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Open-Ended Math Questions?

It is NOT a formal proof!

if you want to take it far enough, no proof is "formal" unless it's written in set theoretic notation.

From a theoretical and pedagogical perspective, the question is whether there is any need to justify the underlying assumption that the addition of infinitely long decimal expansions is not substantially different from the addition of finite ones.

Obviously, you believe there is. And I'd probably agree with you in a graduate mathematics class. I'm not sure I'd agree with you in secondary school.
 
I wish I could remember that problem. It wasn't as simple as the one I presented. It may have been one where you are given 4 sets of numbers and asked which one was certain series. One was obviously correct and another was correct on a logarithmic scale.

I hate forgetting this kind of stuff. I always expect it them to show up as a Jeopardy question.
 
It is NOT a formal proof! The reason being that you're assuming that you know what it means to subtract infinite decimal expansions from each other. Mathematicians tend not to be very happy when you invoke infinity, because it can be very counterintuitive.

Please let me preface this response with an acknowledgement that I know we are both engaging in quite a derail from the OP, but unless someone tells us to stop, I don't see anything wrong with it, as long as we are willing to continue the discussion.

OK, onto the discussion.

First, no one is particularly comfortable discussing infinity. You're right; it's highly counterintuitive and gives everyone fits. It's a terrifically handy concept, however, so we mathematicians and our academic cousins, those delightful oddballs who call themselves "physicists," cannot get around it. At some point, we simply have to dig in our teeth and choke down the concept, then attempt to digest it a bit (see, I really am a mathematician rather than a verbally talented person; I'm not very good with metaphors).

Second, you don't really have to discuss infinity too deeply to grasp on a tangible level the concept of a repeating digit after the decimal that goes on forever. Seventh grade math students tend to accept and grasp it on a level not much higher than a visceral one.

Third, you don't have to assume that "you know what it means to subtract infinite decimal expansions from each other," as you say. Any 7th grade math student is supposed to understand that if you subtract 3 from 3 (or X from X), you get 0. Once a student grasps that, and you have explained what it means for a decimal to repeat indefinitely ("forever" might be the most effective way to describe it in secondary school), then it follows easily that subtracting .999... from .999... equals 0. That, multiplication by 10, and dividing 9 by 9 are the only concepts a student needs in order to balance each side of the equation.

You want the proof to be more formal? Jeez, do I have to state my axioms for you? Do we really do that in the 7th grade, or are they more implicitly understood? To the best of my recollection, we don't have to state them in our proof in the 7th grade. I don't recall hearing the term "axiom" in a mathematics class until college calculus (I must have slept through it in AP high school calculus). Of course, from then on, it was a regular part of every proof.

I think it's possible that you may be burdened in a sense by your mathematics education in approaching these problems, and in particular in your approach to my proof. I get why or how someone might be "too educated" in a sense to relate to it on a simpler level. Nevertheless, I recall how I responded to this proof when I first saw it in the 7th grade, and it was not hard to grasp then. I had never been exposed to the more advanced concepts of infinitely repeating series (I know, I know, that's what .999... is, but at the 7th grade level a student can accept "repeats forever" as a shorthand definition and understand it) converging series, or limits. We didn't go into a lengthy discussion or explanation of what "infinite" means either. I think it's entirely unnecessary at the 7th grade level.

Take 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... . (Ok, first I should prove that this sequence converges, but I'm too lazy to do it right now :D) It has a certain value. But, by rearranging the terms, I can make this sequence converge to whatever the heck I like. (see end of article "Alternating series" on Wikipedia.) And it's knowing about infinity-related loopholes like this that would make a mathematician lose sleep at night having seen your first "proof". You don't know what precisely this 0.999... object is.

So, for that reason, you have to clarify exactly what you mean by 0.999..., without using infinity, to formalise the proof.

Please see my burdened by education comments above. Understand please that I do not mean them to be insulting or denigrating.

My proof is sound, solid, and formal enough for the appropriate and intended audience. I agree with Dr.Kitten that if you want to dissect the hell out of it in a post-graduate seminar, you can spend a great deal of time examining and defining the concept of 0, for instance. That could be the subject of an entire semester's worth of study, and it might prove very interesting. You simply don't need that level of analysis in order to apply it in the setting of my proof.

AS
 
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To be honest, you won't find many "open-ended" maths questions, because if a question has more than one way of being interpreted then it's usually considered a bad question, for being imprecise.

