Is 0.9 recurring equal to 1?

Question

Someone told me that 0.9 recurring is equal to 1. Is this right? I would be eternally grateful if someone could tell me the answer. I just dont think it can possibly be right as it sounds so inplausible. However the person who told me has a maths degree from a top British university so may be right. Thanks!

Philip Chandler

Answer 1

it's as close as you can get but it doesn't equal it.

Answer 2

The answer to your question is generally yes,

0.9 repeating = 1.

Only in very specialized (or specialised, for your UK friend)
situations involving infinitessimal arithmetic would there be
a different answer. Infinitessimals can be useful, but they
would not be taken as the normal mode of operation.

We define 0.9 repeating to be the limit as N goes to infinity
of 0.9999...9 (string of N 9's), which is a finite length decimal
equal to (without controversy)

1 - 1 / 10^N.

By the basic limit theorem, if lim N-> infinity f(N) exists and
is finite and lim N->infinity g(N) exists and is finite, then
lim N->infinity ( f(N) - g(N) ) is simply the difference.
This is pedantry, sorry, but the objectors demand it.

In our case, f(N) = 1, the constant, and the limit is obviously
1. Our g(N) = 1/ 10^N.

lim N->infiinity 1/10^N = 0, because, for any epsilon greater
than zero, (chosen by an adversary), I can produce an A,
depending on epsilon, such that for any N > A, the
inequality 1/10^N < epsilon holds. It is left to the reader to
find one of many schemes for choosing A that works.

I will now cease writing N->infinity with my lim's,
because they are all the same but you should
imagine that it is written.

This is the definition of the limit, and the result is a static
quantity. We may be sloppy and say, "the limit approaches
such a value, " but we mean, "the expression next to the
limit symbol approaches such a value as the variable
underneath does what the arrow says it does."

So, 1/10^N approaches 0 as N approaches infinity, AND
lim 1/10^N is equal to 0, full stop.

Note that it is possible to tinker with the definition of
limit, and get a different number system involving infinitessimals.
Then one has to resolve many consequences for
calculus, analysis, and topology. As mentioned initially,
for specialized pursuits, this may be the way to go, but
as an objection to the question at hand, it is sophistry.


Applying the basic limit theorem,

.9 repeating = lim 1 - /10^N = lim 1 - lim 1/10^N = 1 - 0 = 1.

QED

Answer 3

Well, it depends on how specific you want to be. If you want to be VERY exact, then no, it is not equal, but if you round it, it would be a yes. For example, if you had .99999999999999 cents, then that would mean you have a cent.
If you are trying to make it exact, IE for a science measuring, and you need to be very accurate, then no, .9 recurring does not equal 1. In other words, .9 recurring is the closest number to 1, but, no matter how close, it still does not equal 1. In money and other events, like having 99.999999cents (which would be a dollar), it does seem as if it equals 1

Answer 4

All the answers except perhaps first, are inaccurate to some extent. Rounding off is an approximation procedure applied to decimals which may or may not be terminating. 0.3333..... 3 to any finite number is always < 1/3, but 0.33..... up to infinity non stopping = 1/3. How? 0.33......→ ∞ = (3/10)+(3/100)+((3/1000)+------.→ ∞[It is a geometric infinite series with common ratio 1/10 and sums to: 0.33......→ ∞ = [(3/10)/{1 - (1/10)}] = [(3/10)/(9/10)] = (3/9) = (1/3) Similarly, 0.99......→ ∞ = (9/10)+(9/100)+((9/1000)+------.→ ∞[It is a geometric infinite series with common ratio 1/10 and sums to: 0.99......→ ∞ = [(9/10)/{1 - (1/10)}] = [(9/10)/(9/10)] = (9/9) = 1 So it is exactly equal to one no approximation! No one can humanly sum the infinite series so one has to use special techniques of limit to get the formula for sum, S of infinite series as S = [a/(1-r)], where a is the first term and r is common ratio. when r <1 the sum converges to finite value. The answer of Lucas C s simpler and illuminating and accurate too!

Answer 5

Well saying it is theortically equal is untrue. However, image for a moment a measurement of 1, vs. a measurement of .9999 ... to a thousand places.

Practically, no human being, or tool made by man, could tell the difference between the too. Whether it be the height of a building, the width of a screw, or the standing weight of a bridge, the measurements are identical in their application.

If you had .999999999999999 million dollars, that's a million bucks.

I think that's what your source means.

Answer 6

If you have 0.99999.... as you add the digits it will approach closer and closer 1. So if you have infinite number of digits, it will equal to 1. However, practically speaking, you cannot have infinite numbers and still express it.

Be careful though, you can write down one million 9s and say it is not equal to 1.... always remember you can always add more 9s and you can repeat this all over again.

When you get to calculus level, they will discuss this....

Answer 7

Philip Chandler Wow!Your Extra News here!
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Answer 8

yes, it's right. If you've had infinite geometric series in Alg 2, your 0.9 recurring is

0.9 + 0.09 + 0.009 + ....

which has a starting value of 0.9 and a common ratio of 0.1, so the sum is
0.9 / (1 - 0.1) = 0.9 / 0.9 = 1

Answer 9

First responder is right. Technically, it approaches 1 as a limit, but is never equal.

Answer 10

according to standard it is
as in Texas INstruments calculations

and 1/9 is equal to 0.1 recurring

and so 9/9 woudld be 0.9 recurring

but 9/9 is equal to 1

so it is a bit contradictory

Answer 11

It does not equal one. It has a mathematical limit of one.