To recap:

x = .999…

10x = 9.999…

9x = 9

x = 1

The answer I’ve given for this is that the .999… after the 9 in the second step is not as long as x actually is, and subtracting x from both sides should actually read 9x = 8.999…91 instead. But why, when we do the math incorrectly, does it give us 1? Consider this: what proponents of the theory that .999… = 1 suggest is that when you multiply .999… by 10 you should get the same result that you’d get from adding 9 (9 + .999… equals 9.999… agreed?). Is it possible to get the same result by both adding 9 and by multiplying 10 to .999…? What is x given that 10x = x + 9?

10x = x + 9

9x = 9 (subtract x from both sides)

x = 1 (divide both sides by 9)

“Wow!” you exclaim. “See? x does equal both 1 and .999 because you can do the same math to both and get the same results!” If you really think this, then you have confused cause and effect — by multiplying one side of the equation by 10 and by adding 9 to the other side, you’ve actually then changed the answer to 1. To prove this, let’s change x to 2 in a digit manipulation by the same method, which is by adding 2 to one side while multiplying the other side by 2 (because given 2x = x + 2, x = 2).

x = .999…

2x = 2.999… (after multiplying by 2 on the left and adding 2 on the right)

x = 2 (after subtracting x from both sides)

See how we made x equal what we wanted it to equal? Let’s try to make x = 3 by multiplying one side by 2 and adding 3 to the other side (because given 2x = x + 3, x = 3).

x = .999…

2x = 3.999… (after multiplying by 2 on the left and adding 3 on the right)

x = 3 (after subtracting x on both sides)

Do you see the pattern here? This is why x = 1 in the original equation… because we actually added 9 to .999… instead of multiplying by 10. It gave us an end result that we can predict. Let’s try this one last time, except instead of .999… we’ll use .777… to show that this isn’t a trick that is only true with .999…

x = .777

10x = 7.777…

9x = 7

x = 7/9 which almost equals .777… (see earlier post on why it “almost” equals that).

Given 10x = x + 7, what is x?

9x = 7 (after subtracting x from both sides)

x = 7/9 (after dividing both sides by 9)

See? It’s like a self-fulfilling prophecy.

]]>x = .999…

10x = 9.999… (both sides of the equation are multiplied by 10)

9x = 9 (“x” is subtracted from both sides)

x = 1 (both sides of the equation are divided by 9)

Ta da! What could be more foolproof than that? The starting premise is obviously correct, and the same thing is done to both sides of the equation until it spits out a conclusion that even a guy like me would have to admit is true. Right? Right???

Well, obviously not, otherwise I wouldn’t have written this blog. While it appears that the same x is being subtracted from both sides in the third line, it’s actually two different x’s. When you multiply a number by 10, you don’t simply add 9 to it (more on this in part 2)… you get that 9 before the decimal point by moving the decimal place over to the right one space. This is true of any real number… multiplying by 10 moves the decimal place over once. By doing this, there is one less digit to the right of that decimal point. That’s true of every finite number, and should also be considered true of infinite numbers. Therefore, when one subtracts x in the equation above, you’re subtracting an infinite series of 9’s from another infinite series of 9’s that is one digit shorter. If you want to see me do all of the actual math to show you that this checks out, view my YouTube video on the subject.

A final note: While I welcome criticism, please don’t tell me .999… multiplied by 10 still has the same number of 9’s after the decimal point unless you can demonstrate why this is so. I’ve already talked about how mathematicians accept different size infinities, and I explained in my first blog in this series why inductive reasoning suggests that if any real number multiplied by 10 has one less digit after the decimal point then it should still be true when dealing with infinite digits. If you’re going to argue with me, please present a point that I haven’t already refuted.

]]>I think the proof was well-explained in a “Yahoo! Answers” forum, and so I’m just going to quote it verbatim:

Using calculus (an infinite series to be more exact), we know that .999 repeating can actually be written as The sum from 0 to infinity of(.9(.1^n)), which is a geometric series of the form a=.9, r=.1. (Meaning that it follows the form a+ar1+ar2+ar3… etc.) Using the geometric series convergence rule, we know that any geometric series converges to a/(1-r). So, the series converges to .9/(1-.1), or .9/.9, which obviously equals 1.