If you're looking for fallacies like the Monty Hall Problem, however - my favourite such fallacy is the "All Triangles are Isoceles" proof, which you can find halfway down the page at http: //euler.slu.edu/~clair/puzzlers .html (I can't post links, so you'll have to delete the spaces). You may also be interested in proofs that 1=2; google "1=2 proofs" and you'll get a whole stackfull of 'em.
Perhaps the price of the horse went up due to inflation? Hmm?

Also, proof that 1= 0.9999999999.....(to infinity, not beyond)
 
My proof is sound, solid, and formal enough for the appropriate and intended audience.
I derailed further than you; I wasn't talking about a 7th grade audience. You're helped by the fact that the "intuitive" meaning of those infinite (there's that word again) digit expansions and operations happens to be right. If you are dealing with a more advanced audience (undergraduate, really, not post-doc), however, dealing with the rigorous fiddly bits properly is rather illuminating and you learn a decent amount about real numbers that way.
 
It's like saying if there are an infinite number of universes, there has to be one where purple monkeys are the dominant lifeform on earth.
No, because even infinite sets can be limited. For instance, the set of integers is infinite, but it does not contain the letter H.
 
Not to open up a can of worms, but isn't 0.9999...=1 proven by the equation 1-0.9999...=0.0000....=0?

He he. On practically every internet discussion forum I've seen, merely mentioning that .999... = 1 is opening a can of worms.

We've seen a little bit of evidence of that in this very thread. It's Pandora's Theorem.

AS
 
I'm just starting a Bachelor of Education and my teacher for Gr. 1-8 Mathematics told us in the first class that she had a math problem with no answer next week. She shared the anecdote that a student from last year had written her afterwards asking what the "real" answer was, which was laughable because there wasn't one. Now I'm sure there are good questions like that, but I was shocked when this turned out to be the question:

I saw it and immediately saw the answer was +20. Because of what she'd said in advance, I tried numerous strategies and always got the same answer. So I was confused. Most people in the class ended up getting +20 as well, which puzzled the professor until one girl admitted to getting +10. Apparently all her other sections got several different answers to this question (scary!!). She tried to explain how she'd done it, and she'd used a confusing verbal approach, which to me isn't a valid strategy when you talk yourself into a wrong answer. So I asked what exactly what was open-ended about this question, because she'd been presenting it as though a variety of answers was a good thing. Apparently she'd done it herself individually and with a team of other adults and gotten different answers. But I argued that since we were dealing with concrete numbers, this was not a question with different possible answers. I recognize the value of questions with multiple solution strategies, but I told her I was sure I could find some better "open-ended" problems. I mentioned the Missing Dollar and Monty Hall and forwarded those to her, but I was wondering if anyone else knew some good ones that aren't so pathetically easy. Or even better, that have multiple valid answers. Is that even possible in say, high school level math?

Welcome to the intellectual rigor of the School of Education.
 
I derailed further than you; I wasn't talking about a 7th grade audience. You're helped by the fact that the "intuitive" meaning of those infinite (there's that word again) digit expansions and operations happens to be right.

True. I suppose that's why my math teacher taught us the proof in the 7th grade, when most of my classmates and I were 12.

If you are dealing with a more advanced audience (undergraduate, really, not post-doc), however, dealing with the rigorous fiddly bits properly is rather illuminating and you learn a decent amount about real numbers that way.

I don't doubt it. I'm sure teaching a topic is quite an education in itself.

Actually, I'm embarrassed at how little number theory and upper level university math I recall now. Much of it I can recognize and put into context and grasp superficially when I see it, but I have simply forgotten way more than I would have expected back then when I was learning it in university. Of course, it has been 21 years since I was graduated with a B.S. in math, and I have exercised my math muscles very seldom and under very limited circumstances since then. I suppose it's use it or lose it when it comes to very specific, high-level cognitive functioning and understanding.

Either that, or I'm simply getting feeble and forgetful in my middle age.

AS
 
Okay, I thought of one. What is the next number in this series?

2, 4, ...

is it a) 5, b) 6, c) 7 or d) 8

I do remember a problem like this on a college test. There are two correct answers. The next number is 6 if the series is just adding 2 to the previous number. The next number is 8 if the series is multiplying the previous number by 2.

