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My teacher claims that when an equation in variables $x_1,x_2,\ldots,x_n$ has no solutions, you should denote this fact with $(x_1,x_2,\ldots,x_n)\in\varnothing$.

An empty set can't have an element in it, so this can't be right.

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    $\begingroup$ Your teacher's idiosyncratic notation means exactly what is desired, namely that there are no elements in the set of all solutions. $\endgroup$ – vadim123 Nov 20 '14 at 19:43
  • $\begingroup$ To be an element of an empty set is rather strange. We just had to write down $\varnothing$ for no solution but we were not allowed to write the {} symbols around it. $\endgroup$ – imranfat Nov 20 '14 at 19:44
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    $\begingroup$ @imranfat The sets $\varnothing$ and $\{\varnothing\}$ are different. One of them has an element, and the other doesn't. $\endgroup$ – Arthur Nov 20 '14 at 19:45
  • $\begingroup$ "should denote"? I'd say "can denote". It's not wrong, but it's not notation that is used for this very often. $\endgroup$ – Simon S Nov 20 '14 at 19:46
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    $\begingroup$ An equation (or a system of equations) in so many variables has a solution set $S$ which is a subset of the universe $\Omega$ for which the equations make sense. When there are no solutions we write $S=\emptyset$. A statement as $(x_1,x_2,\ldots, x_n)\in\emptyset$ seems pretty forlorn to me. $\endgroup$ – Christian Blatter Nov 20 '14 at 20:22
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You can form the set $L$ of all solutions. So iff $(x_1, \cdots, x_n)$ is a solution, then $(x_1, \cdots, x_n)\in L$. Now, if there are no solutions, then $L = \varnothing$. Insert that, and you get $(x_1, \cdots, x_n)\in \varnothing$.

In other words, the sequence $(x_1, \cdots, x_n)$ is a solution to an impossible set of equations iff $(x_1, \cdots, x_n)\in \varnothing$.

However, saying that "$(x_1, \cdots, x_n)\in \varnothing$ means there are no solutions" is taking it a bit far. I would rather say "$L = \varnothing$ means there are no solutions".

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You are correct that the empty set is the set with no members. The solution set of an equation is, however defined as the set of all solutions that satisfy the equation and since that set has no members the solution set is empty.

Thus, saying that the equation has no solution, and saying that the solution set of the equation is empty mean the same thing.

Recall that the solution set is the set of all values that turn an open sentence into a true statement.

Thus, the membership symbol is an abbreviation for the fact that the variable can be replaced by a value that is "in", "belongs to", or "is a member of" the set. If the set is empty, then the variable can not be replaced by any value. This is what it means for an equation to have "no solution", there is no value that when substituted for the variable will result in a true statement.

I agree with you that the notation does lead to some confusion when dealing with the special case of the solution set being empty. In this case "what the variable can be replaced by belongs to" a set with no members. But, if you like, you can simply say "No solution." Note that the meaning here is still the value that the variable can be replaced by belongs to the replacement set, which happens to be empty so there is no value.

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  • $\begingroup$ Of course, surely the set of all solutions is empty in this case, but how can an object exist in a fully empty set? If you think this is correct notation, then should $A\in\varnothing$, where $A=\{A\}$ also be correct? In both cases, the object is impossible and if you agree with my teacher's claim, then you agree with this claim, thus overall agreeing that all impossible objects exist in the empty set. $\endgroup$ – user26486 Nov 20 '14 at 21:31
  • $\begingroup$ A variable is not an object, it is a symbol that represents one or more numbers. There are no numbers that make the equation a true statement, thus the solution set that the variable belongs to is empty. $\endgroup$ – skullpetrol Nov 20 '14 at 21:44
  • $\begingroup$ Again, yes, the solution set is empty. I don't know how you defined an object, but let it be something that can exist in a set (which is just a collection). $(x_1,\ldots,x_n)$ is an object and thus should be in a set. But $(x_1,\ldots,x_n)\in\varnothing$ implies it is not able to be a member of a set and therefore it is not a possible object. $(x_1,\ldots,x_n)\in\varnothing$ thus implies an impossible object is in the empty set. But that means that all impossible objects exist in the empty set. You agree with my teacher's claim iff you agree with this claim, unless I'm missing something. $\endgroup$ – user26486 Nov 20 '14 at 21:52
  • $\begingroup$ If I have an equation such as x+2=x+1, what does it mean to say that the equation has no solution? $\endgroup$ – skullpetrol Nov 20 '14 at 21:59
  • $\begingroup$ It simply means $\not\exists(x_1,\ldots,x_n)(f(x_1,\ldots,x_n)=0)$, where $f(x_1,\ldots,x_n)=0$ is the equation we're trying to solve. In your case, $f(x)=x+2-x-1=1$ and so it is true that $\not\exists x(f(x)=0)$, since $\forall x\in\mathbb R, f(x)=1$. $\endgroup$ – user26486 Nov 20 '14 at 22:04
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I would write $card\{X \mid f(X) = 0\} = 0 $ where $card\{S\}$ is the number of elements in a set $S$ and $X$ is a vector $(x_1, x_2, ..., x_n)$.

Another way would be $\{X \mid f(X) = 0\} =\varnothing $.

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  • $\begingroup$ I didn't know that $f(\vec{a})=f(a_1,a_2,\ldots,a_n)$, where $\vec{a}=\{a_1,a_2,\ldots,a_n\}$. $\endgroup$ – user26486 Nov 20 '14 at 20:17
  • $\begingroup$ That is a convenient and standard way of writing vectors. $\endgroup$ – marty cohen Nov 20 '14 at 20:22
  • $\begingroup$ You could also simply write $\not\exists (x_1,\ldots,x_n)\left( f(x_1,\ldots,x_n)=0 \right)$. $\endgroup$ – user26486 Nov 20 '14 at 20:47
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A more efficient way to denote that the equation has no solution, is by a Boolean "false", "F", or "$\bot$", consistent with your observation that the empty set cannot have any elements.

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