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On a practice final exam for my Discrete Math class, I've been asked to prove that the sum of two positive integers is positive. I've been pulling my hair out over how to prove this, as it seems so obvious. I even asked some of my friends taking proof-based calculus, and they said that this was the type of thing that is just assumed to be true. Anyone have an idea of how to prove this? Any help is appreciated.

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closed as unclear what you're asking by Grigory M, egreg, Branimir Ćaćić, Davide Giraudo, Namaste Dec 19 '13 at 22:11

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    $\begingroup$ Define "positive integer" . $\endgroup$ – DonAntonio Dec 19 '13 at 21:02
  • $\begingroup$ How exactly have you defined these objects, and what are some basic facts about them? $\endgroup$ – user61527 Dec 19 '13 at 21:02
  • $\begingroup$ An integer greater than zero $\endgroup$ – tylucaskelley Dec 19 '13 at 21:03
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    $\begingroup$ You're having trouble because you don't have a definition of $>$ or $+$. Without that, you can't get started. You need to find the place in your discrete math notes or text where these are defined, then write down the definition, and start with that. Otherwise you have nothing to work with. $\endgroup$ – MJD Dec 19 '13 at 21:06
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    $\begingroup$ It's all about the definitions. We can't help without knowing which definitions you are using. $\endgroup$ – Thomas Andrews Dec 19 '13 at 21:10
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What's important about the Integers is that they are an totally ordered commutative ring with identity, though more precisely they form a totally ordered integral domain.

There are two components here: being a commutative ring with identity, i.e. an integral domain refers to the algebraic properties of $\mathbb{Z}$: commutativity and associativity of addition and multiplication, distributivity, cancellation for multiplication, etc. These don't really matter to the question. You can look them up if you're interested (you should).

What does it mean for $\mathbb{Z}$ to be totally ordered? It means that the following four axioms of order hold for $\mathbb{Z}$:

  • Trichotomy: if $a,b \in \mathbb{Z}$, then one and only one of the following holds: $a<b$, $a=b$, $a>b$.
  • Transitivity: If $a,b,c \in \mathbb{Z}$ with $a<b$ and $b<c$, then $a<c$.
  • Addition: If $a,b,c \in \mathbb{Z}$ and $a<b$, then $a+c<b+c$.
  • Multiplication by Positive Elements: If $a,b,c \in \mathbb{Z}$, $a<b$ and $c>0$, then $ac<bc$.

You need to use the addition rule to complete your proof. Take a positive integer $a>0$. Take a positive integer $b>0$. So $a+b > b+0$. So $a+b > b$. And $b>0$. So by transitivity, $a+b>0$.

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  • $\begingroup$ Thank you! It seems so obvious now. $\endgroup$ – tylucaskelley Dec 19 '13 at 21:17
  • $\begingroup$ Very instructive answer. $\endgroup$ – John Dec 19 '13 at 21:19
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If you have the more or less usual definitions and operations:

$$a,b\;\;\text{positive}\iff a\stackrel 1>0\,,\,\,b\stackrel 2>0$$

and from here

$$a+b\;\;\text{non-positive}\;\iff a+b\le 0\iff a\le -b\stackrel 2<0\implies a<0\;\;\text{contradiction to}\;\;1$$

Thus is must be $\;a+b>0\implies\;\text{the sum is positive}\;$

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