Jim Hays et. al. reminded me of this old classic from 8th grade math:

Consider $\dfrac{9}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000}+ ...$, which can also be written as $0.\overline{9}$ or $0.999999 ...$?

Is that less than 1, or equal to 1?

Naturally, you'll need to show them this method [from Constance Mueller (and several others) via email]:

If N = 0.9999999 ... and 10N = 9.99999 ...

10N= 9.99999...

- N = 0.99999...

--------------------------

9N = 9

N = 1

If 1/3=.3333... And 2/3=.6666... Then 3/3=?

ReplyDeleteI've shown the 1/3 = 0.3333... => 3/3 = 0.9999... method.

ReplyDeleteI've shown the N = 0.9999.... => 10N = 9.9999.... method.

I've shown the geometric series with a = 9/10 and r = 1/10 method.

Usually, in my experience, one of these methods ends up making sense for a kid.

Now, something else that kids struggle with: 0! = 1. I teach AP Calc BC, and I've seen some write this as 0. Never mind that in a Taylor series, you would end up with your initial term being divided by "0." They just say that's 0. :-p

If 0'9999... isn't equal to 1, find a number between them. After a while, most pupils agree (somehow) that they are the same number

ReplyDeleteJason: Here's one of the perils of trying to write general formulas. The n'th term (n = 0, 1, 2, ...) of the Taylor series for e^x about x=0 is (x^n)/n!, and I'm guessing your comment has to do with interpreting this for n = 0. Have you ever noticed that when n = 0 we get 0^0 to show up when x=0 is plugged into the expansion? Certainly e^0 is defined, but we don't reproduce e^0 when we plug x=0 into the Taylor expansion (about x=0) unless we agree notationally (just for this, if nothing else) to set 0^0 equal to 1.

ReplyDeleteConsider 9+90+900+9000 which can also be written as a 9 with a bar over it (like above just without the decimal point) or as ....9999999. My friend says this is just another way to write -1. She claims that if N=....999999999 then 10N=.....9999999990, so N-10N=9 so -9N=9 so N=-1. So I guess then that .....999999999 = -1. That doesn't seem right, but she has a proof. Let's try a geometric series with a = 9 and r = 10. That gives ....9999999 = 9/(1-10) = 9/-9 = -1. Hmmm..maybe it is right. That was using a formula and that can't be wrong. Maybe I can show there is no number between them. If I take ...999999 - - 1 that would be the same as ...999999 + 1 . So the last digit is 0 and I carry the 1 which gives another 0 and carry the 1, and so on and so on. So I just get a whole string of 0's. which is of course 0. So if the difference of ....9 and -1 is 0 they MUST be equal!

ReplyDeleteThe problem is not in trying to convince students that .9999999.... is equal to 1. The problem is that the students SHOULDN'T be convinced. They should reject these repeating decimals all together. The number .32 has meaning. The 32 is telling us we have a fraction with numerator 32 and the location of the decimal point tells us the denominator is 10 the 2nd, or 100. So .32 is BY DEFINITION 32 hundredths. But what does .3333..... mean. It should mean a fraction whose numerator is ....33333 and whose denominator is 10 to the infinity. That's nonsense. I would rather live in a middle school and high school world where we are willing to accept that all decimals can be written as fractions (which can teach us how to do arithmetic with them) but that some fractions (like one-third) cannot be written as decimals. That's one of the reasons we should learn to work with fractions. Not it is true that every number can be approximated by a decimal to as many decimal places as we'd like (sometimes in more than one way!) and that's a useful fact. It allows us to make use of standard calculators if we're willing to approximate. Students who grow up with this view wouldn't be saying that pi goes on forever any more than the number one-third goes on forever. An irrational number is not one that goes on forever, it's a number that can't be written as a fraction.

Galois: The problem that is not made explicit in your first paragraph is that there is a limit involved, and in the case of geometric series the limit exists when the common ratio is between -1 and 1, but not if the ratio is outside these two values. However, there are other ways of *interpreting* infinite sums than as limits of partial sums (google "summability theory"), and so some of those seemingly nonsensical manipulations do have a place (just not here).

ReplyDeleteI'm not sure what the problem is in your second paragraph. The definition of the value of an infinite decimal is that it is the limit of the finite decimals obtained by taking initial segments of the infinite decimal, so getting arbitrarily close just means the limit is what it's supposed to be. Now there is the problem of trying to define *in advance* what numbers these infinite decimals are supposed to represent (note that saying the finite decimals approach the infinite decimal as a limit assumes that the infinite decimal is a number that is already available and waiting for us to approach), and this is a topic taken up in advanced undergraduate mathematics (the construction of the real numbers).

I understand the logical flaw in the first paragraph and the construction of real numbers. My concern is with the order in the curriculum. Convergence of series is a topic generally covered in a Calc II or an AP Calc course. Repeating decimals and decimal expansions of irrationals are covered in middle school! I don't believe one should try to convince a middle school student with arguments that depend very much on the notion and significance of convergence. Nor is there a need to do so.

DeleteI think a better approach in middle school is not trying to claim that .333.... or 3.141.... is a number. Rather introduce decimals as an alternate notation for fractions whose denominator is a power of 10 (see above). After seeing what that notation does for us, one might get into a really good discussion about what numbers can be written as decimals, why, and how. This may naturally lead to a discovery that all numbers (ie. points on the number line) can be APPROXIMATED by decimals to as close as we'd like by going out sufficient decimal places. When exploring long division, students may discover a way to obtain these great approximations for rational numbers. At that point one can even introduce .3 repeating but at this point the symbol it is NOT a number. Rather it is an indication of how to find APPROXIMATIONS for one-third. This agrees with how I would expect students to use .3 repeating in problems. I would not want them to directly compute with it or use it as an exact number. Rather, depending on context and tools, either round to an appropriate decimal place OR use the exact number one-third. Later, when introduced to irrational numbers perhaps the question comes up of how to find decimal approximations for them. The significance of being irrational, though, is not about what form those approximations look like. It is the inability of an irrational number to be written as a fraction. In this approach there is no argument about whether .9 repeating equals 1, only on whether it provides good approximations for 1. The consensus on that is likely to be that it provides some pretty good approximations, but since 1 can be written as a decimal it's unlikely we would use those approximations in place of just using 1.

Nope. Even though 0.999999.... times 10 SEEMS like 9.999999...., it actually isn't. The RULE OF INFINITY states that any number put after the repeating digit, in this case, will STAY and stay visible. In this case, 10(0.99999...)=9.99999.....0, and that zero stays! As well, all other numbers will add 2 of the same infinite digit.

ReplyDelete9.999999......9|0

- 0.999999......9|99 is actually 1.000000...11!

Also: 1/3=0.3333......4, NOT 0.333333...3.

Oh yes... you might want to go to themathpro.wordpress.com for more information about this!

ReplyDelete0.99999999... equals 99999999...../1000000....., and the 9s and 0s are of the same quantity. So, denominator minus numerator is 1, NOT 0, the required is 0 to reach 1. So, 0.9999.....<1.

DeleteHere is a number between 0.999999... and 1:

ReplyDelete0.9999999..... 1, 0.999999.......2, etc.