I am asked to prove that when squaring any element of the empty set, one should always get zero.

Of course the empty set is the set which contains no elements. If you square nothing then you should get nothing, or equivalently zero. I am having an issue as to how to show this formally.

I'm not even sure on which form of proof I should try to show this, i.e., direct proof, contradiction, or maybe contraposition.

Maybe for contradiction I could say something along the following:

Let $x \in A :x \; \ne0.$ Then $x^2 \ne 0.$ Hence, $A \ne \emptyset.$ We then have arrived at a contradiction!

Does this even make sense?

Maybe I could say something along the lines like: Let $ x \in A \cap B$, where A and B are two disjoint sets. Then try to prove in a direct fashion.

Any guidance or ideas would be much appreciated. Thank you

  • $\begingroup$ How can you write the statement formally? "For any $x$: If $x\in\{\}$, then $x^2=0$"? $\endgroup$ – Akiva Weinberger Feb 15 '15 at 7:34
  • $\begingroup$ Your second paragraph shows that you don't really understand the question. It is equally true, for instance, that squaring any element of the empty set gets the answer 42. $\endgroup$ – TonyK Feb 15 '15 at 8:13
  • $\begingroup$ @TonyK So by squaring an element in the empty set, it produces the meaning of life? $\endgroup$ – mathamphetamines Feb 15 '15 at 8:15
  • $\begingroup$ Yes. Plus a free crate of beer. $\endgroup$ – TonyK Feb 15 '15 at 8:57

I assume what you mean is that you are being asked to prove the statement $$(\forall x)(x\in\emptyset\implies x^2=0)$$ To prove this formally would depend on your axioms or rules of deduction. Intuitively, we would indeed argue by contradiction. If $x\in\emptyset$ such that $x^2\ne0$, then we have $x^2\ne0$ and $x\in\emptyset$, in particular, $x\in\emptyset$, a contradiction. Hence our supposition of $x^2\ne0$ was wrong. It's a clunky argument to see written in words but works in a formal system.

  • $\begingroup$ Once you have an implication $P \Rightarrow Q$, and you can show that $P$ is false, you have thereby shown that $P \Rightarrow Q$ is true. You do not need to consider the truth value of $Q$ at all. $\endgroup$ – David K Feb 15 '15 at 8:37
  • $\begingroup$ I am not dealing with truth values, I am colloquially stating the argument one would use in a formal proof. $\endgroup$ – Jason Feb 15 '15 at 9:20

It is commonly said that "everything is true of the members of the empty set." So, the statement "If $x \in \emptyset$ then $x^2 =0$" is true simply because there is nothing in the empty set. It is vacuously true.

It comes down to the truth values of conditional statements. When considering the statement "If $P$ then $Q$", it is true in 3 cases: when both $P$ and $Q$ are true, when $P$ is false but $Q$ is true, and when both $P$ and $Q$ are false.

So then since it is false that $x \in \emptyset$ for any $x$ in your domain of discourse by definition of the empty set, the statement "if $x \in \emptyset$, $x^2 = 0$" is true. Similarly for "if $x \in \emptyset$, $x$ is made of chocolate," "if $x\in \emptyset$, $x$ is my aunt Sue," etc.

  • $\begingroup$ so, if $x \in \emptyset, then \; x \in \emptyset?$ Will boolean algebra allow me to say that? $\endgroup$ – mathamphetamines Feb 15 '15 at 7:43
  • 2
    $\begingroup$ I'm not sure about Boolean algebras, but the statement "if $x \in \emptyset$ then $x \in \emptyset$" is a true statement in axiomatic set theory. $\endgroup$ – walkar Feb 15 '15 at 7:45

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