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

$(\forall y\in R)(\exists x \in R)(2x-y \neq 3)$


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 $y$.

  • $\begingroup$ Suppose x=0, then $2(0)-y \neq 3$, unless y = -3. Suppose y =-3, then let x =5, then $2(5)-y \neq 3$ Does this then lead to a proof? $\endgroup$ – AlbertWolfgang Sep 21 '15 at 4:59
  • $\begingroup$ @AlbertWolfgang You cannot suppose $y=-3$. You need to prove that the statement is true for every value of $y$. Your proof needs to start with "Let $y$ be an arbitrary real number" and end with "therefore, for this value of $y$, there exists some $x$ that $2x-y=3$ is false." $\endgroup$ – 5xum Sep 21 '15 at 5:01
  • $\begingroup$ let y be an arbitrary real number, $y_0$. then $\forall x \in R (x= \frac{3-y_0}{2})$ This clearly cannot be hold true for all of x, but I'm again unsure if this is enough proof, or what I need to do to provide enough proof. $\endgroup$ – AlbertWolfgang Sep 21 '15 at 5:21
  • $\begingroup$ say x=2, then 2= $\frac(3-(y_0)}{2}$, now let x =3 then 3= $\frac(3-(y_0)}{2}$. These statements imply that 2 = 3, which is clearly false. So therefore the above proof holds? $\endgroup$ – AlbertWolfgang Sep 21 '15 at 5:49

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