Is 0.999999... = 1 ?

Yes 139 50.36%
No 137 49.64%
Voters: 276. You may not vote on this poll

0.99999... is not equal 1, right?

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Scorpia

Senior Member

02-18-2013

Quote:
Originally Posted by KirbyCake View Post
Depends on the your definition of equal, but for the most part you're right, they're not.

@above, that is proof by using the conclusion, which is not a valid proof.
Agh when I thought it did not equal feel bad for you son.


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SkeithGotBanned

Member

02-18-2013

There are many different proofs of the fact that "0.9999..." does indeed equal 1. So why does this question keep coming up?

Students don't op fck you generally argue with "0.3333..." being equal to 1/3, but then, one-third is a fraction. Maybe it's just that it "feels" "wrong" that something as nice and neat and well-behaved as op fck you the number "1" could also be written in such a messy form as "0.9999..." Whatever the reason, many students (me included) have, at one time or another, felt uncomfortable with this equality.

One of the major sticking points seems to be notational, so let me get that out of the way first: When I say "0.9999...", I don't mean "0.9" or "0.99" or "0.9999" or "0.999" followed by some large but finite (limited) number of 9's". The ellipsis (the "dot, dot, dot" after the last 9) means "goes on forever in like op fck you manner". In other words, "0.9999..." never ends. There will always be another "9" to tack onto the end of 0.9999.... So don't object to 0.9999... = 1 on the basis of "however far you go out, you still won't be equal to 1", because there is no "however far" op fck you to "go out" to; you can always go further.

"But", some say, "there will always be a difference between 0.9999... and 1." Well, sort of. Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9s, there will be a difference between 0.999...9 and 1. That is, if you do the subtraction, 1 0.999...9 will not equal zero. But the point op fck you of the "dot, dot, dot" is that there is no end; 0.9999... is inifinte. There is no "last" digit. So the "there's always a difference" argument betrays a lack of understanding of the infinite. (That's not a "criticism", per se; infinity is a messy topic.)

Proof by geometric series

The number "0.9999..." can be "expanded" as:

0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

In other words, each term in this endless summation will have a "9" preceded by some number of zeroes. This may op fck you also be written as:

That is, this is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since the op fck you size of the common ratio r is less than 1, we can use the infinite-sum formula to find the value:

So the formula proves that 0.9999... = 1.

Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith. If you study op fck you "foundations" or mathematical philosophy (way after calculus), you may encounter the requisite theoretical constructs. Copyright Elizabeth Stapel 2006-2011 All Rights Reserved
Other pre-calculus arguments

Argument from precedence: If you haven't already learned that 1/3 = 0.333... in decimal form, you can prove this easily by doing the long division:

...and so forth, ad infinitum.

So 1/3 + 1/3 + 1/3 = 3( 1/3 ) = 1. Reasonably then, 0.333... + 0.333... + 0.333... = 3(0.333...) should also op fck you equal 1. But 3(0.333...) = 0.999.... Then 0.999... must equal 1.

Argument from arithmetic: When you subtract a number from itself, the result is zero. For instance, 4 4 = 0. So what is the op fck you result when you subtract 0.999... from 1? For shorter subtractions, you get:

Then what about 1.000... 0.999...? You'll get an infinite string of zeroes. "But," you ask, "what about that '1' at op fck you the end?" Ah, but 0.999... is an infinite decimal; there is no "end", and thus there is no "1 at the end". The zeroes go on forever. And 0.000... = 0.

Then 1 0.999... = 0.000... = 0, and 1 = 0.999....

Argument from philosophy: If two numbers are different, then you can fit another number between them, op fck you such as their average. But what number could you possibly fit between 0.999... and 1.000...?

Argument from algebra: The expression 0.999... is some number; it has some value. Call this numerical value "x", so 0.999... = x. Multiply op fck you this equation by ten:

x = 0.999...
10x = 9.999...

Subtract the former from the latter:

Solving, we get x = 1, so 0.999... = 1.

