This is a linear equation.....I can substitute any real value in for "n" (the domain) and get back any real number (the range).
So, rhe domain and range is just (∞, ∞)
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Hang on a minute. how can this have a domain and range?
It is not an equation it doesn't even have a y value.
It is just an expression.
Maybe it is me that is wrong. Can other mathematicians arbitrate please?
Well...if we put in any real number for n...don't we get a real number back?? And since we can return any real number we want...that's what I was getting at.....the "y" is implied..
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Sorry...I meant to type "function" rather than "equation".....I'll stand by the rest...
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Yes maybe - I'm not sure.
I knew that is what the question was implying but I'd still like to know what other mathematicians think.
It is a lot better now that you have changed the word equation to function. You are probably right.
Yes I know functions and relations have domains and ranges but this has no equal sign.
Techically speaking - does it really have a domain and a range. To me it is just an expression.
(I know an equal sign and a y or f(n) is implied BUT it is not really there!)
I know I am being padentic but mathematics is a padentic field of study!
Quiet, whippersnapper!!! Things are what we say they are !!!
I'll get you, my pretty...and your little dog, too!!

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It's when you are reduced to threats and witchcraft that I KNOW I am in the lead!!!!!

Even though the function might not be defined in the sense of f(n) = 80n-275, it still maps R->R which even though it is an undefined function still has a domain and range. The question here is not whether or not this function has a domain and range, but rather whether it's a mathematical fallacy to have an undefined function.