Is there any proposition in real analysis or linear algebra that can only be proved by contradiction? By "only be proved by contradiction", I mean either it's probable that this proposition can only be proved using contradiction, or that no one has ever came up with a direct proof.
An undergraduate level one would be better.
Thanks.
 A: I believe the following proposition cannot be done without using contradiction technique.
Proposition 
For any $x,y \in F$ ($F$ if a field).
If $x\neq 0$ and $y\neq 0$ then $xy\neq 0$
Proof: $x\neq 0$ and  $y\neq 0$ assume that $xy=0$ .Then  $1=\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)xy=\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)0=0$ a contradiction(4th field axiom for multiplication says $1\neq 0$).$\qquad \square$
Note the proof requires another proposition $0x=0$, fifth field axiom for multiplication(Existence of multiplicative inverse) and of course 1st and 2nd field axiom for multiplication(Associative property of multiplication) .
You can see here for list of field axioms and propositions that are used to prove the result.
A: In classical logic, you can use a disjunction as the hypothesis of a conditional without knowing which disjunct is true. A direct proof of $(A\vee B)\wedge (A\rightarrow P)\wedge (B\rightarrow P)$ counts as a direct proof of P even if you can't prove A and can't prove B.  Of course, being able to prove $(A\vee B)$ when you can't prove A and can't prove B is unusual to begin with, but it does happen sometimes, especially with the law of the excluded middle taken as a given -- in classical logic, you can always prove $(Q\vee\neg Q)$, even when you can't prove Q and can't prove $\neg$ Q.
That particular case is precisely the one that becomes relevant for turning any indirect proof direct, because any provable conditional must have a contrapositive that is also provable.  If an indirect proof of $P$ is (for some $Q$) a proof of a conditional $\neg P \rightarrow ( Q \wedge \neg Q )$ , then that proof is transformable into a proof of its contrapositive $(Q \vee \neg Q) \rightarrow P $. The objectionable part of this is that if you have to start with the "correct" $Q$, then you simply have no clue how to find it.  Despite that, however,  it remains possible to prove $P$ directly (to prove the contrapositive of the original indirect proof) by cases -- to prove both  $(Q \rightarrow P) \wedge (\neg Q \rightarrow P)$ -- after you know the right $Q$ (i.e. the one that was derived as the contradiction in the original indirect proof).
For this reason, answers to questions like this elsewhere just begin by presuming that of course we must be using intuitionistic logic -- in which deriving a contradiction from $\neg P$ will not give us $P$, but rather $\neg\neg P$, which, in that context, is not equivalent to P.  In intuitionistic logic, almost every proposition that starts with $\neg$ would have to be proved indirectly, because $\neg P$ is defined as $P\rightarrow \bot$. 