This is an infallible proof, using one of the highest levels of math (differential Calculus), that cannot be disproved…

Basically, calculus states that .999… can be written as (.9(.1^n))… if it followed geometric convergence. It doesn’t. This problem could be graphed like 1/X = Y, where X just keeps getting bigger and bigger as Y approaches 0. So how can I be so sure that these lines don’t converge? Because Y can never actually equal 0, otherwise it would have to be multiplied by some very large X to get 1. It doesn’t — 0 times any number is always 0. It gets infinitely close to 0 without ever actually converging, and I’m positing that this infinitely close sum (the smallest possible Y) is .00…01. You can reach this by dividing by the largest possible number, 1000… with zeroes into infinity, as I guessed in an earlier blog.

I’ve mentioned this before, and it bears repeating: most of these arguments from calculus are egotistical. The author of this Yahoo! question wrote that his proof was “using one of the highest levels of math (differential calculus)”. I’ve tried to prove my own point using subtraction, and none of my critics say that subtraction itself is faulty or inferior — they merely think that I’m misapplying it. That’s what I’m saying is true about the geometric proof; it is misapplied calculus. I’m not saying that calculus itself is untrue or inferior, but that those who turn it towards this problem made an error in application, and those that think they’ve made their point because they’re using “a superior form of math” are too blinded by their egos to spot the proofs made in “simpler forms of math”.

]]>Mathematicians posit that there are more real numbers between 0 and 1 than there are positive integers. If you clicked the links showing the definitions, you’ll quickly see why — “real” numbers include rational and irrational numbers while integers are made up only of rational numbers. This limitation is enough to make all of the positive integers (which are indeed infinite) a smaller set than the real numbers.

The way that we find out which is the “bigger” set is by finding the ratio of one set to another by matching numbers. This is incredibly easy to do when comparing these two sets because if you flip any integer around the decimal point (so that it reads backwards with a decimal point in front of it) you’ll find the real number between 0 and 1 that it matches. For example, 1 matches .1, 10060 matches .06001, 575757577 matches .775757575, etc. This can be done with any real number to match an integer except for the irrational numbers (such as pi and the square root of 2) which have no integer match when flipped backwards around the decimal point — therefore, the set of real numbers has “more” numbers than the set of positive integers.

But mathematicians don’t use a proof such as the one I’ve given here, eschewing it for Cantor’s Diagonal Proof. They posit that, after listing all of the real numbers in a column, you can draw a 45 degree diagonal line through all of the real numbers… through the 1st digit of the 1st number, through the 2nd digit of the 2nd number, 3rd of the 3rd, etc…. and then change each of those digits so that it matches none of the numbers that it travels through in order to come up with a “new” real number that you can then add to the bottom of the column to repeat the process and come up with yet other new numbers.

This thinking is flawed on 2 levels. For one thing, it’s logically impossible to create a “new” real number. Here it is as a logical deduction:

Premise 1: Cantor’s diagonal argument produces a new number, *N*.

Premise 2: *N* is a number not found within the set of all real numbers.

Conclusion: Therefore, *N* cannot be a real number.

Even further, the diagonal argument doesn’t produce a new number because it doesn’t cross through every single number. It can’t. Below is a long and detailed explanation, but I think it’s the best way to explain this complicated concept.

Consider two rockets headed into space. One is traveling at 5 kph and the other is going twice as fast at 10 kph. My question is this: when does the slower rocket catch up to the faster rocket? The obvious answer is that it doesn’t. Even when you consider that the slower rocket has gone an infinitely long distance, it has still traveled only as far as half of the faster rocket’s infinitely long distance… because we’ve already assumed that you can have two infinities of different sizes, haven’t we?

This is what happens when you try to draw a line through the set of real numbers. I’m going to do this by listing the set of real numbers in order, which is something that mathematicians purposely avoid because it allows them to avoid observing that this happens. Here are the 1-digit real numbers between 0 and 1 in order:

.~~1~~

.2

.3

.4

.5

.6

.7

.8

.9

You’ll notice that I drew a line through the first digit (.1) and now Cantor’s argument would suggest that we change it to some other digit in order to make a “new” number. But you can’t. All of the possibilities for what it can change to are listed in that same column, so whatever number you change it to will be underneath it.

“But wait!” you cry. “That’s not the end of the line. You are supposed to keep going infinitely with the line, and that way it doesn’t match any of those numbers once you add more digits.” That’s true, but it will then match other numbers below it. Consider the list of real numbers between 0 and 1 with one or two digits:

.~~1~~0

.2~~0~~

.30

…

.98

.99

Now we’ve got a line going from .1 to .2 (adding in the hidden zeroes behind them to allow for a digit to change) and you’re faced with the same situation — there is no combination of two digits that you can change those digits to that aren’t listed below it. This is true when you add another digit, and another, and ad infinitum… you’ll never hit a point where there aren’t more (and in fact an ever-increasing by a factor of 10) digits below the line. You can’t draw a 45 degree line that passes through “every” real number because the column expands much further and faster than the rows do.

