Are these statements true in the following sets?

Question

I'd like to confirm my logic and better understand these statements. I'm supposed to find out if the following statements are true in $U = \Bbb Z$ and/or $\; U = \Bbb R$:

1. $\forall x \;(\exists y \; x \times y = x + y \leftrightarrow \exists y \; x \times y = y + 1)$

I've written this in English (to make it easier for me) as for all $x$, there exists a $y$ such that $x \times y = x + y$ if and only if there exists a $z$ such that $x \times z = z + 1$. If I'm remembering correctly, the variable there can change and it won't affect the statement? The first y and second y introduced are not the same?

So that essentially means that there is at least one $y$ for which $x \times y = x + y\;$ and $\; x \times z = z + 1$

In the case of $U = \Bbb Z$, when $x = 2$, $y = 2$ and $z = 1$, above equations work, meaning there does exist a $y$ and $z$ and so the entire statement is true.

Since it's true in $\Bbb Z$, wouldn't that automatically make it true in $\Bbb R$? Considering the same values can be used for $x,y,z$?

2. $\exists w \forall u \; (u \in w \leftrightarrow \exists x \; ( x \in u \land \forall y \; ( y \in u \rightarrow y = x)))$

English: There exists a $w$, for all $u$, such that $u \in w$ if and only if there exists an $x$, where $x \in u$ and for all $y$, if $y \in u$, then $y = x$

I'm having more difficulty understanding this one in the first place... does it mean that u only has one element, and that element is x? Following from that, would that mean w is the set of all singletons then?

Answer 1

(Too long for a comment)

First of all, You have not provided proof that $(1)$ is true:

You have proven the statement only for the triple (x,y,z) = (2,2,1); in other words, you've shown that that the following statement is true: $$\exists x \big( \exists y(x\times y = x+y) \iff \exists z(x\times z = z+1)\big)$$

You must show that the statements holds $\forall x \in \mathbb Z$.

Note that a biconditional evaluates true if and only if the two statements on each side are such that BOTH are true, or BOTH are false.

Another existence proof (different that x=2): Let $x = 0$. There exists a $y = 0$ such that $xy=0=x+y$. So one side of the biconditional is true. For the other side of the biconditional, we need to determine whether a $z$ exists such that $0\cdot z = z + 1$: Sure, pick $ z = -1$. Then $0\cdot (-1) = 0 = (-1) + 1$. Since both sides of the biconditional evaluate to true under the assignment $(x,y, z) = (0,0,-1)$, the statement holds true for $x = 0$.

We cannot possible check out an infinite number of integers n such that x= n, to determine whether the statement: For each and every $x$ in $\mathbb Z, \big[$(there is some $y\in \mathbb Z$ such that $(x\times y = x + y))\iff ($ there is some $z\in \mathbb Z$ such that $(x\times z =z +1))\big]$, is true.

So we need to find a counterexample to prove the statement false, or algebraiclly prove the statement holds for all integers $x$, (with $y, z \in \mathbb Z)$.

Let's look at x = 13. We need to determine whether an integer y exists such that $13y = 13+y$. Clearly, $y\neq 0.$ So we can divide $13y$ from each side of the equation to get $\dfrac{13+y}{13y} = \frac 1y+\frac 1{13} = 1$. This will be true only if $y= \frac {13}{12}$ so that $\dfrac 1{13/12} + \dfrac 1{13} = \dfrac 1{\frac {13}{12}} = \dfrac {12}{13} + \dfrac 1{13} = 1$.

But in this case $y= \frac {13}{12}\notin \mathbb Z$!

What about $x=13,$ such that there is a $z$ for which $13z = z+1. $

Clearly $z=0$ will not satisfy the equation. So we can divide both sides by $13z$: which gives us $1 = \frac{z+1}{13z} = \frac 1{13}+ \frac 1{13z}$ so if $z= \dfrac 1{12}$, then $$\frac 1{13} + \frac 1{\frac{13}{12}} =\frac {12}{13}+ \frac {1}{13} = 1$$

But note, $z = \frac 1{12} \notin \mathbb Z$

Answer 2

The first and second occurrences of "$\exists y$" are indeed independent. We can insert some extra brackets to clarify this:$$ \forall x\in U\; [\; (\exists y \in U\; (xy=x+y)\;)\iff (\; \exists y\in U\; (xy=y+1)\; )\; ].$$

This is syntactically unambiguous. It is also bad style for human readers, as they are not computers and find it easy to misunderstand it. Humans prefer $$\forall x\ \in U\; [\; (\; \exists y\in U\; (xy=x+y)\;)\iff (\; \exists z\in U\; (xz=z+1)\; )\; ].$$

When $U$ is a subset of the real numbers the above statement is equivalent to $$\forall x \in U\;\;[x\ne 1\iff x\ne 1\;].$$