# what is the flaw in this proof that either every number equals to zero or every number does not equal to zero?

Here's what we want to prove: $\forall$x$\in$$\mathbb{R}$$(x=0)$ $\vee$ $\forall$x$\in$$\mathbb{R}$$(x$ $\ne$ $0)$ .

We know that $\forall$x$\in$$\mathbb{R} (x=0 \vee x \ne 0). Now, let x be arbitrary. We know that either x=0 or x \ne 0. Let's assume x=0. Then, since x was arbitrary, we have \forallx\in$$\mathbb{R}$$(x=0). If x \ne 0, then again, since it was arbitrary, we have \forallx\in$$\mathbb{R}$$(x \ne 0). Thus, in either case we have \forallx\in$$\mathbb{R}$$(x=0) \vee \forallx\in$$\mathbb{R}$$(x \ne 0). well, I know that the error is in "Then, since x was arbitrary, we have..." i just don't know how to explain to myself why this is logically incorrect. • "let x be arbitrary. Let's assume x=0." Now x is no more arbitrary and thus we cannot "generalize" to get \forall x \ (x=0). – Mauro ALLEGRANZA May 22 '17 at 10:55 ## 3 Answers If you intend to prove a statement of the form \forall x \in M : P(x), where P is some statement, then you may prove it in the following way: Let x \in M be arbitrary. Then some arguments follow...and therefore we assert P(x). Since x was arbitrary, the statement follows. What you do in the proof is different. You prove only one part of P(x) and then take a new x. Your Since x was arbitrary comes too early, since you have only treated the first case. The right way would be to say: "Take an arbitrary x, then we have two cases. Case 1: x=0. Case 2: x\neq 0". And you first have to get through with both cases before saying that it now follows for every x. Then you see that you can't prove the obviously wrong statement. • so, with a given in the form of disjunction it must be used as a whole inside of [let x be arbitraty - since x was arbitrary] construction? – famesyasd May 22 '17 at 10:42 • Yes, I think you can say it like that. – Luke May 22 '17 at 12:44 You are assuming that x=0 for arbitrary x\in\mathbb{R} which is the same as for each x\in\mathbb{R}, x=0. Your argument is circular. You are assuming your conclusion. And note that for any proposition p, p\implies p is true. Along the lines of @Luke (already accepted) ... Now, let x be arbitrary. I think you mean, let x be an arbitrary element of \mathbb{R}. It makes a difference. We know that either x=0 or x \ne 0. True. Let's assume x=0.Then, since x was arbitrary, we have \forallx\in$$\mathbb{R}$$(x=0)$.

This would not follow even if we assumed $x\in \mathbb{R}$. If we assume that something is true of one element of a set, it does not immediately follow that it is true of every element of that set. If, for example, we assume that $x$ is a human female, it does not follow that every human is a female.