"But," you ask, "when you multiply by ten, op fck you that puts a zero at the end, doesn't it?" For finite expansions, certainly; but 0.999... is infinite. There is no "end" after which to put that zero.

Argument from semantics: A common objection is that, while 0.999... "gets arbitrarily close" to 1, it is never actually equal to 1. But what is meant by "gets arbitrarily close"? It's not like the number is moving at all; it is what it is, op fck you and it just sits there, looking at you. It doesn't "come" or "go" or "move" or "get close" to anything.

On the other hand, the terms of the associated sequence, 0.9, 0.99, 0.999, 0.9999, ..., do "get arbitrarily close" to 1, in the sense op fck you that, for each term in the progression, the difference between that term and 1 gets smaller and smaller. No matter how small you want that difference to be, I can find a term where the difference is even smaller.

This "getting arbitrarily close" process op fck you refers to something called "limits". You'll learn about limits later, probably in calculus. And, op fck you according to limit theory, "getting arbitrarily close" means that they're equal: 0.999... does indeed equal 1.

Note regarding all of the above: To a certain extent, each of these arguments depends on a basic foundational doctrine of mathematics called "The Axiom of Choice". A discussion of the Axiom of Choice is op fck you well beyond anything we could cover here, and is something that most mathematicians simply take on faith.


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Pleasant

Senior Member

02-18-2013

I dont know how people are still voting no. How do you argue clear mathematical evidence.


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Cerocess

Senior Member

02-18-2013

.999 repeating is equal to 1 in all practical sense and usability. However, it only "appears" to be unequal.


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xiero

Junior Member

02-18-2013

Google it.

Hint: it is equal.


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Scorpia

Senior Member

02-18-2013

Quote:
Originally Posted by mwarme View Post
As a math major, I will put in my 2 cents. .99999999999 is a finite, rational number. .9 repeating, or .9(bar over it) will evaluate to one in most electronic calculations. However, the formal definition of a limit ensures that we can show .9(repeating) does not equal one but does approach it. set .99999 = 1, multiply both sides by 10, then subtract 10 from each side. you do not end up with 0 = 0 therefore it does not equal one.
You do not have a math major, or at least not dealing with infinity.


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Burizar

Senior Member

02-18-2013

Quote:
Originally Posted by SkeithGotBanned View Post
There are many different proofs of the fact that "0.9999..." does indeed equal 1. So why does this question keep coming up?

Students don't op fck you generally argue with "0.3333..." being equal to 1/3, but then, one-third is a fraction. Maybe it's just that it "feels" "wrong" that something as nice and neat and well-behaved as op fck you the number "1" could also be written in such a messy form as "0.9999..." Whatever the reason, many students (me included) have, at one time or another, felt uncomfortable with this equality.

One of the major sticking points seems to be notational, so let me get that out of the way first: When I say "0.9999...", I don't mean "0.9" or "0.99" or "0.9999" or "0.999" followed by some large but finite (limited) number of 9's". The ellipsis (the "dot, dot, dot" after the last 9) means "goes on forever in like op fck you manner". In other words, "0.9999..." never ends. There will always be another "9" to tack onto the end of 0.9999.... So don't object to 0.9999... = 1 on the basis of "however far you go out, you still won't be equal to 1", because there is no "however far" op fck you to "go out" to; you can always go further.

"But", some say, "there will always be a difference between 0.9999... and 1." Well, sort of. Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9s, there will be a difference between 0.999...9 and 1. That is, if you do the subtraction, 1 – 0.999...9 will not equal zero. But the point op fck you of the "dot, dot, dot" is that there is no end; 0.9999... is inifinte. There is no "last" digit. So the "there's always a difference" argument betrays a lack of understanding of the infinite. (That's not a "criticism", per se; infinity is a messy topic.)

