I’m going to devote the rest of my life to proving that wrong.

-A student, reacting to the proof that 0.999… = 1.

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I’m going to devote the rest of my life to proving that wrong.

-A student, reacting to the proof that 0.999… = 1.

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So great.

Thanks, and thanks for the tweet – it took me a while to figure out why my readership had jumped 2000%.

I love it. It also reminds me of the fact that we actually “chose” .9 repeating to be equal to 1, as a convention. It is, not fact, mandated by the mathematics. It’s a highly convenient choice, and not arbitrary, but still just a convention.

This is the kind of thing I wish I knew more about, so I could give students a good way of experiencing that choice … there are advantages and disadvantages to having studied math only in applied contexts. Lots of room to keep learning!

Wait, really? How can they NOT be equal? There can’t be any number between them, and don’t any two rational and/or irrational numbers have another in between them?

Maybe if we decide 0.999… is neither rational nor irrational…?

I’m writing this sentence just to make a declarative statement at least once in this comment.

We’re talking

infinitesimalshere. That whole “There can’t be any number between them” thing? If you assume that there is a number smaller than all positive rational numbers, but greater than zero, and if you call it aninfinitesimal, then there can in fact be a number between them. And a whole different approach to calculus is born. If memory serves, this alternate approach has nearly all the same results but is considered a shaky foundation for the whole affair.The point for the student in question here is that if 0.999…≠1 then he/she has got to accept that idea of a positive number greater than zero but less than all positive rationals. Tough pill to swallow.

I wish I knew more about it, too. I just read about it in a book…I can’t remember the title, and I’m not sure I completely understood it, either. And it certainly doesn’t take away from the greatness of how that students took up the agency to challenge something that didn’t make sense to him.

I’m fairly certain that that’s not true. (The part about the mathematics does not require it.)

If 1/3 = 0.333…, then 0.99999… = 0.333…+0.333…+0.333… = 1/3 + 1/3 + 1/3 = 1

I really don’t see any way around that. (Incidentally, that’s the easiest-to-swallow proof that I know of). I’ve seen several totally independent proofs of the fact. It may be equivalent to another assumption in our number system, but it’s definitely an integrated part.

I wish you remembered the name of the book where you read that.

Here’s my thinking about this:

Saying that “.9 repeating need not be equal to 1” is epistemologically similar to the notion that parallel lines can intersect. In certain choices of geometries, parallel lines can do do intersect. In particular choices of arithmetic systems, .9 repeating doesn’t have to equal one. The point is that it is a choice. I will also, when I get a chance, dig deeper into finding the book I read.

I see. I find calling it a choice a bit disingenuous then. It’s a result of the choice of a base-ten numeral system. But you’ll have an analogous fact in any base: 0.77777… = 1 in octal numerals, and 0.FFFF… = 1 in hexadecimal. And I can’t imagine anyone making the choice of numeral system based on the desire to have 0.999… be something other than 1.

Of course, I’ve also always found it disingenuous when people say “parallel lines actually can intersect” (or similar) without providing context for that statement.

I don’t believe that’s what bw meant by

arithmeticsystem, though. You are totally correct that 0.FFF…=1 hexadecimal, and therefore 0.999…≠1 hexadecimal. But that’s not what he meant. It’s the infinitesimal business, I’m quite sure.Still waiting for that citation, bwfrank!

Apologies if I misunderstood. I think I’ve gotten used to hearing people use arithmetic system to mean numeral system. (Does “arithmetic system” have a widely-accepted meaning? I’m not aware of one.)

Another reason I think I interpreted it that way is that I don’t think that existence of the hyperreals changes the fact that 0.999…=1. If you accept the existence of infinitesimals, then there is a nonreal hyperreal number smaller than 1, but larger than every real number less than 1. But that number is not 0.999… because 0.999… is a real number.

It may be that “arithmetic system” was used to mean “nonstandard model of arithmetic.” Hopefully bwfrank will be able to find that reference.

My earlier examples may be invalid, but the point remains: there are axiomatic choices to be made (whether in arithmetic or geometry) and we get results from those, but that’s not the same as claiming that the result was chosen for convenience.

FYI: http://en.wikipedia.org/wiki/Hyperreal_number

And just to state explicitly what most conversants here have left implicit, Many props to the student in question!

a) Word. That kid rocks.

b) With the caveat that I’ve never studied them, my understanding of hyperreals makes me second Kate, thus disagree with Christopher: in the hyperreals, .9999… = 1. There are hyperreals <1 and greater than any real number <1, but they are not .9999…, which =1.

c) I can think of one (somewhat contrived but mathematically coherent) context in which .9999… does not equal 1. The thing about infinite decimals is that they are shorthand for the limits of convergent series: 0.999… is shorthand for the limit of the series 9/10 + 9/100 + 9/1000 + … This series is convergent if the terms of the series are seen as rational numbers with the usual Euclidean metric d(x,y) = |x-y|. However, if you forget the usual metric and impose the 10-adic metric (which I don’t believe anyone would ever do except to prove this point – people make legit use of p-adic metrics but only for prime p – number theorists, correct me if I’m wrong), then the series is actually divergent. On the other hand, the series 9 + 90 + 900 + … becomes convergent and converges to -1, as discussed in the wikipedia article on p-adic numbers.

d) I suspect that in any field carrying whatever topological structure is needed to define convergent series and guarantee their limits are unique, then

if9/10+9/100+9/1000+… converges, then 1 is the only choice for what it converges to. I feel like the standard proof (0.999… = x; 9.999… = 10x; subtracting, 9 = 9x; then x=1) should go through, but I’m not 100% sure. If a field has a topology, does that mean multiplication by nonzero elements is a homeomorphism? If true, then I think it should work:Say the series converges. Let the limit be x. This means every open set containing x contains all but finitely many partial sums of the series. Consider 10x. Multiplication by 10 is a homeomorphism, so every open set U containing 10x corresponds 1-to-1 with an open set V containing x (got by dividing everything by 10), which thus contains all but finitely many of the partial sums. Then U contains all but finitely many of the partial sums of the multiplied-by-10 series 9 + 9/10 + 9/100 + … Thus this series converges to 10x. But by comparing the nth partial sum of latter series to the n-1th partial sum of the original 9/10 + 9/100 + … we see that the series also has to converge to 9+x. Since limits are unique, 9+x = 10x. x=1 follows by solving the equation.

e) Dan, I hope you share with the kid in question how much discussion they provoked!

One question I’ve asked students when this comes up…what does it mean for two numbers to be different? Usually keeps them busy for a bit.

… It’s actually keeping me busy for a bit!