometimes instead of finding the square root of an irrational number, you need to deal with irrational numbers that are expressed in square root form – one of the most famous you'll learn about is √2.

There's not a lot you can do with √2, aside from approximating its value as described above. But if you get a larger irrational number in square root form, you can sometimes use the fact that

to rewrite the answer in a simpler form.

Consider the irrational square root √32. Although it doesn't have a principal root (that is, a non-negative, integer root), you can factor it into something with a familiar principal root:

You still can't do much with √2, but √16 = 4, so you can take this a step further and write it as

While you haven't eliminated the radical sign entirely, you've simplified this irrational number while also preserving its exact value.

#### References

#### Resources

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