Prove that $\exists y \in \mathbb{R}: \forall x \in \mathbb{R} : x+y =0$ is not correct

Question

Prove that the following statement is not correct $$\exists y \in \mathbb{R}: \forall x \in \mathbb{R} : x+y =0$$ I have difficulties proving this.

I think that the best value $y$ can have is to be equal to $-x$. So I can give a contradiction by setting $x=-y+1$ and so $x+y\neq0$.

Is there a better way to prove this?

Answer 1

The natural way for most mathematicians to prove this is "by contradiction" --- assume that there is such a $y$ and show that this leads to a falsehood. Here's that proof:

Suppose there is such a $y$. Then for every $x \in \Bbb R$, we have $$ x + y = 0. $$ In particular, for $x = 1$ and for $x = 0$, we have $$ 1 + y = 0 \\ 0 + y = 0 $$ Subtracting the bottom equation from the top, we get that $1 = 0$, which is false, so our assumption --- that such a $y$ exists --- must be false.

This may seem unnatural to you, but it's a very useful tool, one that applies to lots of statements that start out claiming that some particular thing exists.

Answer 2

You can try to show that the negation is true; that is $$ \forall y\in\mathbb{R},\exists x\in\mathbb{R},x+y\neq0 $$ Fix $y\in\mathbb{R}$ and choose $x=-y+1\in\mathbb{R}$. Then $x+y=-y+1+y=1\neq0$.

Answer 3

If $x+y=0$, then $(x+1)+y=1\ne0$ and the $\forall x$ clause cannot hold.

Answer 4

Addition is commutative, and so $x+y=y+x$. Therefore, we can write $$x+y=0\iff y+x=0.$$ By the additive inverse axiom, we know that every real number has a unique additive inverse. Therefore, the only $x$ that would satisfy $y+x=0$ is if $x=-y$. Since $\exists x: x\neq-y$, the statement is false.