Proving 0.999... Is Equal To 1 1260
eldavojohn writes "Some of the juiciest parts of mathematics are the really simple statements that cause one to immediately pause and exclaim 'that can't be right!' But a recent 28 page paper in The Montana Mathematics Enthusiast (PDF) spends a great deal of time fielding questions by researchers who have explored this in depth and this seemingly impossibility is further explored in a brief history by Dev Gualtieri who presents the digit manipulation proof: Let a = 0.999... then we can multiply both sides by ten yielding 10a = 9.999... then subtracting a (which is 0.999...) from both sides we get 10a — a = 9.999... — 0.999... which reduces to 9a = 9 and thus a = 1. Mathematicians as far back as Euler have used various means to prove 0.999... = 1."
Re:This is second place (Score:4, Informative)
It is easy to explain.
1. 1/9 = 0.111111111111111111111111111111.....
2. Multiply each side by 9
3. 9/9 = 0.999999999999999999999999999999......
4. Simplify fraction
5. 1 = 0.999999999999999999999999999999......
Monty Hall trips up even serious math enthusiasts.
Re:Oh yeah? Well... (Score:4, Informative)
this probably isn't necessary for most of the Slashdot crowd, but...
(a+b)(a-b) = b(a-b) --> a + b = b
Required division by (a-b) on both sides. Since a = b, this is division by zero.
Re:When you add/subtract/multiply/divide infinite (Score:3, Informative)
Far from true. A rational number is a number you could get by expressing as a ratio (real number divided by real number). Any infinite repeating decimal is easily shown as a ratio (and often of simple integers to boot), i.e., a rational number. 0.22222... is 2/9. 0.456456456456456... is 456/999. And so on.
Re:This is second place (Score:5, Informative)
Unfortunatelly, your proof is not valid. You are trying to prove something which you postulate in your first step.
How do you know 1/9 equals 0.1111111.... ?
He begged the question! For anyone confused about the term "beg the question," this is exactly what it means: assuming the proposition to be proved in the premise.
But that begs the question: is the classical meaning already dead, replaced with the much more easily understood modern usage demonstrated here?
Re:When you add/subtract/multiply/divide infinite (Score:2, Informative)
The number of things wrong with this statement are baffling.
Infinity is not an irrational number. It's not actually a number at all.
0.999... is not an irrational number. It's a rational number, as it can be expressed as a fraction. Irrational numbers cannot be expressed by any repeating set of decimal numbers.
You cannot "set" x to infinity. You cannot multiply infinity by 2. That's like trying to multiply the color red by 2. It simply isn't meaningful. Comparing the sizes of two infinite sets is a very different operation from comparing the sizes of two numbers.
The mathematics of comparing infinite sets does not in any way apply to arithmetical operations on infinitely repeating decimals.
0.999...99 is not the same as 0.999.... The former will, in fact, be less than 1, because it terminates.
Re:Cribbed, Since My Memory for Jokes Sucks (Score:5, Informative)
Actually, a good physicist should have been able to give an answer (or something close to it) as well...
Eventually, they will come to a point where they would be required to move less than 1.616252(81)×1035 meters closer together. From the uncertainty principle, we know we cannot measure position more accurately than that. So either they will not move at all, or they will superimpose at that point.
Re:Finally (Score:5, Informative)
I am compelled to answer...
Divide both sides by (b - c - a) is dividing by zero.
Re:Humans are just biased towards natural numbers (Score:5, Informative)
They're not proving "0.99999 = 1" at all. That's not true. They're proving that "0.999... = 1". One is an infinite sequence of digits, and the other isn't. The distinction is important. The proof of "0.999... = 1" has nothing to do with rounding, and to suggest so indicates a (common) gross misunderstanding of the problem.
First, you only measure things with such poor precision because you're working well above the quantum level.
Second, natural numbers are certainly important. For one, they're critical to our understanding of the rest of mathematics, which is important for fancy things like being able to take measurements and manipulate them at all. For another, we work with whole numbers of objects all the time -- two apples, ten antelope, four huts, etc. It's not "10 +/- 0.01 antelope".
