It’s almost too easy to explain why .999… does not equal 1. Even fools can see the reason right away — it appears to take a college education to get it wrong, as counter-intuitive as that may be.

If .999… = 1, and X – X = 0, then 1 – .999… should equal 0. It doesn’t. You get .000…01. How do I know this? Inductive reasoning.

1.00 – .99 = .01

1.00000 – .99999 = .00001

1.0000000000 – .9999999999 = .0000000001

1 minus point 9 repeating for any length equals point 0 repeating for one digit shorter than the string of 9s followed by a single 1.

There’s no reason to suppose that this pattern suddenly changes just because the string of 9s is infinitely long. For any finite length this is true, so for an infinite length it should still be true.

But let’s not stop there. If .999… = 1 then X squared = X squared. Does it? No.

.999 squared = .998001

.99999 squared = .9999800001

.999… squared = .999…980….0001

None of these equals 1 squared (which is 1). Again, I’ve demonstrated that any finite length of 9’s squared does not even come very close to being 1. It’s true that it gets closer as the string gets longer, but it will never change to become 1.000… No matter how long the number is, you’ll still be multiplying a 9 by a 9, and that will give you a 1, not a zero.

It’s even more revealing when you multiply both sides of the equation by 2.

.999…(2) = 1.999…98

1(2) = 2

Here you see that the distance between both sides doubles, which is consistent with any unbalanced equation. Take any two numbers, multiply both of them by 2, and you’ll always find that the difference between them also doubles (because if X-Y=Z, then 2X-2Y=2Z, where X and Y are any two numbers and Z is their difference).

There is literally nothing you can do to both sides of the equation .999… = 1 and have them still coming out equal. Many people assume that 9.999… = 10 (by multiplying both sides by 10) or that 3.999… = 4 (by adding 3 to both sides), but this is a logical fallacy known as “begging the question“… the conclusion is derived from prior belief that .999… = 1, not because it’s otherwise provable that these equations are true.

One final note: While I welcome criticism, please don’t tell me that I’m doing these problems wrong unless you can explain how they are done right. Don’t tell me that .999… multiplied by 2 does not equal 1.999…98 unless you can explain how those numbers are multiplied correctly and come out to a figure that supports your conclusion. If you can’t offer an alternative explanation, there’s a good chance that there isn’t one.

you should be careful with induction, induction only works for natural numbers… so yes, if there would be an end in .333… you would be right, but it’s not… one way to define .\bar{3} is actually 3/9, that’s if you look at it, as a rational number… Another way to define it, is in a context of limits… your understanding of .333… is more like the inductive class of .33..3, you assume that if for any number .33..3 your assumption is valid it has to for .333.. as well, but that’s not the case… another simple example is (a_n)_{n\in N} with a_n=1/n, it converges against 0 (it’s rather easy to prove). but in fact for all n\in N it is greater as 0.

“Induction only works for natural numbers”… because mathematicians have arbitrarily defined that it does? Inductive reasoning ought to work in any case where a pattern has been established and is expected to continue. I demonstrated not only that the 1 carries when dividing 1 by 3 but also that there is always a remainder. You haven’t given me a reason to assume that it hasn’t except by asserting that it hasn’t.

The only reason to have .9… the same as 1 is convenience. Don’t read more into it. Of course it looks odd. As you say, however long you keep adding 9s to the string, there is no time where the 9s spontaneously change into zeros, and a 1 pops up in front.

All that is said is that, if ever your calculations pop up a .9…, you can replace that by 1.

Why would .9… pop up? Well, for instance when you multiply .3… with 3. You said yourself that you get .3… when you divide 1 by 3, and I am sure you agree that if you do multiply .3… by 3 you will get .9….

You say that 1 divided by 3 is ALMOST .3.., but that isn’t quite right. It is EXACTLY .3…, as long as you don’t stop putting more 3s. As that is what the … mean. Never EVER stop putting them.

You say you get a remainder 1? Nope… you divide again, add an other 3. It never stops.

You had a problem multiplying .4… with 9, as there is a 6 ‘at the end’. But you have to add a 3 to that 6, it carries over from 4*9+3 of the position to the right… remember, there is always an other position to the right, as the the row if 4s NEVER STOPS.

By the way, it is easier to calculate .4… times 9 by first dividing by 4, you get .1…, and if you multiply that with 9, you get .9… and that is the same as 1. Then multiply by 4 to get 4.

Oh, and according to the wiki page you refer to, LOGICAL (as opposed to MATHEMATICAL) induction isn’t as good as logical deduction. Conclusions of induction can just be said to be likely, albeit VERY likely in case of strong induction, while deduction delivers certain conclusions.

Inductive arguments can never “necessarily” be true, but they become “stronger” with each example. When we’re talking about a pattern continuing on into infinity, we have an infinite number of examples to strengthen our argument, so I’m on pretty solid footing.

To answer this problem of “there is always another position to the right” of an infinitely long string, I’ll ask you to read the post about reification. There are cases where we don’t have problems containing an infinity (such as assuming that an infinite number of real numbers exists between 0 and 1) because numbers are ideas rather than physical objects. Please stop treating them like physical objects. We can’t get to the “end” of a real infinity, but we can imagine what we’d find at the end of a hypothetical infinity because such infinities of the mind can have an end.

