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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?

NasuSama
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1 Answers1

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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.

Stefan4024
  • 35,843
  • No, I mean: show this by setting a theorem on inequality/variable. For instance, I prove the statement by induction on the variable. – NasuSama Oct 12 '13 at 17:16