Prove that $mn < 0$ if and only if $m > 0$ and $n < 0$ or $m < 0$ and $n > 0$. $m,n$ element of integers
Just starting out teaching myself discrete math still really bad at proofs, any help/advice on how to think/go about this would be greatly appreciated.
On
1) If m < 0 and n >0 or m>0 and n < 0,then mn < 0.
2) If mn < 0, suppose m > 0 and n > 0,this gives mn > 0 which is contradiction.Hence ,one of them should definitely be less than 0.If both are,then contradiction arises again as product would be greater than 0
On
Consider the possible cases. There are three choices for $m$ and three for $n$, though it turns out five of them amount to the same thing, and there are two further that equal each other.
The first five are easy. If one or both numbers are zero, so is their product.
Suppose both are positive. Then $0<m\cdot n$. That's six down.
Suppose both are negative. Then $m=-|m|$, $n=-|n|$, so $$mn=-|m|\cdot -|n|$$$$={(-1)}^2\cdot{|m|}{|n|}$$$$=|m||n|>0$$ Seven down.
Suppose exactly one is negative. Without loss of generality$^1$, $m<0<n$. Now apply a similar argument to the above case using modulus and bringing out the negation.
Eight and nine down. All cases down, and the hypothesis is perfectly true, so we're done!
$^1$ That means we can pick either one, so despite two cases existing, we can treat them like they're the same case.
On
It depends entirely upon what axioms and definitions we are given.
I am going to assume you given a+0 = a for all a, and that for each there is a unique -a such that a+-a=0. And that you were given that if x < y then x+z < y +z for all z.
From that we can prove if x > 0 then -x < 0. And if x < 0 then -x > 0 and that -(-a) = a Pf: x >0 then x+-x > 0+ -x so 0 > -x. If x < 0 then x+-x < 0+-x so 0 < -x. -a + -(-a) = 0 and -(-a) is unique but -a + a = 0 so -(-a)=a.
I'm going to assume you've been given that a (b+c)=ab+ cd.
From that you can prove i) 0a =0 and ii) (-a)b=a (-b)=-ab.
Pf. i) 0a=(0+0)a=0a+0a so 0a+-0a = 0a +0a+-0a so 0=0a
ii) ab + (-a)b=(a+-a)b=0b =0 so (-a)b=-ab. ab + a (-b)=a (b+-b)=a0=0 so a (-b)=-ab.
I'm going to assume you've been given if x > 0 and y > 0 then xy > 0. From that we can prove your result.
Case 1: m < 0 n > 0. Then -m > 0 so -mn > 0. So mn < 0.
Case 2: m > 0 n < 0. Then -n > 0 so -mnn > 0 and mn < 0
Case 3-6: m = 0 then mn =0. If n=0 then mn = 0. If m > 0 n > 0 then mn > 0. If m < 0 and n < 0 then -m > 0 and -n > 0 so -(-mn) = mn > 0.
But this all assumes you were given certain axioms. If you were given different axioms, it'd be different.
On
The proofs given here can be simplified.
Do one direction: Assume $mn < 0$ then $m\neq 0 \neq n$ because otherwise the product would be zero. Thus assuming without loss of generality that $m>0$ we get by dividing both sides
$$\frac{mn}{m} < \frac{0}{m}$$
or in other words, $n <0$.
The other direction: Without loss of generality assume that $m>0$ and $n<0$. Since the product of a positive number and a negative number is negative we obtain
$$mn <0$$
as desired.
We need to make a few assumptions about how you have integers defined, but a very safe assumption is that you have defined $(\Bbb Z, +, \times, \leq)$ as an ordered commutative ring with identity constructed from the natural numbers in the usual way.
In particular, as part of the very definition of $(\Bbb Z,+,\times,\leq)$, we have the following (among many more properties, these are just the important ones necessary for my proof):
(In all of the following, $a,b,c$ and $x$ are arbitrary integers)
Properties of $+$
Properties of $\leq$ (and $<$)
Properties of $-$
Properties of $\times$
Starting the desired proof, we approach via contrapositive for the forward implication. That is, we wish to prove that if ($0\leq m$ and $0\leq n$) or ($m\leq 0$ and $n\leq 0$) then $0\leq m\times n$. In doing so, we will have proven that $m\times n<0$ implies $m<0$ and $0<n$ or $0<m$ and $n<0$.
Suppose $0\leq m$ and $0\leq n$. Then by definition $0\leq m\times n$, so we are done.
Suppose $m\leq 0$ and $n\leq 0$. Then $m+(-m)\leq 0+(-m)$ and therefore $0\leq -m$. Similarly $0\leq -n$. Then $0\leq (-m)\times (-n) = ((-1)\times m)\times ((-1)\times n) = (-1)\times (-1)\times m\times n = m\times n$.
Thus if $m\leq 0$ and $n\leq 0$ or if $0\leq m$ and $0\leq n$, we have $0\leq m\times n$
For the remaining direction for the proof, we wish to show that if $0<m$ and $n<0$ or if $m<0$ and $0<n$ then $m\times n < 0$
Suppose $0<m$ and $n<0$. Then $n+(-n)<0+(-n)=-n$. Thus $0<m\times (-n)$. Thus $0+(m\times n) < m\times (-n)+(m\times n) = m\times (-1)\times n + m\times n = -(m\times n)+(m\times n) = 0$. Therefore $m\times n<0$
The proof for $m<0$ and $0<n$ is identical by relabeling which was $m$ and which was $n$.