Math question

[Tonylu1595]

So I noticed that .99 repeating is equal to 1 because the limit is 1. So therefore, is every integer and fraction just a limit of a decimal? Such as .50000000 repeating 0 has a limit of 1/2, therefore 1/2 is the limit of the decimal. 

[Statik]

My head.

Yes

[Sauron]

Well, the math behind 0.9 recurring being equal to 1 is much simpler:

 

1/3 = 0.3 recurring

 

0.3 recurring * 3 = 0.9 recurring.

 

Therefore 0.9 recurring must be equal to 1.

 

Repeating 0s count as null, read up on taylor's polynomials if you want to go deeper in this

[DimasRMDO]

My head.

^

That's me right now

[Red Dragon]

Well, the math behind 0.9 recurring being equal to 1 is much simpler:

 

1/3 = 0.3 recurring

 

0.3 recurring * 3 = 0.9 recurring.

 

Therefore 0.9 recurring must be equal to 1.

 

Repeating 0s count as null, read up on taylor's polynomials if you want to go deeper in this

 

but its not, its just so very close that its is considered 1

[Tonylu1595]

but its not, its just so very close that its is considered 1

Actually, no. .9999 repeating IS equal to 1.

 

Students of mathematics often reject the equality of 0.999... and 1, for reasons ranging from their disparate appearance to deep misgivings over the limit concept and disagreements over the nature of infinitesimals. There are many common contributing factors to the confusion:

• Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.^[35]
• Some students interpret "0.999..." (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".^[36]
• Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999..." as meaning the sequence rather than its limit.^[37]

http://en.wikipedia.org/wiki/0.999...#Skepticism_in_education

[Tonylu1595]

Well, the math behind 0.9 recurring being equal to 1 is much simpler:

 

1/3 = 0.3 recurring

 

0.3 recurring * 3 = 0.9 recurring.

 

Therefore 0.9 recurring must be equal to 1.

 

Repeating 0s count as null, read up on taylor's polynomials if you want to go deeper in this

That would lead people to distrust 1/3, as 1/3 is also the limit of .3 repeating.

[ThePhysicist]

So I noticed that .99 repeating is equal to 1 because the limit is 1. So therefore, is every integer and fraction just a limit of a decimal? Such as .50000000 repeating 0 has a limit of 1/2, therefore 1/2 is the limit of the decimal. 

 Personally I don't think that the limit is the right way to comprehend that .9.... = 1

I find it very clear thinking about the axiom of completeness of real number: There is no number between .999... and 1, and this is easily comprehensible, but because the axiom says that there are no "gaps" between real numbers that implies that .999... is exactly equal to 1  

[Red Dragon]

Actually, no. .9999 repeating IS equal to 1.

http://en.wikipedia.org/wiki/0.999...#Skepticism_in_education

lol Wikipedia

[Tonylu1595]

lol Wikipedia

A very good source for maths and science, not so good for politics and people. 

 

And, according to limit theory, "getting arbitrarily close" means that they're equal:0.999... does indeed equal 1.

http://www.purplemath.com/modules/howcan1.htm

[GrimNeo]

lol Wikipedia

Yup, I went to Wikipedia college. Can validate that its a solid source.

 

/s

[Tonylu1595]

Yup, I went to Wikipedia college. Can validate that its a solid source.

 

/s

Fine 

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, 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 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 refers to something called "limits". You'll learn about limits later, probably in calculus. And, 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 well beyond anything we could cover here, and is something that most mathematicians simply take on faith.

http://www.purplemath.com/modules/howcan1.htm

[ThePhysicist]

Yes, 0.999... is not very close to 1, it's exactly 1! Is another way to say 1 like 4/2 is another way to say 2! there is not any approximation involved! 

[Sauron]

but its not, its just so very close that its is considered 1

 

I just proved to you that it is 1

 

1 = 1/3 * 3 = o.333... * 3 = 0.999999...

 

1 = 0.9999...

[SIGSEGV]

0.0 mind blown

[-TesseracT-]

Well, the math behind 0.9 recurring being equal to 1 is much simpler:

 

1/3 = 0.3 recurring

 

0.3 recurring * 3 = 0.9 recurring.

