 # When multiplying two decimal numbers, you first ignore the decimals, find the product, then count the number of decimal places that need to be in the answer by taking the sum of the original decimal places. Why exactly does this work? Kendra Hudson 2022-09-11 Answered
When multiplying two decimal numbers, you first ignore the decimals, find the product, then count the number of decimal places that need to be in the answer by taking the sum of the original decimal places. Why exactly does this work?
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Step 1
Suppose we want to multiply 1.2 with 0.5. Your method would work this way:
"Annihilate" the decimal point of each number, getting 12 and 5
Multiply those numbers, getting $12×5=60$
Add many decimal places to this result as the sum of decimal places of each original number, since each one has one decimal position, we have as final result 0.60.
The reason why this works, is the following.
You can write 1.2 as $12×{10}^{-1}$ and 0.5 as $5×{10}^{-1}$ (in general, the number $n×{10}^{-k}$ is obtained by displacing the decimal point of n by k positions to the left), thus, using properties of exponentiation:
$1.2×0.5=\left(12×{10}^{-1}\right)×\left(5×{10}^{-1}\right)=\left(12×5\right)×\left({10}^{-1}×{10}^{-1}\right)=60×{10}^{-2}=0.60$
So note where the sum of the decimal positions appear ( $-1-1=-2$ ).

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Step 1
$\left({10}^{m}x\right)\cdot \left({10}^{n}y\right)={10}^{m+n}\left(x\cdot y\right)$
In words, if x has m decimals then ${10}^{m}x$ has none. Ditto for y. To get $x\cdot y$ you need to take the product of the decimal-free numbers and put the decimal point after $m+n$ digits, counted from the right. This is the same as dividing by ${10}^{m+n}$ .

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