not true.
0/0 = indeterminate .. it's even worse than 1/0 .. dun dun dun!!!!
however, you can have 0/1 = 0
edmundoab
Diamond Member
well when a 0 value is placed into division
then the number divided becomes undefined because anything that multiplies 0 = 0
hence u get like 0.000000000 ...... and so on to infinite,
which makes it undefined
then the number divided becomes undefined because anything that multiplies 0 = 0
hence u get like 0.000000000 ...... and so on to infinite,
which makes it undefined
DrPizza
Administrator Elite Member Goat Whisperer
Originally posted by: reverend boltron
not true.
0/0 = indeterminate .. it's even worse than 1/0 .. dun dun dun!!!!
however, you can have 0/1 = 0
No, 0/0 isn't indeterminate.
0/0 is undefined.
What you're referring to, is obviously a part of limit problems.
example,
what is the limit of (x^2 - 4) / (x-2) as x approaches 2.
The numerator approaches 0 and the denominator approaches 0.
So, the fraction is APPROACHING a value of 0/0. Neither the numerator nor the denominator actually get there. This is an indeterminate form of a limit. 0/0, however, is quite clearly defined as undefined.
And, 8/2 can be interpretted 2 ways:
1 way is divide 8 into groups of two. How many groups? 4
The other way is to interpret it as "divide 8 into 2 groups. How many in each group?
In this context, divide 8 into 0 groups makes absolutely no sense. Thus, division by 0 is undefined. It is not the case that 8 divided by 0 = positive infinity.
When you're thinking about negative numbers, don't think about using negative numbers. Draw a number line, and then look at each different number not really as a number, but as a unit. So, when you have 5, you are starting from 0, and moving positively 5 units, landing on 5. So, it's the same thing, when you're adding 5 to 10, you start at the 10 unit, and go forward 5 units, giving you 15 units total. Well, think about doing that with negative units. Instead of moving forward 5 units, you're moving backwards 5 units. When I think about negative numbers that way I can think about moving down the number line in the opposite direction that you would normally go. Ya know? So like, if you want to get crafty, you can just cut out 5 pieces of paper that are 5 units in length, and draw arrows making them go the negative direction, and show how they keep on moving, so when you're dividing numbers by that negative value, you'll be dividing it by those units, moving in the opposite direction on the number line.
Think about that when you multiply and divide negative numbers. It's neat
Originally posted by: DrPizza
Originally posted by: reverend boltron
not true.
0/0 = indeterminate .. it's even worse than 1/0 .. dun dun dun!!!!
however, you can have 0/1 = 0
No, 0/0 isn't indeterminate.
0/0 is undefined.
What you're referring to, is obviously a part of limit problems.
example,
what is the limit of (x^2 - 4) / (x-2) as x approaches 2.
The numerator approaches 0 and the denominator approaches 0.
So, the fraction is APPROACHING a value of 0/0. Neither the numerator nor the denominator actually get there. This is an indeterminate form of a limit. 0/0, however, is quite clearly defined as undefined.
And, 8/2 can be interpretted 2 ways:
1 way is divide 8 into groups of two. How many groups? 4
The other way is to interpret it as "divide 8 into 2 groups. How many in each group?
In this context, divide 8 into 0 groups makes absolutely no sense. Thus, division by 0 is undefined. It is not the case that 8 divided by 0 = positive infinity.
I'm going to have to disagree with you. Every book that I've just checked tells me that 1/0 is undefined and 0/0 is indeterminate.
JohnCU
Banned
- Dec 9, 2000
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Originally posted by: reverend boltron
Originally posted by: DrPizza
Originally posted by: reverend boltron
not true.
0/0 = indeterminate .. it's even worse than 1/0 .. dun dun dun!!!!
however, you can have 0/1 = 0
No, 0/0 isn't indeterminate.
0/0 is undefined.
What you're referring to, is obviously a part of limit problems.
example,
what is the limit of (x^2 - 4) / (x-2) as x approaches 2.
The numerator approaches 0 and the denominator approaches 0.