Re:But what you did is flawed (Score:3, Informative)
No, you're missing the whole point. 1/3 is exactly equal to 0.333... with an infinite number of trailing digits. It's not an approximation or an estimate, it is two ways of representing the exact same real number.
Here's how you convince yourself: If 1/3 was really close but not quite 0.333..., then we could split the difference between those numbers and find another real number between them. But we can't, which means we were wrong to assume that 1/3 and 0.333... were distinct.
More fun... (Score:4, Informative)
Re:I went one further (Score:4, Informative)
People assuming they did something wrong when the result "doesn't make sense" isn't the problem.
People failing to distinguish between a notation and a number, creating the belief that "0.99(9)=1" doesn't make sense, is the problem.
Consider this proof, which follows simple steps to reach a conclusion that doesn't make sense:
i^2 = -1 (definition of i)
i^2 * i^2 = -1 * -1
i^4 = 1
sqrt(i^4) = sqrt(1)
i^2 = 1
-1 = 1
Then if you want you can add 1 to both sides and divide by 2, to find 0 = 1.
Now, do you know why this proof is bogus? When I was in high school, we were introduced to imaginary numbers, and I drew up a slightly more obfuscated version of the above; it had a lot of people (including a couple relatively sharp teachers) in "I know you did something wrong because the result doesn't make sense" mode for a long time.
The fault, of course, lies with the sqrt() step. For a=a to imply sqrt(a)=sqrt(a), we have to interpret sqrt(a) as the pricple square root function, so sqrt(x^y) = x^(y/2) doesn't necessary work when x isn't a real number.
Without the motivation of "this result cannot be right", I wouldn't have puzzled this out. More than that, the solution comes from understanding that rules we take for granted only apply to certain types of number. Applying that to 0.99(9), it's easy for people to convince themselves that repeating decimals are a special class of number subject to "some rule I just don't know".
But in this instance, that reasoning is flawed, because .99(9) really is just a regular real number in a weird notation.
Re:I went one further (Score:5, Informative)
Surely the problem is that you're assuming sqrt(1) = 1 when actually it is +- 1? You're throwing away the sign change in that step :)
Re:This is second place (Score:3, Informative)
0.999... is not a "decimal expansion" of some number, but rather it denotes a sequence of numbers: 0.999, 0.9999, 0.99999, ....
Talk about having difficulty understanding the concept. 0.9999... is not a sequence, it is a series:
...
9/10 + 9/100 + 9/1000 + 9/10000 +
Which anyone who paid attention in middle school will recognize as a geometric series (well, actually, it is a multiple of a geometric series).
The fact that "the real numbers can be defined as equivalence classes of Cauchy sequences of rational numbers"...
Also, one should not just casually accept the ideas of infinite series and sequences, because counter-intuitive things DO happen with them.
There is nothing counter-intuitive in this case, it is just a geometric series. More generally, decimal expansions are special cases of infinite series, and there is a well developed theory on infinite series.
Re:I went one further (Score:3, Informative)
Surely the problem is that you're assuming sqrt(1) = 1 when actually it is +- 1? You're throwing away the sign change in that step :)
Yeah, that's the problem, but for those interested...
The base problem with this is that unlike the logarithm for real numbers, the logarithm for complex values is not a function (or, if you like, it's a "multi-valued function"). This comes from the interesting fact that x^1 has 1 solution, x^2 has two solutions, x^3 has 3 solutions, and so on. We kind of fudge around it in reals, because x^n will only ever have one or two solutions, but in the complex plane it has n solutions, and things are much more complicated.
The end result of the multi-valued logarithm is that the normal rules for exponentiation and logarithms can break down in ways that may be unexpected. In this case, yeah, it's confusing sqrt(1) = +- 1, but I've seen more subtle proofs (similar to the GP's) that use cube roots to avoid the math plus-or-minus square root gut reaction.