Let’s look at this problem with a simpler infinite set: Flipping a coin any number of times. The only possibilities allowed are H (heads) and T (tails) and 0 (null, a non-flip). Let’s draw a line through that set.

~~H~~

T~~0~~

HH~~0~~

HT

TH

TT

HHH

…

You’re left with only one possible thing to change each result to… the H becomes a T (which is below it) and all of those null results become either H’s or T’s, the results of which are all found below it on the column. *You can’t make a new result that isn’t found below it*, no matter how far you go.

Let’s look at this one last time with a set that actually does produce a square for that line to go through at a 45 degree angle. Here is a set of zeroes (and only zeroes) taken to any length.

~~0~~

0~~0~~

00~~0~~

000~~0~~

…

Finally, the line passes through each and every result! However, what can you change each of those zeroes into? Only one possibility is allowed, so there’s literally nothing to make those zeroes into that follows the rule for making the set. Thus, even this example doesn’t allow us to create a new number with Cantor’s diagonal.

So I hope by now you can readily see why Cantor’s diagonal argument is not a paradox — it doesn’t do what it proposes to do. I’ve had mathematicians argue that this isn’t what it proposes to do, arguing that this argument isn’t supposed to create a “new” real number, but there are issues with that. For one thing, those who bring forth this argument always suggest that the newly minted number can be “added to the column to make another number”… in which case, why use the diagonal instead of just adding an infinite number of your choosing? Either way, you’ll have a duplicate number in that column. For another thing, the supposed reason given by Wikipedia and others for this argument is that it demonstrates that the real numbers are “uncountable”… but isn’t an infinite set’s uncountability, immeasurability, and incalculability true of infinite sets by definition? And while it may be seen as begging the question to assume that the real numbers between 0 and 1 are infinite without the diagonal argument, there are other ways to figure out if a set is infinite… after all, did we need the diagonal argument to prove that the positive integers make up an infinite set? Of course not

]]>Let’s consider the Sorites paradox, also known as the “paradox of the heap”. If you have a pile of sand containing 10,000 grains of sand, it fits the definition of a “heap”. If you remove a single grain, isn’t it still a heap? Intuitively, it’s still as much a heap as it was before. How about if you remove another? And another? At some point you’ll be left with a single grain of sand and still referring to it as a heap. You could arbitrarily set a point at which it passes from heap to non-heap, but that doesn’t feel right.

The Sorites paradox has been answered multiple ways, but my favorite solution is fuzzy logic in which the heap becomes less of a heap as grains are taken away, and no true boundary has to be set but rather you could describe it with terms such as “small heap” or “medium heap” or “somewhere in between a small heap and medium heap”. Because we have answers like this, the paradox is no longer truly a paradox but rather one that only seems like a problem with mutually exclusive conclusions.

Another set of paradoxes to consider are Zeno’s paradoxes, such as Achilles and the Tortoise. Achilles chases a tortoise but can never catch it because every time he has gone half the distance the tortoise has still gone further, and once he has traveled half of that new distance between them the tortoise has still traveled further so he can never catch up (even though we know he obviously can in real life). We also have the paradox of the arrow in which an arrow can never hit its target because in any instant of time the arrow is motionless (motion occurs over a change in time), and the elapsed time between the arrow leaving the bow and hitting the target is just a series of “instants of time” in which the arrow is motionless. And yet, arrows obviously do travel and hit targets.

A good answer to Zeno’s paradoxes is the geometric series that calculus improved upon with the convergent series, and we’ll be visiting and expanding upon that in the near future (because it is an alleged method for proving that .999… = 1).

So my paradox (which is strangely not seen as a paradox) is this: the ratio of all integers to even integers is inconsistent. Mathematicians argue that there are just as many integers as there are even integers because you can multiply any integer by 2 to get an even integer. So if 1 x 2 = 2 and 2 x 2 = 4 and 3 x 2 = 6, etc then the set of integers [1, 2, 3…] and the set of even integers [2, 4, 6…] have the same number of numbers in them. But that’s obviously not the case. If you match the numbers among all integers rather than multiply them, you can match every other integer but all of the odd ones have no partner. In this instance, there are twice as many integers as there are even integers (which follows intuitively).

In other words, you can match all integers to even integers on a 1:1 ratio by multiplying them by 2 (2X = Y) or you can match half the integers to even integers on a 2:1 ratio by multiplying them by 1 (X = Y). Either these sets are equal or one of them is larger, but they can’t simultaneously be both — that would give us mutually exclusive conclusions, otherwise called a paradox.

