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© 2016 by One on Epsilon PTY LTD

Do you pay 99 cents or 1 dollar?

June 2, 2017

It happens so often that we see price tags at shops with prices such as $1.99 or $36.99 as opposed to $2.00 or $37.00. Such prices are designed to make us feel that we are paying a bit less. 

 

I used to think the same thing until I bought a bottle of water that was priced $0.99 at the shop and the friendly shopkeeper asked "one dollar please". Shouldn't she have said "99 cents" instead? I started questioning her in my head while handing over a dollar.

 

Here in Australia, there aren't any 1 cent coins. So when I pay by cash, I actually have to pay a full dollar. However, when I pay by card, charging me the exact 99 cents is definitely doable. So why did the shopkeeper ask me for a dollar instead of being more precise? Is it because $0.99 is so close to a dollar that she just automatically rounded 0.99 to 1? Similarly, for an item that is priced $1.99, would I be charged $2 cash?

 

 

Having experienced this at the shop got me thinking about an interesting and related mathematical fact:

 

0.99999...  = 1
(when the digits go on forever)

 

You can see that in 0.99999..., the digits go on forever after the decimal point. We call this type of decimal representation of a number 'recurring decimal'. The question is how can I explain to my students that 0.99999... in fact equals to 1?  Can I use rounding?

 

Being a high school math teacher in Australia, I found that rounding is a necessary aspect of the maths curriculum. Students are expected to be fluent in rounding by the time they reach high school. If students know how to round, should I use rounding as a generalized explanation for other content? 

 

For example, when I introduce the relation between 0.99999... and 1, wouldn't it be easy to just say "0.99999… can be treated as 1 because 0.99999 (without the ...) can be rounded to 1". Notice here 0.99999... is a recurring decimal whereas 0.99999 (without the ...) is an exact value. Use rounding, we can say that "0.99999 can be rounded to 1" or "0.99999 is approximately 1". 

 

Is this an acceptable explanation?

 

If having a university mathematics degree means being trained to be precise, I certainly have carried on this habit into my teaching career. When I am teaching, I always try to see myself through the eyes of my university professors and try to speak the correct mathematical language. Would they approve of "0.99999… can be treated as 1 because 0.99999 (without the ...) can be rounded to 1"? Or is there a sharper explanation?

 

As a maths loving person, I believe that there is always a pursuit of better and sharper explanations for a question. In the following, I am going to try to convince you that 0.99999... = 1 in more rigorous ways. Here is what I can do.

First, we know that 1 divided by 3 can be represented as both a fraction and a recurring decimal, shown below:

We also know that when we have an equation, we can apply the same operations on both sides and obtain a new equation. Here if we multiply both sides of the equation by 3, we get:

This is what we want, isn't it?  

 

Well it is, if you accept playing around with the forms used to represent numbers. The trick here is to view the fraction 1/3 as a recurring decimal, 0.33333.... And then to accept that multiplying this recurring decimal by 3 gives us another recurring decimal, 0.99999....

However, what if you don't wish to represent 1/3 as a recurring decimal? Let me show you an alternative way that doesn't involve fractions. For the fun, let's use a slightly different example, proving 1.99999... = 2. Here is what I can do:

 

First, let X be the recurring decimal:

So,

Did you notice that both X and 10X have the same repeating digits after the decimal point? Why? Because when we multiply 1.99999... by 10, we move the decimal point of 1.99999... to the right by one position, and end up having 19.99999....

 

Now that we have two equations, we can merge them into a single equation. In this case, let's subtract the first equation from the second. This gives us,

Therefore, 

Now solving for X, we get:

So we started with X=1.99999... and ended up with X=2. Therefore, it must be that,

 

1.99999... = 2

 

This method is a neat way to prove 1.99999...=2 for those of you who don't believe 1/3 can be represented as 0.33333....

In fact, 1/3 and 0.33333... are indeed equivalent because 0.33333... can be written as the sum of an infinite geometric series, whose limit is 1/3. Let's leave geometric series for another day. But if you want to find out more, explore this video. Here Burkard Polster illustrates these concepts marvellously.

 

Let's go back to the price tag of the water bottle. If I were not a teacher but working in the business profession, I could come up with ways to legitimately charge my customer $1 instead of $0.99.

 

How? I can just write the recurring decimal, $0.99999..., on my price tag.

 

However, this may seem a bit ridiculous to do as there are too many digits and dots on the tag.

 

But no worries, I know that 0.99999... can be written in simpler ways, such as:

These forms are the most commonly used representations of recurring decimals. You can choose to place either a line or a dot over the digit where it starts to repeat. Probably, if I rewrite the price tag in the shop like this below, customers will never question the precision of the money they get charged again.

 

 

 

 

 

 

 

 

 

 

        

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