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I am doing some mathematics, and I am currently stuck on something. I do not understand this part at all, how can two slightly, minor different functions, but near identical—evaluate so differently.

Exercises: enter image description here

Do not understand part:

$$(a)\enspace f:R\to R, f(x)=\frac{x}{3},\enspace \it{Well\enspace Defined} $$ $$(b)\enspace g:J\to J, g(x)=\frac{x}{3},\enspace \it{Not\enspace well\enspace defined} $$

Why is $(a)$ well defined, whereas, $(b)$ is not?

Interesting explanation found here, however, I cannot get any smarter.

Reference:

  • Discrete Mathematics for Computing, 3rd Edition by Peter Grossman

EDIT 8/1/2023 4:50 PM Clarity. $R$ and $J$ are sets.

EDIT: 8/1/2023 5:25 PM From the book:

  • $N$ is the set of natural numbers (or positive integers): $\{1,2,3,4,...\}$.

  • $J$ is the set of integers: $\{...,-3,-2,-1,0,1,2,3,...\}$.

  • $Q$ is the set of rational numbers: $\{x:x=m/n\enspace for\enspace some\enspace integers\enspace m\enspace and\enspace n\}$.

  • $R$ is the set of real numbers.

Alix Blaine
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    What is $J$ in this question? – Chris Lewis Aug 01 '23 at 14:33
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    Find $x\in J$ such that $x/3\notin J$. Can't say much else without knowing what $J$ is. – Jakobian Aug 01 '23 at 14:35
  • @ChrisLewis, please see the photo, it comes from my Book. I haven't created the J or R in this context. They exists in the Exercises section. – Alix Blaine Aug 01 '23 at 14:39
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    I've seen $J$ used to denote the positive integers. I've not sure if that's the case here, but if it were, notice that the second function wouldn't be well-defined. For example, we'd have $g(2) = \frac{2}{3}$, which is definitely not a positive integer, so the codomain of that function wouldn't make sense. – Cardinality Aug 01 '23 at 14:42
  • @Cardinality, you are the only one making sense to me now. Can you please, please expand and elaborate on this? Please kindly. – Alix Blaine Aug 01 '23 at 14:43
  • Look earlier in your book for the correct definition of $J$. You've written $J={}$, but that doesn't make sense. – Karl Aug 01 '23 at 14:55
  • For a function $f: A \to B$ to be well-defined, it must be the case that for every $a \in A$, there exists a unique $b \in B$ with $f(a) = b$. There are two ways this can go wrong immediately. First, it might be that we're trying to send some $a \in A$ to two distinct elements of $B$: that's not allowed. Second, it may be that the rule attempts to an assign an element of $a$ to an element outside of $B$. In the case above, $J$ consists only of the positive integers, but $\frac{2}{3}$ is most certainly not a positive integer. The function isn't well-defined because the codomain doesn't work. – Cardinality Aug 09 '23 at 20:34

1 Answers1

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$\bf{J}$ is Grossman's notation for the set of positive and negative integers (see beginning of the chapter on sets). So obviously $g$ isn't well-defined over that set!

Peter Smith
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  • Thanks. Why isn't g() well defined over that particular set? – Alix Blaine Aug 01 '23 at 15:04
  • Not all integers are divisible by 3 are they? – Peter Smith Aug 01 '23 at 15:07
  • Of course not, so, essentially, the assumption is that when you perform division over the denominator 3, you should end up with a reminder of $1$, denoting that the numerator is correct one. So, the set $R$ in this context does that, where as, $J$ does not because of what you wrote. – Alix Blaine Aug 01 '23 at 15:10
  • Peter, am I correct in my thought here? See previous comment from me. – Alix Blaine Aug 01 '23 at 15:16
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    You are overthinking. A correctly defined function $g\colon J \to J$ will for any input from $J$ OUTPUT A VALUE IN $J$. Trivially division by three doesn't do that over the integers. – Peter Smith Aug 01 '23 at 15:16