 

Therefore 0.9 recurring must be equal to 1.

 

Repeating 0s count as null, read up on taylor's polynomials if you want to go deeper in this

But 1/3 isn't 0.3 recurring which can't be equal to 1 if you multiply by 3. 1/3 is simply 1/3, it is impossible to exactly write in decimal form.

[-TesseracT-]

Accidental double post 

[Sauron]

But 1/3 isn't 0.3 recurring which can't be equal to 1 if you multiply by 3. 1/3 is simply 1/3, it is impossible to exactly write in decimal form.

 

1/3 is 0.3 recurring, it is by definition a rational number, so we know what it is even if we're physically incapable of writing it somewhere. It's a common misconception that there is a missing 1 at the end of 0.9 recurring, but the fact is that there is no "end". The number goes on forever. And it is equal to 1. It's not an approximation, they are the same number. 0.9 recurring is also a rational number, meaning it is equal to a/b where a and b are integers. Can you find two integers different from 1 that give 0.9 recurring?

[-TesseracT-]

1/3 is 0.3 recurring, it is by definition a rational number, so we know what it is even if we're physically incapable of writing it somewhere. It's a common misconception that there is a missing 1 at the end of 0.9 recurring, but the fact is that there is no "end". The number goes on forever. And it is equal to 1. It's not an approximation, they are the same number. 0.9 recurring is also a rational number, meaning it is equal to a/b where a and b are integers. Can you find two integers different from 1 that give 0.9 recurring?

My mind cannot comprehend that 0.9 recurring would equal 1.

[Sauron]

My mind cannot comprehend that 0.9 recurring would equal 1.

 

[Stefken89]

Well, 0.9999... is not 1. 

 

the limit of n->+inf for   0,(n times 9) is 1. That is not the same, right?

Having a limit of a function would mean that it can never exactly reach that limit but it approaches that limit infinately close.

[ThePhysicist]

Well, 0.9999... is not 1. 

 

the limit of n->+inf for   0,(n times 9) is 1. That is not the same, right?

Having a limit of a function would mean that it can never exactly reach that limit but it approaches that limit infinately close.

 

Nope, is a misconception! 0.999.... is the limit of what you have wrote! in 0.999... there is not a final 9 so there cannot be a number between it and 1 so it must be equal to one (this because there are no gaps between real numbers and if you cannot find a number between two numbers those two numbers must be equal)

[-TesseracT-]

 

I asked my maths teacher if 1/3 in decimal form multiplied by 3 was 1. He said it will get very close to 1, but never get there. I think he knows what he's talking about.

[Tonylu1595]

I asked my maths teacher if 1/3 in decimal form multiplied by 3 was 1. He said it will get very close to 1, but never get there. I think he knows what he's talking about.

I think he didn't learn limits yet. Or he didn't understand the question. The problem isn't that it's not proven, it is. 

[Toddskins]

Your math teacher was wrong.

 

There are actually 4 proofs to this, two of which have been mentioned in this thread already, and I do not remember the 4th.  Here are the three I know, 1 & 2 already stated in this thread:

 

1) 1/3 = .3 repeating (I'm unable to draw the line over the .3, so we accept the word "repeating").

 

and 2/3 = .6 repeating (no problem here.  It exactly IS .6 repeating)

 

and 3/3 = .9 repeating = 1

 

 

2) Fact: If two numbers are NOT equal, then there is an infinite set of numbers BETWEEN those 2 numbers.  There are no numbers between .9 repeating and 1, so therefore we know they are equal.

 

3) From algebra days, if you do the same thing to an equation, and then subtract those two equations, you are left with the same thing.  Thus:

 

Let x = .9 repeating

 

Multiply both sides by 10, and we now have 10x = 9.9 repeating.

 

Subtract the original equation from the new equation:

 

10x = 9.9 repeating

-   x =   .9 repeating

________________

 

 9x = 9  (the .9 repeating has been subtracted leaving just 9)

 

Now, solve for x

 

x = 1