How to prove n>0⟹n≥1 for integers n by constructing a truth table

Question

How to prove $n>0\implies n\ge1$ for integers $n$ by constructing a truth table?

I know the some basic concepts of the truth tables like $p,q,r$ but I have no idea how to prove the above statement true using a truth table.

      p     q     r
      T     T     T 
      T     T     F
      T     F     T
      T     F     F
      F     T     T
      F     T     F
      F     F     T
      F     F     F

Answer 1

Truth tables really are just for proving tautologies with statements, i.e. things with a true/false value. That way you can enumerate all possibilities, and verify that they are true in all situations.

In this case, we're dealing with a predicate in $n$. To enumerate all possibilities, you would have to list all integers $n$, and assign a true/false label to the statement to each of them. This, of course, would have to be some kind of infinite table, of which we'd only be able to see a finite number of rows anyway, which defeats the purpose of a truth table, in that you're able to see that the statement is always true.

So, basically, you cannot prove this with a truth table.

Answer 2

Ah, the book asks you to do this by exhausting all possible values of $n \in \mathbb Z$.

$\begin{cases}n&||& n > 0 & n \ge 1 & (n>0)\implies (n\ge 1) \\\\n < 0&||& F & F & T\\ n=0&||& F & F & T\\n = 1&||& T & T & T\\ n>1&||& T & T & T\\\end{cases}$

If you like you can combine lines 1 and 2 with $n \le 0$ and lines 2 and 3 with $n \ge 1$.

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That's a weird question: I can only suppose it is meant to be so thusly:

$\begin{cases} n > 0 & n \ge 1 & \text{is this possible?} & (n>0)\implies (n\ge 1) \\ T & T&\text {yes} & T\\ \color{red}T & \color{red}F & \color{red}{\text{no}}& \color{red}F\\ \color{red}F&\color{red}T&\color{red}{\text{no}}& \color{red}T\\ F&F&\text{yes} & T\end{cases}$

So the only case where $(n>0)\implies (n\ge 1)$ could be false, (If $n > 0$ is true and $n\ge 1$ is false) is impossible (because $n \not \ge 1 \implies n <1 \implies n \le 0\implies n \not > 0$) so of all the rest of the possible options (and one of the impossible ones) $(n>0)\implies (n\ge 1)$ is true.

If this feels silly and slightly circular and as though you are assuming what you have to prove mathematically before you prove it in pure logic.... well, I agree....