## 187 – Quiz 4

Here is quiz 4.

Problem 1 asks to determine (with brief justifications) the truth value of the following statements about integers:

1. $\forall x\,\forall y\,(x>y).$
2. $\exists x\,\forall y\,(x>y).$
3. $\forall x\,\exists y\,(x>y).$
4. $\exists x\,\exists y\,(x>y).$

1. is False. To show this we provide a counterexample: Specific integers $x,y$ such that $x\not>y.$ For example, $1\not>23.$

2. is False. To show this we need to exhibit for each integer $x$ an integer $y$ such that $x\not> y.$ For example, $x\not>x+1.$ Note that, although $y$ is a fixed integer once we know $x,$ we are not giving a fixed value of $y$ that serves as a simultaneous counterexample for all values of $x.$

3. is True. To show this we exhibit for each integer $x$ a specific integer $y$ such that $x>y.$ For example: $x>x-1.$ Note that, although $y$ is a fixed integer once we know $x,$ we are not giving a fixed value of $y$ that works simultaneously for all $x.$

4. is True. To show this, we exhibit specific values of $x,y$ such that $x>y.$ For example: $1777>-52451256.$

Problem 2 asks to show by contradiction that no integer can be both odd and even. Here is the proof: Suppose otherwise, i.e., there is an integer, let’s call it $x,$ such that $x$ is both odd and even. This means that there are integers $y,z$ such that $x=2y+1$ (since $x$ is odd) and $x=2z$ (since $x$ is even).

Then we have that $2y+1=2z,$ or $1=2(z-y).$ But this is impossible, since 1 is not divisible by 2. We have reached a contradiction, and therefore our assumption that there is such an integer $x$ ought to be false. This means that no integer can be both odd and even, which is what we wanted to show.

Note that we have not shown that every integer is either odd or even. We will use mathematical induction to do this.

Problem 3 asks for symbolic formulas stating Goldbach’s conjecture and the twin primes conjecture (both are famous open problems in number theory).

Goldbach’s conjecture asserts that every even integer larger than 2 is sum of two primes:

$\forall x\,([{\tt Even}(x)\land(x>2)]\Rightarrow\exists p\,\exists q\,[x=p+q\,\land$ ${\tt Prime}(p)\land{\tt Prime}(q)]).$

Here, ${\tt Even}(x)$ is the formula asserting that $x$ is even, namely, $\exists y\in{\mathbb Z}\,(x=2y),$ and ${\tt Prime}(n)$ is the formula (given in the quiz) asserting that $n$ is prime. Note we had to add existential quantifiers in order to be able to refer to the two prime numbers that add up to $x.$

The twin primes conjecture asserts that there are infinitely many primes $p$ such that $p+2$ is also prime.

The difficulty here is in saying “there are infinitely many,” since the quantifier $\exists$ only allows us to mention one integer at a time, and writing something of infinite length such as $\exists x_1\,\exists x_2\,\exists x_3,\dots$ is not allowed.

We follow the suggestion given in the quiz, and represent “there are infinitely many $n$ with [some property]” by saying “for all $m$ there is a larger $n$ with [some property].”

$\forall m\in{\mathbb Z}\exists n\,(n>m\land {\tt Prime}(n)\land {\tt Prime}(n+2)).$