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How to prove these 2 (one is true, the other one is false):

  • ∃n∈ Z: ∀k ∈ Z: n < k

  • ∀n∈ Z: ∃k ∈ Z: k < n

where Z = {0,±1,±2,...}

EDIT1: What I have so far:

.

∀n ∈ Z: ∃k ∈ Z: k < n

Pose n = x et k = x - 1,

For any x ∈ Z, there will always be a term inferior to x such as: if n = -665, k = -665 - 1 =-666 where k < n

Indeed, ∀n ∈ Z: ∃k ∈ Z: k < n is true.

.

∃n ∈ Z: ∀k ∈ Z: n < k

Pose n = x and k = y for any x,

There will exist x < y

But, limite of Z is +-∞

Then, for any x, there will always exist y = x-1 < x.

This proposition is false.

.

I know this is not working, i'm new to this

EDIT2:

To prove the second part, i would have to prove the opposite.

∃n ∈ Z: ∀k ∈ Z: n < k would be ¬(∃n ∈ Z: ∀k ∈ Z: n < k) = ∀n∃k: k < n

nnnn
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  • Welcome to MSE, we often ask for context and what you have tried with every question you ask – person Sep 30 '20 at 23:35
  • @person I added what I have so far, still I have no clue how to proceed – nnnn Sep 30 '20 at 23:46
  • I upvoted to partially reverse the downvotes, because you have now shown good work. I agree with both your reasoning and your conclusion in the 1st part. With the 2nd part, I found your reasoning hazy. I suggest you take another crack at the 2nd part. You might come to the same conclusion, but with clearer analysis. It seems to me that you have failed to focus on a key issue: what is the difference between $[\forall n \exists k]$ and $[\exists n \forall k]$? I suggest that you take your time considering the difference in these two ideas. – user2661923 Sep 30 '20 at 23:55
  • @user2661923 thank you for your feedback! I really don't know on the second part how you can prove that there is a n that is smaller then all the k numbers at once. It's basically infinity but it's the formulation that is hard – nnnn Oct 01 '20 at 00:01
  • @Nico there is no n smaller than all the k numbers at once. The second part is false, like you said. How do you prove something is false? –  Oct 01 '20 at 00:12
  • @SenZen I will say you prove something is false by proving that the opposite is true. – nnnn Oct 01 '20 at 00:29
  • ∃n ∈ Z: ∀k ∈ Z: n < k opposite would be ¬(∃n ∈ Z: ∀k ∈ Z: n < k) = ∀n∃k: k < n – nnnn Oct 01 '20 at 00:31

1 Answers1

1

Observe that $$t\in \Bbb Z \iff \{t+1 \in \Bbb Z\} \cap \{t-1 \in \Bbb Z\}\tag 1$$

An implication of this is that $\Bbb Z$ has no greatest or least element.

We can see this by assuming $t$ is the greatest (*least) element of $\Bbb Z$, and noticing that $(1)\implies t+1(^*t-1)\in\Bbb Z$ and that $t+1>t \ (^*t-1<t)$ and so $t$ is not the greatest (*least) element. By contradiction, there must be no such element.

The first statement asks whether there exists an $n\in\Bbb Z$ such that every $k\in\Bbb Z$ is greater than it. Such an $n$ would be the least element of $\Bbb Z$, but as stated earlier, it doesn't have one, and this statement is false!

The second statement asks whether for every $n\in \Bbb Z$, there exists a $k\in\Bbb Z$ less than it. Indeed, since there is no least element of $\Bbb Z$, we may easily find such an element. $(1)$ identifies that picking $k=n-1$ works fine here.

Rhys Hughes
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