It’s also important to note that (like in every post) I ask not just for you to explain why you think I’m wrong but an argument about how your position is right. You don’t think that multiplying .444… by 9 gives us a “6” at the end, but you haven’t put forth an argument for why it ought to have yet another “4” at the end, as you believe it does.

Inductive reasoning tells you that 2(.999…)=1.999… if you look at the steps in algorithm instead of looking at what your algorithm tosses out. I quickly figured this out when I was first learning how to do arithmetic with repeating decimals.

I think you’ll need to explain this a bit better. Could you explain the “steps in [the] algorithm”?

Computer programming stuff. I know a little. Not sure what to call this stuff. Like I said in the comment to your video

.4444…. + .8888…. = 1.3333…

When you add 8 to 4 you get a 1 to carry over. If it is infinite it doesn’t matter where you start because 8+4 will always give you 12. You carry the 1 over and your 2 becomes a 3 by the one carried over from the right.

However if you’re only looking at answers

.44 + .88=1.32

.444 + .888=1.332

.4444 + .8888=1.3332

You can’t see how changing the input will change the output. You have to carefully examine the rules and making sure you’re applying them when generating an output. Patterns are fine but sometimes there’s a specific reason why you get a result in one situation and not another even though they look similar.

In this case the with finite numbers you can’t take into consideration the numbers that will always be to the right of you.

“There’s no reason to suppose that this pattern suddenly changes just because the string of 9s is infinitely long. For any finite length this is true, so for an infinite length it should still be true.” Yes there is. If there’s infinitely many 0’s, you will never reach the 1. There is no 1 at the end. The 1 ceases to exist. Infinity doesn’t mean “a really large number, bigger than you can possibly imagine and if you can imagine a big enough number, imagine one bigger”, infinity means “without end”, and that means that 1 at the end doesn’t exist.

You don’t have to agree with my math or my concept of numerical infinity to get my point. Can you demonstrate that .999… squared equals 1? I offer a solution to the problem, but even if you don’t like my solution, it still presents a problem.

Using your induction method, yes.

0.9² = 0.81.

0.99² = 0.9801

0.999² = 0.998001

And so the pattern continues. When we extend it out to infinity, there no longer is an end, which means there’s nowhere to stick the 80…01 at the end (And no, you can’t bookend and infinite number with an 80…01 at the end, because if there’s an end, it’s not an infinite decimal). We just get another endless stream of 0.999…

Jeremy, I imagine that a computer program will give you that kind of output because it is forced to round off, not because that’s the literal answer. But I don’t know for certain because I have no idea what method the program actually uses to get that answer.

Wani, you haven’t actually showed me how you get to .999… repeating by squaring .999… repeating except to say that it starts with an infinite string of 9’s and you still deny that anything can follow them. The only way I see you getting to that answer is using my induction method and then editing my result. I don’t get it… do you believe that I did the math right or not? This looks like the worst form of cherry-picking, taking only the parts of my argument that support your conclusion and ignoring the parts that don’t.

I’ll try write it a bit more vigorously then.

Let x be 0.999…(n)…9 with n 9’s after the decimal place. We’re not looking at the infinite string of 9’s yet, just at a finite, by variable length string of 9’s. It’s also worth nothing that if we look up the nth decimal place, there’s the last 9, then an infinite string of 0’s.

From the induction I did above, we can see that the general form of x² is 0.999…(n-1)…9800…(n-1)…01. Or, (n-1) many 9’s after the decimal place, then, starting from the nth decimal place, an 8, then a (n-1) string of 0’s and then, finally, 1.

If we make x an infinitely long string of 9’s, there’s no longer a point, n where we have the last 9. Which means, in x², there’s no longer a point, n, where we stick on the 8, then the (n-1) string of 0’s, then 1. Only more 9’s. If x = 0.999… infinitely repeating, then for any decimal point m, there’s a 9. Likewise, for x², for any decimal point m, there’s a 9. There’s nowhere to start sticking the 800…1 string.

I can prove it by contradiction if you like too, because that’s always a fun was to demonstrate things. Assuming that x=0.999…, where, for any decimal place n, there’s a 9. Again, this is the definition of 0.999… when mathematicians talk about it. Now let’s take x² and assume that, at some point m, there’s 8, plus an infinitely long string of 0’s and then a 1, because that makes it follow the same pattern as any other finitely long 0.99(lots more 9’s here)9², so like you’ve said, it makes sense that x² will follow the same pattern. As I showed above, the point m happens at the point where, in x, there’s a final 9, then an endless string of 0’s. Except this is a contradiction, because we’ve defined x such that that point does not exist in x. This is a contradiction, so something we’ve said is wrong. It can’t be our definition of x, so it must be false that there’s some point m, where there’s an 8 and then other stuff. In other words, for any decimal place n within x, there’s a 9, and the same winds up being true of x² (Or else the contradiction above). In other words, x² = x, which is only true for 1. Therefore, 0.999… = 1.

Er, it’s true for 0 too, but it’s easy to show that 0.999… isn’t 0.