LOCK I DUN LIKE ZERO

[ragnarok1945]

it doesn't, but the difference is so small in practical applications today, you can say they're the same thing

[Guest]

You can't divide by zero.

 

Chuck Norris can.

 

Thus' date=' 0/0 is undefined.

 

I want to change the topic, the question was already answered.

[/quote']

 

Yes you can, the standard answer is undefined, but a theoretical answer is actually different (it's just that virtually no one sees such this answer)

 

Unless your theoretical answer is an approximation of what it ought to be based on the use of limits, your theoretical answer is trash and ignores the literal meaning of "undefined". It is "undefined" in the sense that it is literally not defined; the definition of division specifically disallows it.

 

' pid='1442117' dateline='1228702688']

But I still don't understand why 0/0 is undefined.

 

I posted a proof in my last post. Read it.

 

Sorry' date=' this is a math topic.

 

What rangarok1945 said, but I don't know anything about the subject.

 

How about this: Why does 0.999999999999999999999.....=1? Supposedly, 0.999.... would be a very tiny bit less than 1, but that is not the case...

 

I just feel like testing people.

[/quote']

 

Here's a simple explanation:

 

Define x where x = .999...

 

x = .999...

 

Multiplying both sides of the equation by ten:

 

10x = 9.999...

 

By definition of x:

 

10x = 9 + x

 

Subtracting x from both sides of the equation:

 

9x = 9

 

Dividing:

 

x = 1

 

This is by no means a rigorous proof, but it's an easy way of giving you a general sense of why it is true. The more rigorous way of doing it is to simply state .999... as a geometric series of first term .9 and common ratio 1/10, then simply calculate the convergent sum - it's 1.

 

it doesn't' date=' but the difference is so small in practical applications today, you can say they're the same thing

[/quote']

 

No. .999... = 1 in the same way that 1 + 1 = 2. They are not just really really close; they are precisely equal.

[Quicksilver]

[ragnarok1945]

You can't divide by zero.

 

Chuck Norris can.

 

Thus' date=' 0/0 is undefined.

 

I want to change the topic, the question was already answered.

[/quote']

 

Yes you can, the standard answer is undefined, but a theoretical answer is actually different (it's just that virtually no one sees such this answer)

 

Unless your theoretical answer is an approximation of what it ought to be based on the use of limits, your theoretical answer is trash and ignores the literal meaning of "undefined". It is "undefined" in the sense that it is literally not defined; the definition of division specifically disallows it.

 

' pid='1442117' dateline='1228702688']

But I still don't understand why 0/0 is undefined.

 

I posted a proof in my last post. Read it.

 

Sorry' date=' this is a math topic.

 

What rangarok1945 said, but I don't know anything about the subject.

 

How about this: Why does 0.999999999999999999999.....=1? Supposedly, 0.999.... would be a very tiny bit less than 1, but that is not the case...

 

I just feel like testing people.

[/quote']

 

Here's a simple explanation:

 

Define x where x = .999...

 

x = .999...

 

Multiplying both sides of the equation by ten:

 

10x = 9.999...

 

By definition of x:

 

10x = 9 + x

 

Subtracting x from both sides of the equation:

 

9x = 9

 

Dividing:

 

x = 1

 

This is by no means a rigorous proof, but it's an easy way of giving you a general sense of why it is true. The more rigorous way of doing it is to simply state .999... as a geometric series of first term .9 and common ratio 1/10, then simply calculate the convergent sum - it's 1.

 

it doesn't' date=' but the difference is so small in practical applications today, you can say they're the same thing

[/quote']

 

No. .999... = 1 in the same way that 1 + 1 = 2. They are not just really really close; they are precisely equal.

 

Their definition of precise is actually just an approximation. These numbers are not the same, but the difference is too small to matter, so they say it's precise.

[Guest]

You can't divide by zero.

 

Chuck Norris can.

 

Thus' date=' 0/0 is undefined.

