How to prove that $x \gt 0, y \gt 0 \Rightarrow xy \gt 0$

Question

Show that, for every $x,y \in \mathbb Z$, we have:

$x \gt 0, y \gt 0 \Rightarrow xy \gt 0$

I've tried this way:

supose $x= \overline {(a,b)}, y= \overline {(c,d)}$. Then, $x \gt 0 \Rightarrow \overline {(a,b)} \gt \overline {(0,0)} \Rightarrow a \gt b$ and $y \gt 0 \Rightarrow \overline {(c,d)} \gt \overline {(0,0)} \Rightarrow c \gt d$.

I must prove that $xy \gt 0 \Rightarrow \overline {(a,b)}. \overline {(c,d)} \gt \overline {(0,0)} \Rightarrow \overline {(ac+bd,ad+bc)} \gt \overline {(0,0)} \Rightarrow ac+bd \gt ad+bc$.

Does anyone have any idea about how to connect hypothesis and thesis?

why "$x = {\overline {(a,b)}}$"? This is so obvious it's hard to prove. How about induction? First induct on $x$, then on $y$? — Stefan Smith, Nov 21 '13 at 23:20
I think he means the construction of $Z$ from $N$ http://en.wikipedia.org/wiki/Integer#Construction — TBrendle, Nov 21 '13 at 23:24
Yes, that's it. Puting in other words: how to prove that $ac+bd \gt ad+bc$ follows from $a \gt b$ and $c \gt d$? — Walter r, Nov 22 '13 at 01:02
Walt: Note if you already read my answer that I've included just now an alternate approach not using subtraction. — coffeemath, Nov 22 '13 at 03:49

coffeemath · Accepted Answer · 2013-11-22T03:44:43.233

$ac+bd>ad+bc$ iff $ac-ad>bc-bd$ iff $a(c-d)>b(c-d)$ and you can cancel $c-d$ here without reversing the inequality since you know $c>d$, giving $a>b$ which you also have as an assumption.

Of course this assumes one has subtraction defined in $\mathbb{N}$ provided one only subtracts a smaller from a larger natural number, but I think that is OK in some treatments.

Another approach: From $a>b$ there is a (positive) natural number $p$ with $a=b+p,$ and similarly from $c>d$ a positive natural number $q$ with $c=d+q.$

Then $$ac+bd=(b+p)(d+q)+bd=2bd+bq+dp+pq, \\ ad+bc=(b+p)d+b(d+q)=2bd+bq+dp.$$ Thus the first is greater than the second by the positive quantity $pq.$

How to prove that $x \gt 0, y \gt 0 \Rightarrow xy \gt 0$

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