Proposition: $\exists n \in \mathbb{Z} $ such that $\forall m \in \mathbb{Z}, m<n $.

a) Negate the statement of proposition.

b) Disprove proposition

Here is what I wrote:

a) $\forall n \in \mathbb{Z}, \exists m \in \mathbb{Z}, m >= n$.

b) The statement says "there exists an integer n, such that for every integer m, m is less than n." As a counterexample, let n = 1. For every element in the integers, m, must bust be less than n. This may be true for some cases, but not all cases of m. Let us show that, when m=2, 2 < 1 is not a true statement. Hence, disproven.

What did I do wrong here? I only received partial credit. How do I disproove this? Did I get part A correct at least?


2 Answers 2


Part (a) is correct.

Your proof of (b) is completely incorrect. You want to disprove a statement of the following form:

There exists an $n \in \Bbb Z$ such that blah blah blah.

In order to prove a "there exists" statement, it suffices to find an example. But in order to disprove a "there exists" statement, you have to show that every possible example does not work. In other words, you have to show that no matter what $n \in \Bbb Z$ you pick, the statement is false.

Instead, you just disproved a single counterexample (where $n=1$.) That doesn't count as a proof. You just showed that $n \not= 1$. You didn't show that no such $n$ exists.

However, the general idea behind your proof was correct. This is what a full proof of (b) could look like:

Let $n \in \Bbb Z$ be an arbitrary integer. Then $n+1$ is an integer such that $n+1 \geq n$. This is a contradiction. Thus, the original statement is false.


It's wrong since it's not because it's doesn't work for $n=1$ that it will never work. Your proposition says that $\mathbb Z$ is upper bounded. So suppose it's true and let denote $M=\sup \mathbb Z$. In particular $\lfloor M\rfloor +1>M$ and $\lfloor M\rfloor +1\in \mathbb Z$. Contradiction.


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