I need to either prove or disprove this statement.
$(\exists y\in R)(\forall x \in R)(2x-y=3)$
I want to provide a proof by contradiction, so I'm proving the negation
$(\forall y\in R)(\exists x \in R)(2x-y \neq 3)$
Proof:
Suppose y = 1 and x =3 then $2(3)-3 =5 \neq 3$
therefore $(\exists y\in R)(\forall x \in R)(2x-y=3)$ is false.
From what I'm reading this is sufficient proof, but I can't convince myself this is totally correct, but know of no other way to go about it. Any help would be great. Thanks!
If you want to prove the negation $(\forall y\in R)(\exists x \in R)(2x-y \neq 3)$, you have to stick to the statement. The negation claims that for each real number $y$ there is $x$ such that $2x-y \neq 3$. So, to prove this, for every $y$ you have to find an $x$, not only for one special $y$.