 

I want to change the topic, the question was already answered.

[/quote']

 

Yes you can, the standard answer is undefined, but a theoretical answer is actually different (it's just that virtually no one sees such this answer)

 

Unless your theoretical answer is an approximation of what it ought to be based on the use of limits, your theoretical answer is trash and ignores the literal meaning of "undefined". It is "undefined" in the sense that it is literally not defined; the definition of division specifically disallows it.

 

' pid='1442117' dateline='1228702688']

But I still don't understand why 0/0 is undefined.

 

I posted a proof in my last post. Read it.

 

Sorry' date=' this is a math topic.

 

What rangarok1945 said, but I don't know anything about the subject.

 

How about this: Why does 0.999999999999999999999.....=1? Supposedly, 0.999.... would be a very tiny bit less than 1, but that is not the case...

 

I just feel like testing people.

[/quote']

 

Here's a simple explanation:

 

Define x where x = .999...

 

x = .999...

 

Multiplying both sides of the equation by ten:

 

10x = 9.999...

 

By definition of x:

 

10x = 9 + x

 

Subtracting x from both sides of the equation:

 

9x = 9

 

Dividing:

 

x = 1

 

This is by no means a rigorous proof, but it's an easy way of giving you a general sense of why it is true. The more rigorous way of doing it is to simply state .999... as a geometric series of first term .9 and common ratio 1/10, then simply calculate the convergent sum - it's 1.

 

it doesn't' date=' but the difference is so small in practical applications today, you can say they're the same thing

[/quote']

 

No. .999... = 1 in the same way that 1 + 1 = 2. They are not just really really close; they are precisely equal.

 

Their definition of precise is actually just an approximation. These numbers are not the same, but the difference is too small to matter, so they say it's precise.

 

No, you don't understand. They're not infinitesimally close to one another; they are literally exactly the same.

 

Consider this: .999... can be defined as the sum of a geometric series whose first term is .9 and whose common ratio is .1. As we all know, the formula for the convergent sum of a geometric series whose common ratio has an absolute value less than one is given by the formula t/(1-r), where t is the first term and r is the common ratio. Plugging in our values, we have .9/(1-.1) = .9/.9 = 1. QED.

 

Learn math.

[ragnarok1945]

but the very first term is not 0.9, it's 0.999999999999999..............

[Guest]

but the very first term is not 0.9' date=' it's 0.999999999999999..............

[/quote']

In this case, Crab is right:

 

http://en.wikipedia.org/wiki/.999_%3D_1

 

 

My question, though, is: how many 9's does it take to be considered the rounded number? As in, .999 = 1, but would .99 = 1?

[ragnarok1945]

No, a 0.01 difference is still too great to be considered equal to one. It takes at least 3 decimals

[Guest]

but the very first term is not 0.9' date=' it's 0.999999999999999..............

[/quote']

 

It's a geometric series. Essentially, we are re-expressing 0.999... as the sum of .9 + .09 + .009 + .0009 + ... and so on ad infinitum. The first term of this sum is .9, and each subsequent term is the term preceding it multiplied by .1.

 

My question' date=' though, is: how many 9's does it take to be considered the rounded number? As in, .999 = 1, but would .99 = 1?

[/quote']

 

It depends on the case in question. In pure math, you would need an infinite series of nines for it to be rounded (although you should still check to make sure your calculator didn't make a minor rounding error when doing calculations). In a science, it depends on how many sig figs you have.

[ragnarok1945]

but the very first term is not 0.9' date=' it's 0.999999999999999..............

[/quote']

 

It's a geometric series. Essentially, we are re-expressing 0.999... as the sum of .9 + .09 + .009 + .0009 + ... and so on ad infinitum. The first term of this sum is .9, and each subsequent term is the term preceding it multiplied by .1.