All of these are correct :cool:

a) 2, 4, 5 ... Includes (to name a few) the sequences (1) Numbers that are the sum of 2 squares. (2) Not divisible by 3 (3) Numbers of the form 2^i*5^j

b) 2, 4, 6 ... Includes (to name a few) the sequences (1) Differences between consecutive primes. (2) Primes minus 1 (3) Numbers n such that n^2 + 1 is prime

c) 2, 4, 7 ... Includes (to name a few) the sequences (1) Tribonacci numbers (2) Fibonacci numbers - 1 (3) Odd number of 1's in binary expansion

d) 2, 4, 8 ... Includes (to name a few) the sequences (1) Expansion of (1-x)/(1-2x) in powers of x (2) Number of configurations of the sliding block 8-puzzle that require a minimum of n moves to be reached, starting with the empty square in one of the corners. (3) Palindromic in bases 3 and 10

Yes.. if a test has a "whats the next number in the sequence?" question then the results of that test is ******** unless all answers are acceptable for that question.
 
My proof is sound, solid, and formal enough for the appropriate and intended audience. I agree with Dr.Kitten that if you want to dissect the hell out of it in a post-graduate seminar, you can spend a great deal of time examining and defining the concept of 0, for instance. That could be the subject of an entire semester's worth of study, and it might prove very interesting. You simply don't need that level of analysis in order to apply it in the setting of my proof.

AS
Let N = 9+90+900...

10 N = 90+900+9000+... (multiple both sides by 10)

10N - N = (90+900+9000+...) - (9 + (90+900+9000+...)) (subtract N from each side)

9N = -9

N = -1

9+90+900... = -1

I'd argue that even in 7th grade, if a student asked why the above doesn't work just as well as the other proof, some discussion on well-defined terms might be warranted.
 
Let N = 9+90+900...

10 N = 90+900+9000+... (multiple both sides by 10)

10N - N = (90+900+9000+...) - (9 + (90+900+9000+...)) (subtract N from each side)

9N = -9

N = -1

9+90+900... = -1

I'd argue that even in 7th grade, if a student asked why the above doesn't work just as well as the other proof, some discussion on well-defined terms might be warranted.

Your proof contains an egregious algebraic equation error. Please look at your step 2 and tell me it's the same as mine.

I'll wait.























OK. Time. I hope you found it. Here it is in case you didn't.

10N - N = (90+900+9000+...) - (9 + (90+900+9000+...)) (subtract N from each side)


What you have done is simply add 9 to your series. I didn't do that in my step 2. What I did is multiply BOTH sides by 10.

In an algebraic equation, each time you perform an arithmetic function on one side of the equation, you must perform the same function on the other side.

What you have done in step 2 is to multiply by 10 on the left side, but ADD 9 to the other. Oops. It's incorrect because you violated the most basic rule of algebra. Please try again.



There is actually another major problem with your proof, and it is in step 1.

The algebraic proof that .999... = 1 is not the result of plugging in some randomly chosen and ill defined series as N. It's a proof that .999... is exactly, not approximately, equal to 1. Yes, much later we learn why that is so in terms of limits of converging series. That's not why it's true in simple algebraic terms, and it's no criticism of the proof to try to point out that .999... is ill defined. That's because it isn't.

Most students capable of earning an A or a B in algebra in 8th grade or pre-algebra in the 7th should be able to grasp that .999... is well enough defined such that subtracting it from itself yields the result of 0. The proof wouldn't work if that were not so. It is so, however, because X - X = 0, regardless of how X is defined.

Thus 9.999..., which is .999... TIMES 10 in my proof, (not .999... PLUS 9, which is the result of your incorrect application of two different arithmetic functions on each side of an equation, which is a violation of basic algebraic rules -- i.e., in order to keep the EQUATION balanced, one must perform the same function on each side), is an application of the simple fact that N - N = 0. In step 2, it's simply equivalent to noting that 10N - N = 9N.

I kept both sides of the equation balanced. You didn't.

AS
 
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It is NOT a formal proof! The reason being that you're assuming that you know what it means to subtract infinite decimal expansions from each other. Mathematicians tend not to be very happy when you invoke infinity, because it can be very counterintuitive.

Take 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... . (Ok, first I should prove that this sequence converges, but I'm too lazy to do it right now :D) It has a certain value. But, by rearranging the terms, I can make this sequence converge to whatever the heck I like. (see end of article "Alternating series" on Wikipedia.) And it's knowing about infinity-related loopholes like this that would make a mathematician lose sleep at night having seen your first "proof". You don't know what precisely this 0.999... object is.

So, for that reason, you have to clarify exactly what you mean by 0.999..., without using infinity, to formalise the proof.

This has been great - This problem was supposedly discovered/thought up by one of a group of friends of (that was his story) in high school in 1962 or 3. We asked multiple math teachers about it and I asked college math teachers about it through '67 with none of them mentioning the items explained here. Gave up on finding out it's flaw/reason for working at that point (not a mathematician). Just been demonstrating the basic form since. Thank you kindly for the info!!
 