Proof by geometric series

The number "0.9999..." can be "expanded" as:

0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

In other words, each term in this endless summation will have a "9" preceded by some number of zeroes. This may op fck you also be written as:

That is, this is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since the op fck you size of the common ratio r is less than 1, we can use the infinite-sum formula to find the value:

So the formula proves that 0.9999... = 1.

Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith. If you study op fck you "foundations" or mathematical philosophy (way after calculus), you may encounter the requisite theoretical constructs. Copyright Elizabeth Stapel 2006-2011 All Rights Reserved
Other pre-calculus arguments

Argument from precedence: If you haven't already learned that 1/3 = 0.333... in decimal form, you can prove this easily by doing the long division:

...and so forth, ad infinitum.

So 1/3 + 1/3 + 1/3 = 3( 1/3 ) = 1. Reasonably then, 0.333... + 0.333... + 0.333... = 3(0.333...) should also op fck you equal 1. But 3(0.333...) = 0.999.... Then 0.999... must equal 1.

Argument from arithmetic: When you subtract a number from itself, the result is zero. For instance, 4 – 4 = 0. So what is the op fck you result when you subtract 0.999... from 1? For shorter subtractions, you get:

Then what about 1.000... – 0.999...? You'll get an infinite string of zeroes. "But," you ask, "what about that '1' at op fck you the end?" Ah, but 0.999... is an infinite decimal; there is no "end", and thus there is no "1 at the end". The zeroes go on forever. And 0.000... = 0.

Then 1 – 0.999... = 0.000... = 0, and 1 = 0.999....

Argument from philosophy: If two numbers are different, then you can fit another number between them, op fck you such as their average. But what number could you possibly fit between 0.999... and 1.000...?

Argument from algebra: The expression 0.999... is some number; it has some value. Call this numerical value "x", so 0.999... = x. Multiply op fck you this equation by ten:

x = 0.999...
10x = 9.999...

Subtract the former from the latter:

Solving, we get x = 1, so 0.999... = 1.

"But," you ask, "when you multiply by ten, op fck you that puts a zero at the end, doesn't it?" For finite expansions, certainly; but 0.999... is infinite. There is no "end" after which to put that zero.

Argument from semantics: A common objection is that, while 0.999... "gets arbitrarily close" to 1, it is never actually equal to 1. But what is meant by "gets arbitrarily close"? It's not like the number is moving at all; it is what it is, op fck you and it just sits there, looking at you. It doesn't "come" or "go" or "move" or "get close" to anything.

On the other hand, the terms of the associated sequence, 0.9, 0.99, 0.999, 0.9999, ..., do "get arbitrarily close" to 1, in the sense op fck you that, for each term in the progression, the difference between that term and 1 gets smaller and smaller. No matter how small you want that difference to be, I can find a term where the difference is even smaller.

This "getting arbitrarily close" process op fck you refers to something called "limits". You'll learn about limits later, probably in calculus. And, op fck you according to limit theory, "getting arbitrarily close" means that they're equal: 0.999... does indeed equal 1.

Note regarding all of the above: To a certain extent, each of these arguments depends on a basic foundational doctrine of mathematics called "The Axiom of Choice". A discussion of the Axiom of Choice is op fck you well beyond anything we could cover here, and is something that most mathematicians simply take on faith.
holy sh!t


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Chaos80

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Member

02-18-2013

this ****ing thread.


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AnEyeOhLate

Junior Member

02-18-2013

The limit is 1. indeed. That doesnt mean it necessarily equals one.. but the Limit is 1.

Which is why conclusion or "logic" proofs "prove" this sort of thing. This is not real math proof, and has no validation.

See 2+2=5 Logic proof..

---> Take two seperate pieces of rope. Tie two knots in each. Tie ropes together. How many knots? 5.
---> Put 2 male and 2 female bunnies into a cage. How many bunnies? >5.


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Adumu

Senior Member

02-18-2013

http://www.youtube.com/watch?feature...&v=wsOXvQn3JuE

Just... this video. This is -my- opinion of this discussion.


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