999 Is Equal To 1

.999999... = ane?

Not that long agone, a company left a note:

Just this weekend, I tried to explain to my father-in-law the problem with the base 10 numbering system that we accept in this world today.

For case:

ane divided by iii is .3333 repeating
one third times 3 is 1
.3333 repeating times three is .9999

Let me say upward front end, I sincerely believe that the base 10 (decimal) organisation is very good. I utilise it all the time, discover it convenient and handy, would heartily recommend information technology to everyone else. I would non hesitate to cast a "No" vote on whatever referendum suggesting to change our number system to another.

This said, I must admit that the old skilful decimal arrangement is not much different from a few others. The basic reward of base 10 is that this is a positional system. The Roman number 12, XII, contains two symbols of units that both add together exactly the aforementioned quantity (one) to the total. In the decimal arrangement, eleven means 10 + i, then that the same digit i plays unlike roles depending on its position in the number representation. The Romans had to invent new symbols for bigger numbers, which sometimes they did and sometimes they did non. David Wells has this entry for the number two,300,000

The primeval inscription in Europe containing a very big number is on the Columna Rostrata, a monument erected in the Roman Forum to commemorate the victory of 260 BC over the Carthaginians. The symbol for 100,000 was repeated 23 times, a full of 2,300,000.

In the decimal system, we only need 7 symbols to express the same quantity. In base 20, this takes just 3 digits: D9A. In base 20 system, all numbers go shorter. The downside is that the number of digits grows to xx. In base 5, there are only v digits but numbers become longer: (2300000)₁₀ = (1042100000)₅. In base of operations 12, (2300000)₁₀ = (92B028)₁₂ - just two more digits to remember but somewhat shorter numbers.

The reasons our club has settled on base 10 may exist explained by our anatomy rather than past whatsoever mathematical reasoning: 10 fingers per private make a expert reference point. Also, memorizing 10 digits does not seem to exert an undue force per unit area on the mental abilities of an average individual.

The problem addressed past the visitor I mentioned at the beginning of this page pops upwardly in all bases. For example, 1/iv = (0.111...)₅. And four·(0.111...)_five = (0.444...)₅. Is this 1? In every base of operations, dividing i by the largest digit of the system gives 0.111... As another example, (1/F)₁₆ = (0.111...)₁₆. (F)_xvi·(0.111...)₁₆ = (0.FFF...)₁₆. Are we getting 1 dorsum? Yes we do:

0.999... = i
(0.444...)_five = 1
(0.FFF...)₁₆ = ane

Of grade, I might have used other identities: 1/2 = (0.222...)₅ and, equally before, two·(0.222...)_v = (0.444...)₅. Likewise, 1/3 = (0.555...)₁₆, and multiplying back past 3 gives 3·(0.555...)₁₆ = (0.FFF...)₁₆. Equally nosotros run into, it is even so in other bases. The question whether 0.999... = i or not can't be resolved past turning to other systems. Then how?

Those who ask this question should as well pause to ponder another question, What is a number? For, if 0.999... is but a symbol, a name (like Seven in the Star Expedition series or i of its derivatives) or is it a number? Names are private: each exists in its own right. Numbers exist all together, so we can apply various operations to one or more numbers to go other numbers. My visitor obtained 0.999... by multiplication (3·0.333...). I can then safely assume he thinks of 0.999... equally a number subject to other arithmetic operations. For instance, 0.999... = 0.9 + 0.099... = 0.99 + 0.0099..., and and then on. If we concord on interpretation of ellipses "...", then I would expect that nosotros should likewise concord that

0.999... + 0.1 = 0.9 + 0.i + 0.099... = 1 + 0.099... > ane
0.999... + 0.01 = 0.99 + 0.01 + 0.0099... = 1 + 0.0099... > 1
0.999... + 0.001 = 0.999 + 0.001 + 0.00099... = 1 + 0.00099... > ane

All the same small fraction we add to 0.999..., we always become a number greater than ane. To me, this implies that the difference between 1 and 0.999... is less than any positive number and definitely information technology is not negative. Thus, it must be zilch: 0.999... = 1. Is it surprising that a number can be written in several means? No, information technology should not be. 1.v = 3/2 = six/4 = (1.one)₂. This also true that i.5 = ane.4999... The question is actually,

What stands behind the ellipses "..." ?

As earlier, 0.999... = 0.9 + 0.09 + 0.009 + ... = 9/10 + nine/100 + ix/thou + ... The expression 0.999... implies an infinite sum, i.due east., a sum of an infinite number of terms. In mathematics, such infinite sums are called series. Series are introduced and studied rigorously in Calculus, where a distinction is made: some series are convergent, some are divergent. Every convergent series has a unique number associated with information technology, its sum. Divergent series are not so lucky. All infinite decimal fractions, similar 0.999..., are shown to represent to convergent series (which converge to their corresponding sums.) 0.999... converges to 1 which is expressed simply as 0.999... = one. Similarly, 0.333... is a convergent serial whose sum is 1/3: 0.333... = 1/three.