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AS, I think the error isn't exactly as you've presented it.

If N = 9+90+900...

Then 10 N = 90+900+9000+...

Then 10N - N = (90+900+9000+...) - (9 + 90+900+9000+...)

The error is that when GA renders the above step as
10N - N = (90+900+9000+...) - (9 + (90+900+9000+...))
it appears to yield -9 only because of the bracketing and linear representation.

In order to carry out the subtraction, however, that leftmost 9 would be paired with the 90, the 90 with the 900, and so forth.

In other words, the "repeat" to the right of the series is being used to disguise a displacement of the terms in the equation.

At least, that's how it appears to a word nerd.
 
AS, I think the error isn't exactly as you've presented it.

If N = 9+90+900...

Then 10 N = 90+900+9000+...

Then 10N - N = (90+900+9000+...) - (9 + 90+900+9000+...)

The error is that when GA renders the above step as
10N - N = (90+900+9000+...) - (9 + (90+900+9000+...))
it appears to yield -9 only because of the bracketing and linear representation.

In order to carry out the subtraction, however, that leftmost 9 would be paired with the 90, the 90 with the 900, and so forth.

In other words, the "repeat" to the right of the series is being used to disguise a displacement of the terms in the equation.

At least, that's how it appears to a word nerd.

That's because he's equating the function of expanding his series backwards to show the hidden 9 with the function of multiplying by 10, and they are manifestly not the same arithmetic functions. He's taking a series he chose that wouldn't work in the proof (and the only one that will is .999..., because that's another way of defining 1 using limits, but his series simply isn't) and expanding the series by showing an additional member in the series.

The point you and he and everyone else criticising the algebraic form of the proof seem to miss is that you don't need to spend an inordinate amount of time exploring series in order to demonstrate or understand the proof. You just need to grasp the rather simple concept of a repeating decimal, without the necessity of breaking it into its component parts. The pre-algebra student doesn't need to regard .999... as the sum of .9 + .09 + .009 + .0009, etc. She can understand that .999... repeats forever. Simple enough.

Now subtract it (.999...) from itself. 0.

Multiply it (.999...) by 10. That means moving the decimal point one place to the right. That's the crucial part of the proof that is rather simple to understand in the 7th grade. It's not a bit of trickery. It's a simple application of what it means in base-10 to multiply any number by 10. It doesn't matter if the number is the sum of a series, the number 3, or 1,000,000. Multiplying by 10 always means moving the decimal one place to the right. For positive integers, we think of it as adding a 0 to the end, but that's just another way of moving the decimal one place to the right.

It's when you subtract N from 10N that the proof gets interesting, because that's where the simple dropping of the repeating decimal to the right of 9.999... yields the result that there is no more repeating decimal. That result is not because of limits of converging series, or the continued expansion of the sum of a series; it's because X - X = 0.

KISS. Keep it simple, Stupid.

(You realize I'm not calling you Stupid, of course).

AS
 
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Gave up on finding out it's flaw/reason for working at that point (not a mathematician). Just been demonstrating the basic form since. Thank you kindly for the info!!

His info is wrong. There is no flaw in my proof (it's not "mine," as in I'm not claiming credit for originating it). Sorry to disappoint you.

.999... is exactly, not approximately, equal to 1. Period. No tricks, no sleight of hand, no word games, no flaw.

My proof is the simplest algebraic form of demonstrating (proving) that .999... is in fact 1. Other mathematicians in this thread who are criticising it are trying to remove it from the simple 7th grade level and place it in the context of a calculus course examining series. That's roughly like trying to take an explanation to kindergarteners that "cat" is spelled C-A-T, and rendering it into a discussion of the etymology of the word, taken from the Latin cattus.

