Formalize word statements with mathematical notation

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: $$\forall m,n\in M:m \neq n$$ $$\exists z\in M: mz>n$$ 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 $$\forall m_1,m_2 \in M : m_1=m_2$$ $$\exists m\in M: m_1\cdot m_2\neq m$$

• Not a bad start. There is a missing implication arrow in (2). The negations for (2) and (4) both have serious problems with quantifier order and primitives like $m<z<n$ are incorrectly negated. – Erick Wong Jul 29 at 15:14

(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: $$\forall m,n\in M:m \neq n$$ $$\exists z\in M: mz>n$$ 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. $$\lnot (\varphi\land\psi)\equiv (\lnot \varphi\lor\lnot\psi)$$

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:

1. $$\exists 𝑚,𝑛 \in 𝑀 : 𝑚≠𝑛 \land \forall 𝑧∈𝑀: (𝑚\not<𝑧 \lor z\not<𝑛) \land (𝑚\not>𝑧 \lor z\not>𝑛)$$

and

1. $$\exists m \in 𝑀 : \forall 𝑚_1 , m_2 \in M : (𝑚_1 = 𝑚_2) \lor (𝑚_1⋅𝑚_2 \neq 𝑚)$$.

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

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

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

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