For more information, see the Powers of Complex Numbers [wikipedia.org] and Complex Logarithm [wikipedia.org] pages on Wikipedia.
Re:I went one further (Score:4, Informative)
Re:This is second place (Score:3, Informative)
Re:This is second place (Score:3, Informative)
This is Slashdot. Better explained this way:
When the decimal repeats, you enter a loop. The ...9991 is unreachable code.
Re:I went one further (Score:1, Informative)
Surely the problem is that you're assuming sqrt(1) = 1 when actually it is +- 1?
Technically, sqrt(1) = 1.
sqrt is a function, and a function is not allowed to have more than one range value for a given domain value.
By convention, we define sqrt(x) to be the non-negative root.
However, in algebra, when taking the square root of both sides of an equation, you must split the analysis into two parallel tracks -- one track uses the +sqrt and the other track uses the -sqrt. This is required in order to do a thorough analysis. Don't confuse this two-track algebra technique with the definition of the sqrt function.
Re:Corrected, Since My Memory for Jokes Sucks (Score:1, Informative)
1.616252(81)×10^-35 that is
Re:This is just faulty math (Score:2, Informative)
The only reason infinity might come into play is because 1/9 has no finite representation in base 10. But this is not a problem. Consider the number 1/10. Would you say that you can multiply 1/10 by 10? Assuming that you would agree to this, consider what would happen if you performed this operation in binary? In base 2, 1/10 has no finite representation, so by your logic, we cannot multiply it by 10. In fact, given any finite decimal there is another base in which it has only an infinite representation, and hence by your logic we cannot multiply numbers by decimals ever.
You have imposed an artificial limitation on the basic concepts of arithmetic by limiting yourself to floats in base 10.
I also have a problem with you stating that you can't use the value of pi in math when you absolutely can. The inability to write down infinitely many digits (by which I mean write down more digits than any finite number of digits) does not preclude the fact that we can do math with pi. Math is in no way limited to calculations which can only be done with finite precision. Take this example from calculus: find the area under the curve 4/(1 + x^2) from x = 0 to x = 1. We can show that the antiderivative of 4/(1+x^2) is 4arctan(x), and so the answer is 4arctan(1) - 4arctan(0). By the definition of the tangent function, tan(pi/4) = 1 and tan(0) = 0, so 4arctan(1) - 4arctan(0) = pi. That is, the area described above is exactly pi, or exactly the same as the area of half a unit disk (even though these regions look nothing alike). Notice that the answer is not that these are pretty close, they are in fact so close that given a set of tools with any arbitrarily small margin of error, we could not tell them apart. This is just one example of doing math with pi and not a definable approximation of pi.
In base 10 it is 0.14285714285711. (Score:4, Informative)
Also a terminology issue, the number of ones in 0.111111... is indeed "Countable" [wikipedia.org].
Re:It is not true and I can prove it. (Score:2, Informative)
You fail at subtraction.
1 - .111... is not equal to .999...
Try this: .1 = .9 .11 = .89 .111 = .889 .111... = .888...
1 -
1 -
1 -
1 -
Re:I went one further (Score:3, Informative)
I'm probably too late to get modded up, but since none of the existing responses gave the exactly correct explanation, I'll have to post rather than moderated.
sqrt(1) is 1. It's not -1. By definition.
A list of transformations of an equality like the one given in the grandparent's "proof" is shorthand for a list of "implies" statements. For example, a proof like this:
is actually shorthand for:
When you rewrite the shorthand proof in the grandparent post in full form, the mistake becomes (more) obvious: a^2=b^2 does not imply that a=b. But this has nothing to do with the sqrt function, it is because of the square function; because it is not an injective [wikipedia.org] (one-to-one) function.
To illustrate by taking it to an extreme - instead of f(x)=x^2, let's take a different non-injective function: f(x)=0. Would you have any trouble realizing that f(a)=f(b) does not imply a=b?
As an amusing curiosity, one way to define |x| (the absolute value of x) is sqrt(x^2). |x|, as you may guess, is also a non-injective function.