O.E. (c.700), from W.Gmc. (c.400-450), from P.Gmc. *kattuz, from L.L. cattus. The near-universal European word now, it appeared in Europe as L. catta (Martial, c.75 C.E.), Byzantine Gk. katta (c.350) and was in general use on the continent by c. 700, replacing L. feles. Probably ult. Afro-Asiatic (cf. Nubian kadis, Berber kadiska, both meaning "cat"). Ar. qitt "tomcat" may be from the same source. Cats were domestic in Egypt from c.2000 B.C.E., but not a familiar household animal to classical Greeks and Romans. The nine lives have been proverbial since at least c.1562. Extended to lions, tigers, etc. 1607. As a term of contempt for a woman, from c.1225. Slang sense of "prostitute" is from at least 1401. Slang sense of "fellow, guy," is from 1920, originally in U.S. Black Eng.; narrower sense of "jazz enthusiast" is recorded from 1931. Catcall first recorded 1659; catnap is from 1823; catfish is from 1620; catwalk is from 1917. Cat's-cradle is from 1768. Cat-o'-nine-tails (1695), probably so called in reference to its "claws," was legal instrument of punishment in British Navy until 1881. Cat's paw (1769, but cat's foot in the same sense, 1597) refers to old folk tale in which the monkey tricks the cat into pawing chestnuts from a fire; the monkey gets the nuts, the cat gets a burnt paw. To rain cats and dogs (c.1652) is probably an extension of cats and dogs as proverbial for "strife, enmity" (1579). Cat-witted "small-minded, obstinate, and spiteful" (1673) deserved to survive. For Cat's meow, cat's pajamas, see bee's knees.

Source: http://www.etymonline.com/index.php?term=cat

True, but irrelevant and inappropriately and unnecessarily confusing to the student.

AS
 
That's because he's equating the function of expanding his series backwards to show the hidden 9 with the function of multiplying by 10, and they are manifestly not the same arithmetic functions. He's taking a series he chose that wouldn't work in the proof (and the only one that will is .999..., because that's another way of defining 1 using limits, but his series simply isn't) and expanding the series by showing an additional member in the series.

No. The series expansion of N contains, as a subset, the series expansion of 10N formed by the simple expedient of moving the decimal point one place to the right (hence producing the extra 0 on the end).

(1)
N = 9/10 + 9/100 + 9/1000 + 9/10000 ... (AS's decimal, as a series)
10N = 10*9/10 + 10*9/100 + 10*9/1000 ... (multiply both sides by 10)
10N = 9 + 9/10 + 9/100 + 9/1000 ... (each term can be simplified)
10N = 9 + N (I recognise that series)
9N = 9 (subtract N from each side)
N = 1 (QED)

(2)
N = 9 + 9*10 + 9*100 + 9*1000 ... (GA's series)
10N = 10*9 + 10*9*10 + 10*9*100 ... (multiply both sides by 10)
10N = 9*10 + 9*100 + 9*1000 ... (each term can be simplified)
10N = N - 9 (I reconise that series)
9N = -9 (subtract N from each side)
N = -1 (QED)

In (1) the series is convergent, N therefore has a finite value; for finite numbers, 10N - N is a defined mathematical operation. In (2) the series is divergent and N (in this case) will be oo; for infinity, 10oo - oo is an undefined mathematical operation.

AS presented an algebraic recipe that 'proves' 0.99999... = 1, the same recipe was used by GA to show that 99999... = -1. The recipe may simplistically show that .9999... = 1 but it is not mathematically rigorous.

The point you and he and everyone else criticising the algebraic form of the proof seem to miss is that you don't need to spend an inordinate amount of time exploring series in order to demonstrate or understand the proof. You just need to grasp the rather simple concept of a repeating decimal, without the necessity of breaking it into its component parts. The pre-algebra student doesn't need to regard .999... as the sum of .9 + .09 + .009 + .0009, etc. She can understand that .999... repeats forever. Simple enough.

Now subtract it (.999...) from itself. 0.

Multiply it (.999...) by 10. That means moving the decimal point one place to the right. That's the crucial part of the proof that is rather simple to understand in the 7th grade. It's not a bit of trickery. It's a simple application of what it means in base-10 to multiply any number by 10. It doesn't matter if the number is the sum of a series, the number 3, or 1,000,000. Multiplying by 10 always means moving the decimal one place to the right. For positive integers, we think of it as adding a 0 to the end, but that's just another way of moving the decimal one place to the right.

N = ...999999
10N = ...999990
10N - N = ...999990 - ...999999 = -9
9N = -9
N = -1

No component parts there that I can see.
 
The point you and he and everyone else criticising the algebraic form of the proof seem to miss is that you don't need to spend an inordinate amount of time exploring series in order to demonstrate or understand the proof.
Well, like I say, I'm a word nerd. As far as I'm concerned, the matter was settled at 1-0.9999...=0.0000...=0, because if X-Y=0, then X=Y.

And the 10N proof didn't make sense to me b/c it appeared to involve a shifty shift.

Your point may be right, but who knows -- I don't trust you much AS, because while I may not know beans about math, I do know this: Don't put your life in the hands of no backwoods Southern lawyer. ;)
 

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