 

 

ok my bad. It's been a long time since I've done geometric series. But yes, I remember now, you just keep adding the series like this continuously, then with the formula given you can determine what number it will reach at infinity

[Guest]

It depends on the case in question. In pure math' date=' you would need an infinite series of nines for it to be rounded (although you should still check to make sure your calculator didn't make a minor rounding error when doing calculations). In a science, it depends on how many sig figs you have.

[/quote']

Ah, mm'kay. That makes sense.

[ragnarok1945]

Not all science. Certain forms of engineering (engineering is still a science) will only care up to like 6 decimals and that's it.

[Guest]

Not all science. Certain forms of engineering (engineering is still a science) will only care up to like 6 decimals and that's it.

 

Learn what sig figs are.

[ragnarok1945]

I know what they are, Crab Helmet. During the times they apply your answer may have only one (such as 10000) because the problem's numbers have only 1 sig fig

[Guest]

I know what they are' date=' Crab Helmet. During the times they apply your answer may have only one (such as 10000) because the problem's numbers have only 1 sig fig

[/quote']

 

Then you should know that the number of important digits in the result is determined not by the science being used but by the precision to which the input numbers are known.

[ragnarok1945]

I do, but for some cases you use more than number of sig figs given to you from the input (physics stressed that you don't, but later on in engineering they flex it a little)

[Guest]

I do' date=' but for some cases you use more than number of sig figs given to you from the input (physics stressed that you don't, but later on in engineering they flex it a little)

[/quote']

 

There is very little point in using more sig figs than you actually have; by that point, you are basically making up numbers. In contrast, using fewer sig figs than you actually have is a senseless sacrifice of precision.

[ragnarok1945]

that may well be, but in engineering practice sometimes you must make up numbers. Take for example a problem who's lowest sig fig is 1, but all the other numbers given have up to 8 sig figs

 

According to you the answer should have only one sig fig.

 

You do that and your answer could be like 6 times off compared to you using 8 sig figs, where you'd be more accurate.

[Guest]

that may well be' date=' but in engineering practice sometimes you must make up numbers. Take for example a problem who's lowest sig fig is 1, but all the other numbers given have up to 8 sig figs

 

According to you the answer should have only one sig fig.

 

You do that and your answer could be like 6 times off compared to you using 8 sig figs, where you'd be more accurate.

[/quote']

 

Using non-existent sig figs will not improve the accuracy of your answer. The additional digits added to the length of your answer will be as good as meaningless.

[Monkey]

Its undefined, ANYTHING divided by 0 = 0/undefined, even if the numerator is also 0

[ragnarok1945]

that may well be' date=' but in engineering practice sometimes you must make up numbers. Take for example a problem who's lowest sig fig is 1, but all the other numbers given have up to 8 sig figs

 

According to you the answer should have only one sig fig.

 

You do that and your answer could be like 6 times off compared to you using 8 sig figs, where you'd be more accurate.

[/quote']

 

Using non-existent sig figs will not improve the accuracy of your answer. The additional digits added to the length of your answer will be as good as meaningless.

 

And using just one sig fig also worsens the accuracy of your answer.

[Guest]

that may well be' date=' but in engineering practice sometimes you must make up numbers. Take for example a problem who's lowest sig fig is 1, but all the other numbers given have up to 8 sig figs

 

According to you the answer should have only one sig fig.

 

You do that and your answer could be like 6 times off compared to you using 8 sig figs, where you'd be more accurate.

[/quote']

 

Using non-existent sig figs will not improve the accuracy of your answer. The additional digits added to the length of your answer will be as good as meaningless.

 

And using just one sig fig also worsens the accuracy of your answer.

 

If you truly only have one sig fig, then it will not do so in general. By definition, only having one sig fig implies that all other digits are as accurate as random guesses.

[ragnarok1945]

and what if every other input number has up to 8 sig figs? Are we supposed to say they won't matter?

[Jerry OldMaster]

I'm following this thread =)

Keep them coming.

[Guest PikaPerson01]

Sorry to backtrack to that .999... = 1 thing but:

 

http://qntm.org/?pointnine