I am quite lost on proving or disproving these type of statements, if anyone can guide me for a way to approach this problem that would be great.
Determine whether the statement is true or false, and prove or disprove it: $(\forall x \in \mathbb{R})(\exists y \in \mathbb{R})(\forall z \in \mathbb{R})[xy = xz]$
This is how I went by trying to see if its true or false: I picked an x = 2, and y = 3, and a z = 3. Then I got $(2*3)=(2*3) \rightarrow 6 = 6$, which is true?
$\endgroup$3 Answers
$\begingroup$I am going to show that this is false. Take the negation of the original statement $(0)$: $$\exists x\in\Bbb R\forall y\in\Bbb R\exists z\in\Bbb R xy\ne xz\tag1$$ Now let $x=1$. $$\forall y\in\Bbb R\exists z\in\Bbb R y\ne z\tag2$$ Clearly we can take $z=y+1$ to make $y\ne z$. $(2)$ is thus true, and since $x=1$ is the element for which this was proved true, $(1)$ is true and $(0)$ is false.
$\endgroup$ $\begingroup$Let $x=1$ for a counterexample.
Assume that exists $y \in \mathbb{R}$ from hypothesis such that $y=z,\forall z \in \mathbb{R}$
So for $z=1,2$ we have that $1=2$ which is a contradiction.
So the statement is not true for $x=1$ thus it is not true $\forall x \in \mathbb{R}$
So the statement is not true .
$\endgroup$ $\begingroup$Firstly, with this type of problem, try fixing $x$ and determining $y$ from it.
Given, $x$:
$$\forall z\in\mathbb{R}[xy=xz]$$
Assuming $y$ and $x$ are nonzero:
$$\forall z\in\mathbb{R}[y=z]$$
This is a contradiction, as $y\neq y+1$, though $y+1$ is a real number. Therefore at least one of $x$ and $y$ are nonzero.
If $x$ is nonzero, then that means $y$ is consequently only $0$. Of course, this is also a problem, because $xy=0$ and $x\cdot 1=x\neq0$. So its false.
If $x$ is zero, then:
$$\forall z\in\mathbb{R}[xy=xz]$$
is inhearently true, because $xy=0$ and $xz=0$ for every $y$ and $z$.
$\endgroup$
