Poftethef9t

2022-06-27

Determining the truth value of $\exists x\forall y\left(y={x}^{2}+2x+1\right)$
The domain for x and y are all integers.
$\exists x\forall y\left(y={x}^{2}+2x+1\right)$ 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?

Blaine Foster

Step 1
$\begin{array}{}\text{(*)}& \exists x\forall y\left(y={x}^{2}+2x+1\right)\end{array}$ 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 $y={x}^{2}+2x+2>{x}^{2}+2x+1$.
Step 2
The answer sheet is merely pointing out that the negation $\forall x\exists y\left(y\ne {x}^{2}+2x+1\right)$ of (∗) is True, and thus (∗) itself must be False.

Semaj Christian

Step 1
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 ${x}^{2}+2x+1=\left(x+1{\right)}^{2}$. 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 $y\ne {x}^{2}+2x+1$. Therefore contradiction.
Step 2
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 $\underset{<0}{\underset{⏟}{y}}=\underset{\ge 0}{\underset{⏟}{{x}^{2}+2x+1}}$.
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.

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