So, the fraction is APPROACHING a value of 0/0. Neither the numerator nor the denominator actually get there. This is an indeterminate form of a limit. 0/0, however, is quite clearly defined as undefined.
And, 8/2 can be interpretted 2 ways:
1 way is divide 8 into groups of two. How many groups? 4
The other way is to interpret it as "divide 8 into 2 groups. How many in each group?
In this context, divide 8 into 0 groups makes absolutely no sense. Thus, division by 0 is undefined. It is not the case that 8 divided by 0 = positive infinity.
I'm going to have to disagree with you. Every book that I've just checked tells me that 1/0 is undefined and 0/0 is indeterminate.
DrPizza
Administrator Elite Member Goat Whisperer
Thanks
I have all of my math books on the shelves in my room, so I can check them all pretty quick. But I appreciate your noticing.
IAteYourMother
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JohnCU
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DrPizza
Administrator Elite Member Goat Whisperer
Originally posted by: reverend boltron
Thanks
I will admit I was wrong. However, I have yet to find any reference of 0/0 where it isn't in the context of limits. I even found one site that states "7x/x is equal to 0/0 when x = 0. This is indeterminate because 7x/x is equal to 7 when x = 0." (This from what I'd consider to be a fairly reputable site!) Nonetheless, the site is wrong. The function is "undefined" when x = 0. 0 is not in the domain of the function. Were it defined, then the function would be continuous. However, any math text dealing with continuity will clearly state that such a function has a "removeable" discontinuity at x = 0. (or a point-discontinuity.) A typical calculus text will note that as the left and right hand limits approach the same value, the function can be re-written as a piecewise defined function which is continuous. Still, some people may be convinced that cancelling out the x's is allowable and have the same function. It is not the same, because the function is "undefined" at x = 0, not "indeterminate"
A better example, where clearly, x cannot be cancelled, would be f(x) = sin(x) / x
Again, this function is "undefined" at x=0. However, the left and right hand limits as x approaches zero are 1.
Originally posted by: DrPizza
Originally posted by: reverend boltron
Thanks
I will admit I was wrong. However, I have yet to find any reference of 0/0 where it isn't in the context of limits. I even found one site that states "7x/x is equal to 0/0 when x = 0. This is indeterminate because 7x/x is equal to 7 when x = 0." (This from what I'd consider to be a fairly reputable site!) Nonetheless, the site is wrong. The function is "undefined" when x = 0. 0 is not in the domain of the function. Were it defined, then the function would be continuous. However, any math text dealing with continuity will clearly state that such a function has a "removeable" discontinuity at x = 0. (or a point-discontinuity.) A typical calculus text will note that as the left and right hand limits approach the same value, the function can be re-written as a piecewise defined function which is continuous. Still, some people may be convinced that cancelling out the x's is allowable and have the same function. It is not the same, because the function is "undefined" at x = 0, not "indeterminate"
A better example, where clearly, x cannot be cancelled, would be f(x) = sin(x) / x
Again, this function is "undefined" at x=0. However, the left and right hand limits as x approaches zero are 1.
Word, thanks for explaining it to me. I only knew it because of memorizing and thinking that like, if you're dividing nothing by nothing, then you aren't even able to determine what you would be doing in the first place (ya know?). But like dividing 0 by 1 would be undefined.. but seriously, thanks for explaining it like that because it makes a lot more sense in my head now
DrPizza
Administrator Elite Member Goat Whisperer
I've actually found a few respected sites that do seem to agree with my opinion, that division by zero is *always* undefined in the real numbers.
However, what I found more interesting was a discussion on 0^0 power... does it = 1, or is it undefined?
Good reasons here: http://db.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html
However, what I found more interesting was a discussion on 0^0 power... does it = 1, or is it undefined?
Good reasons here: http://db.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html
Dividing by D means multiplying by the unique number which multiplied to D equals 1, which would be 1/D for any D \neq 0.
So if you are dividing by zero, that means you are multiplying by whatever number you multiply 0 by to get 1.
So... what number is that?
So if you are dividing by zero, that means you are multiplying by whatever number you multiply 0 by to get 1.
So... what number is that?
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