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x + y = x, solve for X (pg. 2)
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| Jer |
| Ok, what I'm curious about is what exactly is an undefined number? Is it a value that they can't agree how to define? Or is it some mysterious form of number yet to be discovered? Or is it literally impossible to define? Sorry for my naivety, I was never good at math... |
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| dj_mdma |
undefined number has no value, but it still exists.
The square root of a minus number doesn't work on a calculator, but it still exists.
Just like minus infinity, etc |
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| whiskers |
GEEK ALERT!! :D
but yes, x is a free variable = it can be anything you want it to be. |
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| zarathustra |
| quote: | Originally posted by Jer
Ok, what I'm curious about is what exactly is an undefined number? Is it a value that they can't agree how to define? Or is it some mysterious form of number yet to be discovered? Or is it literally impossible to define? Sorry for my naivety, I was never good at math... |
It is just something you can't evaluate directly. Take y = x/x. For any non-zero value you get y=1. Once you put in 0, you get 0/0. Now you can't just divide top and bottom by zero and get 1, but you can say that lim x->0 x/x = lim x->0- x/x = lim x->0+ x/x = 1. This just means that as you approach x=0 from the left and right you always get 1 so the value at 0 is 1. |
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| hadi burpee |
| quote: | Originally posted by whiskers
GEEK ALERT!! :D
but yes, x is a free variable = it can be anything you want it to be. |
haha geek alert, this is an easy in problem |
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| whiskers |
| quote: | Originally posted by zarathustra
It is just something you can't evaluate directly. Take y = x/x. For any non-zero value you get y=1. Once you put in 0, you get 0/0. Now you can't just divide top and bottom by zero and get 1, but you can say that lim x->0 x/x = lim x->0- x/x = lim x->0+ x/x = 1. This just means that as you approach x=0 from the left and right you always get 1 so the value at 0 is 1. |
not really, the value at 0 is undefined, therefore by the definition of the limit, lim as x-> 0 doesn't exist
there's a limit as x->0- and x->0+, but not x->0 |
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| whiskers |
| quote: | Originally posted by hadi ******
haha geek alert, this is an easy in problem |
lol, but we expanded the discussion about it to 2 pages! |
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| zarathustra |
| quote: | Originally posted by whiskers
not really, the value at 0 is undefined, therefore by the definition of the limit, lim as x-> 0 doesn't exist
there's a limit as x->0- and x->0+, but not x->0 |
But, by definition, for some number a, if lim x->a- = lim x->a+ then lim x->a exists. |
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| Jer |
| Woah! I didn't realize this was such an interesting topic! Now we're getting into calculus to solve 0/0! So we have one view that says that 0/0=1 and 0/0=undefined. What about 0/0=infinity? Is there an argument for that? Like, because 3/0 = infinity, why shouldn't 0/0 also be infinity? |
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| zarathustra |
| quote: | Originally posted by Jer
Woah! I didn't realize this was such an interesting topic! Now we're getting into calculus to solve 0/0! So we have one view that says that 0/0=1 and 0/0=undefined. What about 0/0=infinity? Is there an argument for that? Like, because 3/0 = infinity, why shouldn't 0/0 also be infinity? |
We have two conflicting definitions here: number/0 = inf and number*0 = 0. One cannot simply evaluate but must instead study the behaviour of the functions and one way to do that is to use limits. |
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| whiskers |
| quote: | Originally posted by zarathustra
But, by definition, for some number a, if lim x->a- = lim x->a+ then lim x->a exists. |
you're right, i think i was thinking of vector-valued functions and their limits or something like that... stupid multivariable calculus... |
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| Jer |
| quote: | Originally posted by zarathustra
It is just something you can't evaluate directly. Take y = x/x. For any non-zero value you get y=1. Once you put in 0, you get 0/0. Now you can't just divide top and bottom by zero and get 1, but you can say that lim x->0 x/x = lim x->0- x/x = lim x->0+ x/x = 1. This just means that as you approach x=0 from the left and right you always get 1 so the value at 0 is 1. |
Ok. What about lim x->0 x/0? Wouldn't that work out to 0/0=0? |
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