Perhaps this is a known paradox and I haven’t come across it yet, or perhaps it has even been solved. In any case, I think modern mathematicians are too dismissive of it and simply deny that there is more than one way to look at it (they see it only as a 1:1 ratio), and I think it deserves some deeper thought.

]]>I ask the question for a reason, though. In a comment to one of my math videos on YouTube, one viewer suggested that the largest number would be an infinite series of nines; and why not? Nine is the biggest digit and surely a string of them would be bigger than a string of any lesser digits. I simply didn’t know what the largest number would be, and this was as logical a guess as any.

However, when I started subtracting large numbers from smaller numbers for the purpose of my previous blog (“A Mind-Blowing Proof”), I found a pattern. The answer to “subtraction done incorrectly” was always a string of nines followed a seemingly random number — but it’s not random. If you added to this answer the actual difference between them you’d get 1000… (one followed by infinite zeroes). This was even true with .999… minus 1, assuming that the answer is what I posited it to be earlier (.00….001). In math terms, if X – Y = Z (where Y > X), Y – X + Z = 1000…. (where Y – X is subtraction done the way mathematicians would never suggest).

Why is that? My conjecture is that, if there is a largest number, 1000… is it. Why should this be so? I honestly don’t know. It just feels intuitively true, although I have no evidence, so take this conjecture with a grain of salt. It’s a good starting point for an investigation, though, and I’ll be sure to give this some more thought. I hope you’ll share your thoughts with me, too.

]]>After a few months of tossing around this idea, I was finally struck by an even better proof that is just as simple as the other one and yet more profound. As far as I can find, it hasn’t yet been explored, and yet it’s so incredibly simple. Not only do I feel stupid for not seeing it sooner, but it’s shocking that no one else has either. Of course, no one saw that energy and mass were directly correlated until Einstein came along and pointed out the obvious.

What is: .999… minus 1? Naturally we all see the 1 as the larger number (if we see either of them as larger), so we’re biased to do the subtraction in only one direction. But if they’re equal, shouldn’t the subtraction work in either direction? When you subtract X from X, it literally doesn’t matter which X comes first because the answer will be zero either way. So let’s see if that’s the case here… but first let’s consider a simpler problem for context in analyzing the result.

What is 2 minus 4? A mathematician will tell you to invert the problem and then make the answer a negative (for an answer of -2). Why? Because subtraction freaks out when it tries to subtract larger numbers from smaller ones. Here’s what I came up with:

You’ll notice that it keeps going to the left… infinitely. The result is an infinite number of 9’s followed by a single 8, which is clearly not the answer to 2 minus 4. This isn’t a special case, either… any time that you subtract a larger number from a smaller number you’ll get an infinite series of 9’s before a wrong answer. Try it yourself by hand with any two numbers.

Now for the problem at hand. What do you get when you subtract 1 from .999…?

There’s something familiar about this result, isn’t there? The answer is important for 4 good reasons:

1. The result is a non-zero number, which proves that .999… does not equal 1. X – X always equals zero.

2. The math is all based around the decimal point, just as other problems that are “doable” such as .888… minus .111… or like .999…. minus .1 — nobody can complain that solving this requires finding the last digit of the series of 9’s (unless they deny the double-standard of doing similar problems that I’ve mentioned here).

3. The pattern of repeating 9’s going infinitely to the left is exactly like the pattern we observe whenever we subtract a larger number from a smaller one, which means that .999… is smaller than 1 as a person would suspect.

4. The answer is internally consistent with my answer to 1 minus .999… as I’ll explain in the next blog.

So there you have it. .999… cannot equal 1 because, no matter how you believe you proved that it worked, the answer is wrong. To quote Richard Feynman, the famous physicist:

“It does not make any difference how beautiful your guess is. It does not make any difference how smart you are, who made the guess, or what his name is – if it disagrees with experiment it is wrong.”

]]>1/3 = .333…

(1/3)3 = (.333…)3

1 = .999…

One divided by 3 equals point 3 repeating. If you multiply both sides by 3, you get 1 equals point 9 repeating.

When I first saw this proof, my mind was blown. I had heard that .999 repeating was exactly equal to 1 rather than almost, but I had never seen it proven. This was proof. I shared this with several people, and I was so proud to be “in the know”. But upon further examination, I see that there are a couple of problems with this seemingly easy math problem as I will demonstrate in this post.

The first is that 1 divided by 3 does not equal point 3 repeating. It’s *almost* equal. The explanation is long but educational. Bear with me on this one.

