Poll: 6÷2(1+2)

[ordinaryjoe]

a number divided by 0 is not infinity... It is undefined. Saying a number divided by 0 is infinity implies that 0 times infinity can be whatever number you want.

Anyway, what about -1/0?

 

[FoxEye]

And yet you say that 0.0r1 is not equal to 0?

0.0r1 doesn't exist. If the 0 is recurring there will never be a 1 at the end of it.

In other words, wherever you chose to put the 1 you are wrong, it should be 0.

 

[ordinaryjoe]

And yet you say that 0.0r1 is not equal to 0?

0.0r1 does not exist. You can not have an infinite number of 0s before the 1 - the 1 would not exist, so it's really just 0

 

[Vonhelmet]

And yet you can have enough 9s that 0.99r is equal to 1?

Admit the contradiction.

 

[Vonhelmet]

a number divided by 0 is not infinity... It is undefined. Saying a number divided by 0 is infinity implies that 0 times infinity can be whatever number you want.

Anyway, what about -1/0?

The answer is infinity, but it is infinity that is undefined, not the answer.

And -1/0 is -infinity.

And yes, before you ask, infinity can be positive or negative.

 

[wmb]

They think its 9 because by convention it is 9

we introduce these conventions in order to remove ambiguity

by having an order of operations you remove the need to use extra parenthesis

the answer is only 1 if you chose to ignore convention and assume that the question is either asking something it isn't or is ambiguous with a possible answer of 1 or 9. With convention the ambiguity is removed and the answer is 9.

But the equation is not presented in a CONVENTIONAL format, therefore it is AMBIGUOUS. If it was presented in a conventional way, there would be no argument one way or the other.

 

[FoxEye]

The answer is infinity, but it is infinity that is undefined, not the answer.

And -1/0 is -infinity.

And yes, before you ask, infinity can be positive or negative.

I believe this is incorrect.

You cannot divide something finite and existing into an infinite number of non-existent parts.

Even if some branch of mathematics says you can, it contradicts reality.

 

[Vonhelmet]

I believe this in incorrect.

You cannot divide something finite and existing into an infinite number of non-existent parts.

In the limit as x->0 that is exactly what you can conceptually do.

 

[marc_howarth]

Basically if you do multiplying before diving you get one answer and if you do dividing before multiplying you get another answer. The question is ambiguous and really both answers are correct.

 

[Castiel]

But the equation is not presented in a CONVENTIONAL format, therefore it is AMBIGUOUS. If it was presented in a conventional way, there would be no argument one way or the other.

+0.99r

 

[mjt]

25 pages? I'm guessing the topic has moved on?

I mean really. The only way to get 1 is 6/(2(1+2))

 

[ordinaryjoe]

And yet you can have enough 9s that 0.99r is equal to 1?

Admit the contradiction.

There is no contradiction. I don't run out of 9s... there is no limited supply of 9s and there is nothing that comes after it. However, you can't have infinity 0s, then add a 1 on after. If you could, the 1 would have a value of 10^-infinity, which is 0.

The answer is infinity, but it is infinity that is undefined, not the answer.

And -1/0 is -infinity.

And yes, before you ask, infinity can be positive or negative.

You said that all numbers divided by 0 are infinity though, so what happens when you multiply infinity by 0?

 

[Killerkebab]

We don't say 1/0 is infinity, we say 1/0 is undefined

However, if you take 1/x, we say that as x -> 0 (i.e: as x gets smaller and smaller), then 1/x -> infinity.

You read "->" as "tends to"

 

[ordinaryjoe]

Exactly, but x tending to 0 isn't the same as dividing by zero. It is finding out what the number approaches as you get closer and closer to 0.

 

[Killerkebab]

Exactly, but x tending to 0 isn't the same as dividing by zero. It is finding out what the number approaches as you get closer and closer to 0.

Yes sir.

 

[FoxEye]

In the limit as x->0 that is exactly what you can conceptually do.

In reality, however, you cannot as I said, turn something that exists into something that doesn't exist by dividing it.

Otherwise, you should also be able to do the reverse. Bring something into existence from nothing.

So infinity lots of 0 should create something, no? If it doesn't, then you have a problem! Things can disappear into non-existence and not come back!

 

[FoxEye]

And yet you can have enough 9s that 0.99r is equal to 1?

Admit the contradiction.

I think I understand where you're coming from.

Put it this way, there is no such thing as "an infinitely small number". Any number that is > 0 is finitely small.

1 - 0.9r does not yield an infinitely small number. It yields 0.

0.9r builds the number 1 using an infinite number of building blocks. But it does build the number 1. We have trouble because we only tend to look at the first n building blocks in our minds. And the first n blocks build a number < 1. But there are always more blocks that we can't see.

Given that there is no infinitely small number, 1/x as x->0 does not define or attempt to define 1/0.

 

[Vonhelmet]

There is no contradiction. I don't run out of 9s... there is no limited supply of 9s and there is nothing that comes after it. However, you can't have infinity 0s, then add a 1 on after. If you could, the 1 would have a value of 10^-infinity, which is 0.

So you can have infinite 9s but not infinite 0s?

If 0.99r=1 then 0.0r1 must =0 if only because 0.99r+0.0r1 must = 1+0.

 

[ordinaryjoe]

So you can have infinite 9s but not infinite 0s?

If 0.99r=1 then 0.0r1 must =0 if only because 0.99r+0.0r1 must = 1+0.

No, that's not what I said... You can have an infinite number of 0s, but you can't then put a 1 on the end of it.

Your 2nd line is right. 0.0r1 is equal to 0.

 

[Blue160]

0.0r1 isn't even a number. The r means the 0's repeat infinitely so writing any number after that is meaningless.