Prove that $j < n$, $k < n - j$ and $(n - j) - k = n - (j + k)$ if we know that $j + k < n$ in $\mathbb{N}$

Question

Suppose $j + k < n$ in $\mathbb{N}$.

1. Show that $j < n$
2. Show that $k < n - j$
3. Show that $(n - j) - k = n - (j + k)$

Thoughts for the Problems

I'm not really good with proof, so I start to have some thoughts about it. I learned addition, subtraction, distributive law and well-ordering in my number system class.

For the first two parts, I believe that well-ordering, trichotomy and induction are needed to show certain statements. I think that for the last part, I need to apply the first two parts for this problem and then, use the elementary additions and subtraction laws.

Any comments or advices?

Answer 1

Because $j,k,n \in \mathbb{N}$ it obvious that:

$$j < j+k$$

From the condition we have:

$$j+k<n \implies j<n$$

The second inequality is just a rearrangment of the initial condition.

$$j+k<n \implies j<n-k$$

And for the equality just get rid of the parenthesis. So we end up with:

$$(n-j) - k = n - (j+k) \implies n-j-k = n-j-k$$, which is obviously true.