result of multiplication having zeroes after the decimal

Question

Given the multiplication $3.25 \times 0.4$,

the primary school students learn that

1. we multiply the digit 4 to the number 325 which result in 1300
2. we count and then add the number of decimal place existing in 3.25 and 0.4 corresponding to 3
3. we move the decimal point 3 times to the right which corresponding to 1.300

My main question regards those two zeroes at the back and how to justify it. Is this incorrect to simplify the number to 1.3? or the opposite is this incorrect to leave the answer as $1.300$?

My approach to this would be to say that if one keep the answer as $1.300$, this number can be interpreted as the result of an approximation such as , for example, $1.2999 \approx 1.300$ correct to 4 significant figure

Answer 1

I think there are two separate issues here:

• the multiplication of two numbers, treated as precise
• the real-world validity of the level of precision.

Treated purely as arithmetic, it's obviously correct to say that the zeros can be discarded without affecting the result.

If you were multiplying two physical measurements, then the number of significant figures you kept would express the precision you believed the result to have, so it would be reasonable to either round the result to the same number of significant figures as the original numbers, or make an actual calculation of the expected error.

The thing with multiplication is that it normally produces an unrealistic number of significant figures.

So I think you try to teach them:

• $3.25 × 0.4$ is $1.3$
• keeping the zeros in, or keeping lots of decimal places, means we're saying that the numbers we're using really are that accurate.

I think it's useful for them to know about the second one, even though they probably won't actually use it until much later. (I don't think I was actually introduced to the concept of significant figures until I was doing science subjects at secondary school.)

Answer 2

Decimal is only shorthand.

$3.25$ is shorthand for $3 + \frac 2{10} + \frac 5{100}$ and $.4$ is shorthand for $\frac 4{10}$.

So $3.25 \times .4 = (3 + \frac 2{10} + \frac 5{100})(\frac 4{10}) = $

$\frac {12}{10} + \frac 8{100} + \frac {20}{1000} =$

$\frac {10}{10} + \frac {2}{10} + \frac 8{100} + \frac 2{100}=$

$1 + \frac {2}{10} + \frac {10}{100} =$

$1 + \frac 2{10} + \frac 1{10} = $

$1 + \frac 3{10} = $

$1.3$.

But $1 + \frac 3{10} = 1 + \frac 3{10} + \frac 0{100} +\frac 0{1000} + \frac 0{10000}$

So it's equally correct to write this as $1.3000$

Or $1 + \frac 3{10} = 0\times 1000 + 0\times 100 + 0\times 10 + 1 + \frac 3{10} + \frac 0{100}$ so it would be okay (albeit weird) to write it as $001.30$.

We don't write those leading $0$s in the beginning because ... we don't have to and it'd be really weird to. Also we intuitively start at the biggest part of a number and work our way to the smaller so we don't usually think about all those (non-existent) powers of $10$ before we start.

Likewise we don't write all the trailing $0$s at the end because we don't have to. However because we started at the biggest part and worked to the smallest we often do think about those (non-existent) negative powers of $10$ after we end.

In fact $1.3$ is actually $1.30000000000.....$ with an infinite number of $0$ after the end. We don't write them because... we don't need to. (And we can't.)

I'd even argue that $1.3$ is actually $......00000001.3000000.....$ but we don't worry about any of the zeros that aren't saving a place between digits that are "doing work".

Answer 3

Keep in mind, if you are going to introduce significant digits to grade school students, you need to better understand them yourself. 0.4 has 1 significant digit. To think that 3.25×0.4 = 1.300 is to think that the number of significant digits in a result is the sum of the two operands' significant digits. The truth is that it is the lesser of the 2.

In my opinion, the concept is best left for high schoolers, and your kids should just use 1.3 as the answer. That said, there's no harm in asking peers how they do it. I am in a high school, as an in-house math tutor, and often have to defer to how a teacher handles a topic or how the department does, regardless of my own preference.