Determining the truth value of $∃x∀y (y=x^2+2x+1)$

Question

The domain for $x$ and $y$ are all integers.

$∃x∀y (y=x^2+2x+1)$ can be interpreted as: There is an $x$ for all $y$ such that it satisfies $y=x^2+2x+1$. There is a single $x$ value which results in all the domain values of $y$ - all the integers.

This can't be true, as the relationship between $x$ and $y$ is given; a single $x$ value cannot result in every elements in the range of integers. Thus, the truth value of the statement is false.

Yet, a different justification for the false value was stated on the answer sheet. Referring to it, the statement's truth value is false because for every $x$, there is always an integer $y=x^2+2x+2>x^2+2x+1$.

Can anyone explain to me regarding its meaning? Thanks.

Answer 1

$$∃x∀y (y=x^2+2x+1)\tag{*}$$

the truth value of the statement is false.

Yet, a different justification for the false value was stated on the answer sheet. Referring to it, the statement's truth value is false because for every $x$, there is always an integer $y=x^2+2x+2>x^2+2x+1$.

The answer sheet is merely pointing out that the negation $$∀x∃y (y\neq x^2+2x+1)$$ of $(*)$ is True, and thus $(*)$ itself must be False.

Answer 2

Suppose that the proposition is true, that is, there is a fixed $x\in\mathbb Z$. It is assumed that when taking any $y\in\mathbb Z$ it must have the form $x^2+2x+1=(x+1)^2$. But for example we take $y=x^2+2x+k$, with $k\in\mathbb Z_{>1}$, in which it does not have the form, since $y=x^2+2x+k>x^2+2x+1$, that is $y\neq x^2+2x+1$. Therefore contradiction.

Although perhaps it would have been easier to take a $y\in\mathbb Z^-$, and for him there is no $x\in\mathbb Z$ that has the form $\underbrace{y}_{<0}=\underbrace{x^2+2x+1}_{\geq 0}$.

Another way to see it can be: Suppose that the proposition is true, then for all $y\in\mathbb Z$, we must have $y=x^2+2x+1$ for example,

• for $y=1$, we have $1=x^2+2x+1$ (remember that $x$ is fixed)
• for $y=2$, we have $2=x^2+2x+1$ Then $1=x^2+2x+1=2$, contradiction.