Formalize word statements with mathematical notation

Question

I am giving the following statements and I have to formalize them into mathematical notation. Moreover I have to negate them. It would be amazing if you could be my second pair of eyes to proofread my answers!

(1) Every Element of $M$ is negative

My response: $\forall m \in M : m<0$ with the negation $\exists m \in M : m\geq 0$

(2) Between two different elements of $M$ there is another element of $M$.

My response: $\forall m,n\in M:m \neq n$ $\exists z\in M: m<z<n \vee m>z>n$ with the negation $\forall z \in M : m=z=n$ $\exists m,n \in M: m=n$.

(3) $M$ contains at least two elements

My response: $\exists m_1,m_2\in M : m_1\neq m_2$ with the negation $\forall m_1,m_2\in M : m_1=m_2$.

(4) Every element of $M$ can be expressed as the product of two different elements of $M$.

My response: $\forall m\in M$ $\exists m_1,m_2 \in M: m_1 \neq m_2 \land m_1\cdot m_2=m$ with the negation $\forall m_1,m_2 \in M : m_1=m_2$ $\exists m\in M: m_1\cdot m_2\neq m$

Answer 1

Your negations of statements (2) and (4) are wrong, essentially for two reasons:

• the negation of a universal quantifier is an existential quantifier, and you cannot invert the order of an existential and a universal quantifier (if you invert their order you change the meaning of the sentence: think of the difference between $\forall m \exists n \, m < n$ and $\exists n \forall m \, m < n$);

• statements of the form $m < z < n$ are actually compound statements that mean $m < z \land z < n$ (conjunction of two facts), so their negation is $m \not< z \lor z \not< n$ (disjunction of the negation of the two facts).

Using your informal notation, the negations of the statements (2) and (4) are:

2. $\exists , \in : ≠ \land \forall ∈: (\not< \lor z\not<) \land (\not> \lor z\not>)$

and

4. $\exists m \in : \forall _1 , m_2 \in M : (_1 = _2) \lor (_1⋅_2 \neq )$.

Moreover, in your formalizations, you keep implicit some connectives. In an informal setting, it is fine, but in a more rigorous setting you have to make explicit all the connectives (this also allows you to avoid mistakes in negating statements). The followings are a more explicit formalization of your statements.

1. $\forall m (m \in M \to m < 0)$.

   Negation: $\exists m (m \in M \land m \not < 0)$. If you know that $<$ is a total order on $M$, then you can equivalently say $\exists m (m \in M \land m \geq 0)$.

2. $\forall \forall ((m \!\in\! \land n \!\in\! M \land \!\neq\! ) \to \exists (z \!\in\! \land ((\!<\! \land z\!<\!) \lor (n \!<\! z \land z \!<\! m))))$.

   Negation: $\exists m \exists n ((m \!\in\! \land n \!\in\! M \land \!\neq\! ) \land \forall (z \!\in\! \to (( \!\not<\! \lor z \!\not<\!) \land (n \!\not<\! z \lor z \!\not<\! m))))$. If you know that $<$ is a total order on $M$, then you can equivalently say $\exists m \exists n ((m \!\in\! M \land n \!\in\! M \land m \!\neq\! n) \land \forall z(z \!\in\! \to ((m \!\geq\! z \lor z \!\geq\! n) \land (n \!\geq\! z \lor z \!\geq\! m))))$.

3. $\exists m \exists n (m \in M \land n \in M \land m \neq n)$.

   Negation: $\forall m \forall n ((m \in M \land n \in M) \to m = n)$.

4. $\forall (m \in \to \exists _1 \exists _2 (m_1 \in \land m_2 \in M \land _1 \neq _2 \land _1⋅_2=))$.

   Negation: $\exists (m \in \land \forall _1 \forall _2 ((m_1 \in \land m_2 \in M) \to (_1 = _2 \lor _1⋅_2 \neq )))$.

Answer 2

(1) Every Element of $M$ is negative

My response: $\forall m \in M : m<0$ with the negation $\exists m \in M : m\geq 0$

Yes.

(2) Between two different elements of $M$ there is another element of $M$.

My response: $\forall m,n\in M:m \neq n$ $\exists z\in M: m<z<n \vee m>z>n$ with the negation $\forall z \in M : m=z=n$ $\exists m,n \in M: m=n$.

The response is mostly okay, apart from the missing implication.   Also recall that $(a\lt b\lt c)$ is a conjunction $((a< b)\land(b\lt c))$.

$$\forall m,n\in M: (m\neq n\to\exists z: (m\lt z\land z\lt n)\lor(m\gt z\land z\gt n))$$

Now, try negating that .

(3) $M$ contains at least two elements

My response: $\exists m_1,m_2\in M : m_1\neq m_2$ with the negation $\forall m_1,m_2\in M : m_1=m_2$.

Indeed.

(4) Every element of $M$ can be expressed as the product of two different elements of $M$.

My response: $\forall m\in M$ $\exists m_1,m_2 \in M: m_1 \neq m_2 \land m_1\cdot m_2=m$ with the negation $\forall m_1,m_2 \in M : m_1=m_2$ $\exists m\in M: m_1\cdot m_2\neq m$

The response is okay , but the negation is bad.   Do not change the order of the quantified terms.   Also remember that the negation of a conjunction is a disjunction of negations. $$\lnot (\varphi\land\psi)\equiv (\lnot \varphi\lor\lnot\psi)$$