Here'southward some other enlightening argument from Burger. I never met anybody who thought 0.999... greater than 1. So, if it's not equal to 1, it is less than 1. Allow's think of the average of 0.999... and one. As an average of any two numbers, information technology'south greater than 0.999... only is less than 1. Tin can we make up one's mind its decimal expansion? Say, what is its integer part. Since it's less than 1 but greater than 0 < 0.999..., its integer function is bound to be 0. What virtually its first decimal digit. Since 0.9 < 0.999..., that digit must be 9. And the 2d ane? Since 0.99 < 0.999..., the 2d digit must likewise exist 9. And then on. It appears like the average of 0.999... and i is 0.999... If the latter is denoted as 10, (X + one)/2 = Ten. 10 + one = 2X. X = 1. The conclusion tin can't be helped.

The is another fashion to exploit the notion of everage. Fifty-fifty intuitively, the boilerplate of 2 unequal numbers is greater than the smaller of the two is less than the larger one. If the average is equal to ane of the numbers then all three are equal. So what is (0.999... + ane)/2? Yous tin can come across that it's 0.999... because

2×.999... = 1.eight + 0.18 + 0.018 ... = 1.999... = 0.999... + i.

but this ways that all three numbers are equal!

Remarks

Ane is led to a curious conclusion in the framework of the non-standard analysis. Fifty-fifty in the non-standard assay, .999... tin't exist said to exist a sum of an infinite number of terms!
Using calculators appears counterintuitive unless ane understands their limitations. The computer plan that comes standard with Windows 95 gives i/ix = 0.1111111111111 (13 i's.) Then that successive sectionalization and multiplication by 9 produce a wrong identity: 1/9·9 = 0.9999999999999 where ellipses may be assumed by the user simply are not actually present. Calculators take stock-still space to shop each number. If there is not enough room to store a number, a calculator stores its approximation instead. The button "=" on a calculator'due south pad means "Compute" or "Calculate" which sometimes results in an actual equality and sometimes in an approximate one. Once 1/9 is stored as 0.1111111111111, a calculator has no retentiveness that the number was obtained as an approximation to ane/9. Multiplication by 9 displays now an exact issue, 0.1111111111111 · 9 = 0.9999999999999.
A elementary example of a divergent serial is the sum of all integers 1 + 2 + iii + ... As more integers are added the sum grows without spring and no number may sensibly be associated with such a sum.
Another example is the famous 1 - 1 + 1 - 1 + ane - ... where units are subtracted and added intermittently. Presume a number S is associated with the series: S = ane - one + 1 - 1 + ... And then we should exist able to write, on the one paw, S = (i - 1) + (1 - 1) + ... = 0, and, on the other, S = 1 - (1 - 1) - (1 - i) - ... = one. Which does non make a lot of sense. Nonetheless, the question is not at all trivial. We may also write Southward = 1 - (1 - one + ...) = 1 - S, from which 2S = 1 and S = 1/2. Many a famous mathematician, L. Euler in particular, believed this to be true.

Euler relied on the formula for the geometric series:

(*)

1/(1 - x) = one + 10 + x^two + x³ + ...,

which he was willing to consider as the definition of the infinite sum on the right for whatsoever 10, for which the left side was defined. In particular, for x = -ane, we obtain

1/2 = ane - 1 + 1 - 1 + 1 - ...

What if nosotros kickoff with S = 1 - 2 + 4 - 8 + ... ? Kickoff we tin can write

South = one - 2(one - 2 + iv - ...),

and so that S = 1 - 2S, from where S = i/3. On the other hand, S = (one - 2) + (4 - 8) + ... and surely can't be positive. We obtain the aforementioned result from (*) with x = -2.

Euler was certain that the series on the correct is uniquely determined by the analytic expression on the left. However, this does non imply that the numeric series obtained by a commutation of a specific value for 10 could not be obtained from a dissimilar formula. For example, Jean-Charles Callet (1744-1799) plant that

(**)

(1 - x^m) / (1 - xⁿ) = 1 - x^g + ten^{due north} - 10^{n + k} + 10²ⁿ...,

For a finite geometric series,

ane + x + 10ⁱⁱ + ... + x^northward-1 = (one - xⁿ)/(i - x),

which yields

(1 - x^m) / (1 - x^northward) = (1 + ten + x^two + ... + x^grand-1)/(1 + x + 10² + ... + x^north-ane)

then that, for x = 1, we get from (**)

m/n = 1 - one + 1 - ane + 1 - ...

Trouble

Ii cars start simultaneously towards each other from two cities located at the distance of 100 miles. Ane auto goes at the speed of xx miles per hr, the other at thirty mph. A fly was sitting on one of the cars before the race started. Startled by a revving engine, the fly alights from 1 car and flies towards the other at the speed of 40 mph. The moment it reaches the other machine, it turns dorsum and proceeds at the same speed and turns back again just a moment before it is squashed past the incoming automobile. Thus it flies dorsum and forth until it meets its sad fate when the cars bump into each other. How big the distance has been covered by the fly earlier its untimely finish?

