# What Is 5/7 as a Decimal + Solution With Free Steps

**The fraction 5/7 as a decimal is equal to 0.714.**

We have all come across **Fractions** at some point in time as they are used for expressing a division operation between two numbers.

But some **Fractions** donâ€™t solve completely and those result in **Decimal Values**, and here we are interested in solving for those.

To solve a division that is not conclusive, we use a method called **Long Division** so letâ€™s look at the solution of our fraction 5 / 7.

## Solution

First, we get started by getting the **Dividend** and the **Divisor** out of our fraction. This is done as follows:

**Dividend = 5**

**Divisor = 7**

Knowing that the numerator is the Dividend, and the denominator is the Divisor. Now, we can smoothly move on to the **Quotient** as well, which is defined as the solution to a division. So, a **Quotient** under the given circumstances would look like this:

**Quotient = Dividend $\div$ Divisor = 5 $\div$ 7**

Here, we have completely transformed the expression for the fraction, and now we are ready to solve this division using the **Long Division Method**.

Figure 1

### 5/7 Long Division Method

We have a starting point here, and it is:

**Â 5 $\div$ 7Â **

Now, this very expression can tell a lot about the nature of the **Quotient**. As it can be seen the dividend is **smaller** than the divisor, so the Quotient will be smaller than 1.

Finally, one last important piece of information is without a doubt the **Remainder**. The number will carry forward an **Inconclusive Division**, and also replace the dividend multiple times.

So, we have 5 smaller than 7 which tells us that we require to introduce a **Zero** to the right of the dividend, and hence a **decimal point** to the quotient. This leads to the dividend becoming 50, and its division is given below:

**50 $\div$ 7 $\approx$ 7**

Where:

**Â 7 x 7 = 49Â **

Which will give us a remainder of 50 â€“ 49 = 1.

Therefore, a **Remainder** of 1 was generated as a result of the incomplete division between our dividend and the divisor. And now it is time for the remainder to become the new dividend, we can see that 1 needs a **Zero** to be solved further. So, we get the new dividend as 10:

**10 $\div$ 7 $\approx$ 1Â **

Where:

**7 x 1 = 7Â **

Hence, we have 10 â€“ 7 = 3 as the remainder.

It is common knowledge that the **Division** is carried out to the third decimal place for accuracy in case of no apparent complete solution. So, we repeat the process one last time, the dividend becomes 30.

**30 $\div$ 7 $\approx$ 4Â **

Â Where:

**7 x 4 = 28Â **

Thus, 30 â€“ 28 = 2 is the remainder.

We conclude our efforts here, therefore we have a **Quotient** of 0.714 and a **Remainder** of 2 after three iterations.

*Images/mathematical drawings are created with GeoGebra.*