UK Secondary (7-11)
Multiply decimals by decimal using algorithm
Lesson

We have used a standard algorithm to multiply whole numbers before, and the process for multiplying decimals is almost the same. The multiplication follows the same steps, then there is one extra step at the end to account for the decimal point.

#### Exploration

Let's follow through the process for the multiplication $4.83\times5.7$4.83×5.7.

To start off with, we simply ignore the decimal points. In this case, we get $483$483 and $57$57. We then multiply these together with the method we are used to using for whole numbers:

 $4$4 $8$8 $3$3 $\times$× $5$5 $7$7 $3$3 $3$3 $8$8 $1$1 (this is $483\times7$483×7) $+$+ $2$2 $4$4 $1$1 $5$5 $0$0 (this is $483\times5$483×5, shifted one place to the left) $2$2 $7$7 $5$5 $3$3 $1$1

Now we need to account for the decimal point. To do so, we add the total number of decimal places in the original numbers together.

In this case the original numbers are $4.83$4.83, which has $2$2 decimal places, and $5.7$5.7, which has $1$1 decimal place. So their product will have $2+1=3$2+1=3 decimal places.

Then to find the final answer, we take the product that we calculated before and insert the decimal point such that there are $3$3 decimal places:

So we have found that $4.83\times5.7=27.531$4.83×5.7=27.531.

### Why does this work?

Remember that we can represent any finite decimal as a fraction by using a power of $10$10 in the denominator. In this case, $4.83$4.83 is equal to $\frac{483}{100}$483100 and $5.7$5.7 is equal to $\frac{57}{10}$5710.

Let's now multiply these numbers by using their fraction forms instead. To do so, recall that we just multiply their numerators together and their denominators together:

 $\frac{483}{100}\times\frac{57}{10}$483100​×5710​ $=$= $\frac{483\times57}{100\times10}$483×57100×10​ $=$= $\frac{27531}{1000}$275311000​

The numerator contains the product $483\times57=27531$483×57=27531, which is what we initially calculated above. The denominator then tells us the place value of the number

In this case, the original denominators were $100$100 and $10$10. The final denominator is their product of $1000$1000. Dividing by $1000$1000 is the same as making the number three place values smaller, and so we get $27.531$27.531 as our final answer.

#### Practice questions

##### Question 1

Consider $0.6\times0.1$0.6×0.1.

1. How many digits after the decimal point will the product have?

2. Complete the vertical algorithm to evaluate the product.

 $0$0 . $6$6 $\times$× $0$0 . $1$1 $\editable{}$ . $\editable{}$ $\editable{}$

##### Question 2

Consider $4.9\times7.4$4.9×7.4.

1. How many digits after the decimal point will the product have?

2. Complete the vertical algorithm to evaluate the product.

 $4$4 . $9$9 $\times$× $7$7 . $4$4 $\editable{}$ . $\editable{}$ $\editable{}$ $+$+ $\editable{}$ $\editable{}$ . $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ . $\editable{}$ $\editable{}$

##### Question 3

Consider $4.87\times9.4$4.87×9.4.

1. How many digits after the decimal point will the product have?

2. Complete the vertical algorithm to evaluate the product.

 $4$4 . $8$8 $7$7 $\times$× $9$9 . $4$4 $\editable{}$ . $\editable{}$ $\editable{}$ $\editable{}$ $+$+ $\editable{}$ $\editable{}$ . $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ . $\editable{}$ $\editable{}$ $\editable{}$