Wikipedia:Reference desk/Archives/Mathematics/2007 November 14

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November 14

Union and Roots

Hello. Why is union inadmissible when stating the two roots of a quadratic function? For example, can x = 3 or x = 5 be stated as x = 3 ∪ x = 5? Thanks in advance. --Mayfare 02:22, 14 November 2007 (UTC)

I see no problem with $x\in \{3\}\cup \{5\}=\{3,5\}$. Just make sure you distinguish union as an operator on sets and x as an element in a set. SamuelRiv 02:53, 14 November 2007 (UTC)

I would find (x = 3)∨(x = 5) acceptable. Replacing the "∨" with a "∪", however, requires that we interpret "x = 3" as a set, which is more abuse of notation than I find comfortable. --KSmrq^T 04:22, 14 November 2007 (UTC)

I would think that notation is totally ok, our set could be defined as $\lbrace S\rbrace =\lbrace x|x\in (x-3)(x-5)=0\rbrace$, so the set $\lbrace S\rbrace$ contains all x that are solution to the given equation since that is the condition on x. A math-wiki 08:42, 14 November 2007 (UTC)

Is it OK to write "x belongs to the equation..."? And why S (the symbol of a set, if I get it correctly) is itself in braces? IMHO that should be written as $S=\{x|(x-3)(x-5)=0\}\,\!$ – or am I wrong? --CiaPan 10:07, 14 November 2007 (UTC)

I'll try to put everything mentioned here in order. The short answer is: $\cup$ is an operation that takes two sets and returns another set: For example, {3} and {5} are sets, and $\{3\}\cup \{5\}=\{3,5\}$ is another set. "$x=3$" is not a set. It is a statement that says something about x. You can't take an operator that works on sets and use it to join two statements, just as you can't write $(x=3)+(x=5)$, since + is an operator that works on numbers.

What you can do without problem is use the operator $\vee$, which works on statements and stands for "or". So $x=3\vee x=5$ is a valid representation of the statement, "$x=3$ or $x=5$".

Now, as a somewhat understandable abuse of notation, the symbol $x=3$ is sometimes used to denote $\{x|x=3\}$, the set of all x's which are equal to 3. This set is of course equal to {3}. So you could write $(x=3\cup x=5)=\{x|x=3\}\cup \{x|x=5\}=\{3\}\cup \{5\}=\{3,5\}$, and so the set of all solutions to the equation $(x-3)(x-5)=0$, which is denoted with this abuse simply $(x-3)(x-5)=0$, would be equal to $x=3\cup x=5$. Again, this is abuse of notation and you should only use it when you know what you're doing and there is no risk of confusion. You shouldn't write it in a quiz, for instance.

Denoting the set of solutions of $(x-3)(x-5)=0$ by S, the correct way to write it is $S=\{x|(x-3)(x-5)=0\}=\{x|x=3\}\cup \{x|x=5\}=\{3,5\}$. -- Meni Rosenfeld (talk) 13:48, 14 November 2007 (UTC)

I don't think sets correctly capture the meaning of the equation. The equation specifies a constraint; the solution depends on what kind of space you are constraining. If, for example, you were looking at points on the real line that satisfies that equation, i.e. $\{x|(x-3)(x-5)=0\}$, then yes, it becomes is equivalent to $\{x|x=3\}\cup \{x|x=5\}$. But what if you were constraining points on the (x,y) plane? Something like $\{(x,y)|(x-3)(x-5)=0\}$; which is equivalent to $\{(x,y)|x=3\}\cup \{(x,y)|x=5\}$. You can see that these are very things. By saying that the solution is some set, you are assuming some set to apply the equation to in the first place. The constraint "$x=3\vee x=5$", however, does capture the meaning of the equation. --Spoon! 22:34, 14 November 2007 (UTC)

I am assuming the OP has been taught to discuss the "solution set" of an equation, and that the nature of the set is the one which most naturally follows from the context. If the equation $(x-3)(x-5)=0$ appears in vaccum, it should be assumed that its solution set is the set of real x's satisfying the equation, without paying attention to additional unmentioned variables, nonreal solutions, etc. -- Meni Rosenfeld (talk) 23:16, 14 November 2007 (UTC)

I always just put x=3,5 for the answer, since that's what my teachers usually use. It seems easier than trying to make union fit, and it's faster than writing out 'x=3 or x=5'. Black Carrot 21:11, 15 November 2007 (UTC)