Solution 1

The cars race towards each other at the speed of 50 (= 20 + 30) mph. Allow, at some moment merely before the fly turns towards some other motorcar, the distance betwixt the two cars be A. Presume it takes fourth dimension T for a wing to attain the other automobile. Nosotros thus get the equation

(1)

A - 50T = 40T,

from which T = A/90. In time T, the fly covers the distance 40·A/90 or 4/9 of A. The cars cover the remaining 5/9 of A. The next leg of the wing'south journey will exist (4/9)A. Taking (4/nine)A instead of A we encounter that the fly will cover the distance (4/ix)²A before he turns once again, and so on. With the original A = 100, the fly will cover the distance given past the series:

(4/9)100 + (4/ix)²100 + (iv/9)³100 + ...

The sum of this geometric serial is known to exist 100(iv/9)/(i - 4/nine) = 80. Answer: fourscore miles.

Solution two

The cars approach each other at the speed of 50 mph. At this speed they will run into in 2 hours (ii = 100/50). Flying at the speed of 40 mph, the fly will cover 80 miles in 2 hours. Respond: 80 miles.

Anecdote

It so happened that a similar problem was once proposed to John von Neumann who immediately gave the right answer. The proposer and so remarked that von Neumann probably knew the trick, to which he is said to accept replied, "What trick? What'south easier than summing the series?"

Sum of the geometric series

With |q| < ane,

(2)

a + aq + aq² + aq³ + ... = a/(one - q)

That the series in (2) is convergent is a issue from the kickoff Calculus. Denote the left-hand side of (2) equally S. It is further shown that the common rules of mathematical operations apply to the infinite sums of convergent serial. In particular, this is true of the distributive police:

(3)

Sq = aq + aqⁱⁱ + aq^three + ... = Southward - a,

from which (2) follows immediately. As an example,

0.999... = 0.nine + 0.09 + 0.009 + ... = (9/10)(1 + (ane/x) + (1/ten)² + ...

This is the same as (2) with a = 9/10 and q = 1/10. This gives

0.999... = (nine/ten)/(1 - 1/10) = one,

in agreement with a more than intuitive derivation above.

Equally we see, the question whether or not .999... equals 1 is thus reduced to evaluating the same quantity (1, in this example) also as a sum of a geometric series. The state of affairs is quite analogous to the problem we discussed above. Scott Brodie describes a "real earth" problem that leads to the sum of a geometric serial but also has a natural and unproblematic direct solution. Computing the area of the Sierpinski gasket serves every bit an additional example. Another convergent series has a squeamish geometric interpretation.

In his volume The Jaguar and the Quark (p 208), The Nobel Price winner Murray Gell-Mann describes a profound state of affairs that leads to the summation of an infinite serial:

... [Richard Feynman] found that it could be done. However, the activeness came out in the form of an infinite series, and summation of that serial was virtually impossible in the absence of the geometrical bespeak of view and the invariance principle. The principle, full general relativity, yields the answer directly, without any need for the brute force method or the infinite series. ... Perhaps the situation in superstring theory is like. If theorists understood the invariance principle of superstring theory, they might be able to write down the formula for the action in short social club, without resorting to the summation of the infinite series. (While we are waiting for it to be discovered, what should nosotros phone call that principle? Field marshal relativity? Generalissimo relativity? Certainly it goes far across general relativity.)

As a marvel, you may desire to check that, as it was observed past Harold B. Curtis (Mathematics Mag, Vol. 42, No. 1 (Jan., 1969), pp. 43-52), 0.666... + (0.666...)² = i.111... .

Remark

Information technology is important to realize that arithmetic manipulations in (3) just worked because the series was convergent. For divergent series such operations make no sense. Consider the series of all the integer powers of 2: ane + 2 + 4 + 8 + ... This is a divergent series. Assume nonetheless a sum S is ascribed to it:

Southward = 1 + 2 + 4 + viii + ...

Then 2S = 2 + 4 + 8 + ... = Due south - 1, from where S = -1 which is surely absurd since all the terms in the series are positive.

Note:The absurdity of the equality between a "big positive" number and a negative i stems from the arithmetic context. In the theory of p-adic numbers the identity -1 = 1 + 2 + 4 + viii + ... is perfectly verifiable.
At least two uncomplicated geometric series accept 1/three as the limit:

one/3 = 1/4 + one/16 + one/64 + ... (a = one/4, q = 1/four) and
1/3 = 1/2 - ane/4 + 1/8 - i/16 + ... (a = 1/ii, q = -1/ii)

Denominators of every term in the serial are powers of 2. As we know, for whatever angle A, we are able to construct (with a straightedge and a compass) A times a sum of any finite number of terms in the series. However, in guild to construct A/3 we need an infinite number of steps, which is forbidden past the rules. The problem of trisecting a general angle is, in principle, not solvable.

References

E. B. Burger, M. Starbird, The Centre of Mathematics, Key College Publishing, 2000
Thou. Gell-Mann, The Jaguar and the Quark, Due west.H.Freeman and Company, 1994
J. A. Paulos, Beyond Numeracy, Vintage Books, 1992
D. Pedoe, The Gentle Fine art of Mathematics, Dover, 1973
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987.