When we were kids first learning about division, we were taught a mathematical model that I’ll refer to as the Remainder Model. When we divided 10 by 3, we got “3 with a remainder of 1” (written as 3R1). If you wanted to check the math, you’d do it in reverse by multiplying by the number you divided by and then adding the remainder, so 3(3) + 1 gives you 10. This allowed us to divide uneven numbers exactly.

When we got older we replaced the Remainder Model with the Decimal Model. Instead of stopping at 3R1, we’d drop a zero from behind our original 3 (because it’s actually 3.000…) and then divide our remainder of 1 with a 0 (10) by 3 again, giving us 3.3. Now while we didn’t do this in school (at least I know *I* didn’t), we *could have* used both models to get an exact answer… all you’d have to do is carry the remainder as many times as you drop zeroes from behind the 3. Dividing 10 by 3 just a couple of times gave us 3.3R.1 (checked by multiplying 3(3.3) to get 9.9 and then adding the remainder of .1 to get 10), and doing it three times gave us 3.33R.01 and doing it ten times gave us 3.333333333R.000000001. Through inductive reasoning, we can see that if you divided 10 by 3 an infinite number of times you’d get 3.333…R.000…01. That remainder may get infinitely small, but it never disappears — no matter how many times you do the division, it will always carry a remainder.

Thus, 10 divided by 3 does not equal 3.333… but rather *almost* equals 3.333… because there’s a remainder. It’s easy to see why this is also true of dividing 1 by 3. In our proof above, we don’t actually get 1 = .999… but rather 1 = .999… with a remainder of .000…01.

I said above that there were “a couple of problems”, and this rounding error is just the first. The second one is with multiplying .333… by 3. “Why would that be a problem?” you may ask. Well, I don’t see it as a personal problem, but rather one for my opponents. There are those who would “reify” .333… (see my last post) and suggest that this problem is unsolvable because multiplication requires that both numbers have a finite end. After all, how else would you even start such a math problem? The answer is easy — through inductive reasoning. We know that any finite string of 3s multiplied by 3 will give us an equally long string of 9s, and there’s no reason to think this will change just because the string of 3s is infinitely long. But this is a problem for some people, and I’ve seen this proof less and less over time because mathematicians have realized that allowing this multiplication creates an obvious double standard. While this may seem a trivial problem compared to the first one, it is major enough to prevent many mathematicians from even offering this proof any more.

One final note: While I welcome criticism, please don’t tell me that my math is wrong unless you can suggest how it could have been done right. Don’t tell me that there is no remainder when you divide 1/3 unless you can demonstrate how it is done evenly. If you can’t offer an alternative explanation, there’s a good chance that there isn’t one.

]]>You may have noticed the title of this post and thought, “What is reification?” Reification is a logical fallacy in which an abstract idea is treated like a physical object. For example, many people talk about feelings being “bottled up”, and suggest that letting out anger through venting will reduce the anger that you have. Studies have proven this wrong, because anger is not a physical object that can be reduced by “getting rid of it” through expressing it.

The problem with this objection to my method is that mathematicians who think that it’s impossible to “bookend” an infinite series of 9’s with a decimal point and an 8 will bookend other infinities without batting an eye. Consider the number of real numbers between 0 and 1. The obvious answer is “they’re infinite“, despite the existence of a floor and a ceiling to this series of numbers. One might argue that I’m arguing apples and oranges — the series between 0 and 1 is made up of numbers, not digits. So let’s convert that to digits. If you took all the real numbers between 0 and 1 and listed them from least to greatest, then drew a line crossing the first digit of each number, your result would be 01234567891, where every number except that final 1 would be repeated infinitely. Would this new number be consider a “real number”? “Sure!”, say proponents of Cantor’s diagonal argument, which creates a real number through almost identical means.

Do I agree that the number created is a real number? No, but that’s not the point. It’s not important to agree that a real number is made here, but instead to focus on the fact that the digits in this number are indisputably infinite, even though they end abruptly with other infinite series of digits and finally with one finite digit. This is because we’re not talking about a physical reality but a thought experiment. Numbers are just abstract ideas that we use to describe the relationships between mathematical models such as graphed lines, sets of objects, and sometimes just other numbers. While it may offend your sensibilities to try to grasp a concept like .000….01 (point zero repeating ending with a single one), there is no logical reason that such a number can’t exist.

One final note: While I welcome comments, please don’t tell me I’m wrong unless you can explain how I could have been right. If you can’t come up with an alternative explanation, there’s a good chance that there isn’t one.

]]>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.

]]>