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Logic notation to find the domain of a piecewise function watch

1. Let's say we're trying to find the domain of the following piecewise function. Is the notation of my resolution accurate? My calculus I teacher said he'd be strict on notation.
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2. (Original post by MartyO)
Let's say we're trying to find the domain of the following piecewise function. Is the notation of my resolution accurate? My calculus I teacher said he'd be strict on notation.
Ruddy hell, why would you want to write such ugly maths? Please don't. It's just horrid.

Anywho... pretty sure you've confused your 1 and -1. Should the first line of the function not be ?

You seem to have screwed up your brackets in the next line. Should be

Screwed up brackets in bit again.

Screwed up brackets in last line as wqell.

Also, for all of them you use comma's not semicolons in the brackets to specify intervals.

I re-iterate, try and avoid horrid notation like this in the future. It's hideous.
3. Sorry about the lack of parenthesis, the brackets for the intervals and the semicolons, I guess in Canada it's different, I don't know. I did make a typo for the restriction of the first piece, sorry if it was confusing. I'll still have to conform to my teacher's expectations though, for the grade, and he exclusively uses semicolons and brackets (or the lack of) for intervals.

Really, what my question was is if I can use the logical disjunction and logical conjunction symbols (that return a boolean) in my steps, since the propositions I make are the equivalent of sets. That notation was not introduced in my class, but I found it helpful to "concisely" express my steps in a linear fashion (stating each conditions that restrict the domain of f, then develop each conditions...).

In propositional logic though, I've never seen the notation used this way apart from Wikipedia (not the greatest resource, but otherwise videos use sets A∧B⇒...) in the list of logic symbols.

"n < 4 ∧ n >2 ⇔ n = 3 when n is a natural number."
"n ≥ 4 ∨ n ≤ 2 ⇔ n ≠ 3 when n is a natural number."

The way I see it right now for the really long line is:
x is in the domain of f if and only if (Condition 1) or (Condition 2) or (Condition 3) is true,
where Condition 1 is (when the function isn't indeterminate and in the boundaries of the part's domain),
where Condition 2 is (when the function isn't indeterminate and in the boundaries of the part's domain)
and where Condition 3 is (when the function isn't indeterminate and in the boundaries of the part's domain).

Meaning that x is in the domain if either of the three conditions separated with ∨ is true, and in each of those conditions, propositions separated with ∧ have to all be true for the whole condition to be true.
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4. (Original post by MartyO)

Really, what my question was is if I can use the logical disjunction and logical conjunction symbols (that return a boolean) in my steps, since the propositions I make are the equivalent of sets. That notation was not introduced in my class, but I found it helpful to "concisely" express my steps in a linear fashion (stating each conditions that restrict the domain of f, then develop each conditions...).
Yeah, the way you've done that is fine.

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Updated: September 4, 2016
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