Can anyone prove for every $a,b \in \mathbb Z^+ < p$ ( $p$ is a prime), $p \nmid ab$? Can anyone prove for every $a,b \in \mathbb Z^+ < p$ ( $p$ is a prime), $p \nmid ab$?
I was trying my best to do the problem but like I don't know where to start or anything!
 A: Assuming you mean: $p$ does not divide $ab$.
It comes from the definition of prime. If $p \mid ab$, then $p \mid a$ or $p \mid b$. But we know that if $x \mid y$, and $x,y > 0$, then $x \le y$. Since $a, b < p$, we have $p \nmid ab$ by contradiction.

So as Andres pointed out below, some people say "prime" where I would say "irreducible". So if your definition is "cannot be divided by anything but $\pm 1, \pm p$", use Bezout's theorem. 
In my opinion, it's most straightforward to prove it for the equivalent form: for any $a,b,c$, if $a \mid bc$ and $(a,b) = 1$, then $a \mid c$. Then push your proof around until it proves the statement you want.
A: Hint $\ $ The set $\,S$ of naturals $\,n\,$ such that $\,\color{#c00}{p\mid nb}\,$ is closed under subtraction  and contains $\ a,p\,$ therefore its least positive element $\,\color{#0a0}{d\mid a,p}.\,$ Since $\,\color{#a0f}{d\mid p\ \ \rm prime},\,$ either $\,\color{#a0f}{d=p}\,$ so $\ \color{#0a0}{p=d\mid a},\,$ or $\,\color{#a0f}{d=1}\in S\ $ thus $\ \color{#c00}{p\mid d b = b},\ $ i.e. $\,\ p\mid \color{}a\,$ or $\ p\mid \color{}b.\ \ $ QED
Note $ $ If we know gcds then we know the gcd  $\, (p,a)\,$ exists, so we can rewrite the proof as
$$ p\mid pb,ab\,\Rightarrow\, p\mid(pb,ab)\overset{\color{brown}{\rm(D\,L)}}= (p,a)b = b\ \ {\rm if}\ \  \,(p,a)= 1,\ \ {\rm i.e.}\ \  p\nmid a$$
where we used $\,\color{brown}{\rm(DL)}$ = gcd Distributive Law. If you do not know that basic law then you can  instead employ the gcd Bezout Identity  i.e. $\, (p,a)=1\,$ so $\,jp\!+\!ka = 1\,$ for $\,j,k\in\Bbb Z,\,$ hence 
$\qquad\qquad\qquad\ \ p\mid pb, ab\,\Rightarrow\, p\mid jpb,kab\,\Rightarrow\, p\mid (jp\!+\!ka)b = b$
A: $a,b < p \Rightarrow ax+py=1 \Rightarrow abx+bpy=b$ and $b<p \Rightarrow$ $p$ does not divide $b \Rightarrow p$ does not divide $ab$.
A: OK, here is as bare bones a proof as I can see at the moment. It uses the well ordering principle and the fact that given any positive integers $n,m$, there are unique integers $q,r$ with $0\le r<m$ such that $n=mq+r$.
Suppose the result is false, so there is some prime for which it fails. There is therefore a smallest prime for which it fails. Call it $p$. So, there are positive integers $a,b<p$ such that $ab=pk$ for some $k$. Pick $k$ as small as possible. I will argue that $k=1$, that is, $a,b<p$ but $ab=p$. This contradicts that $p$ is prime, and we are done.
To prove that $k=1$, we argue again by contradiction: Otherwise, there is a prime factor dividing $k$, call it $q$. Since $a,b<p$, then $kp=ab<p^2$, so $k$ (and therefore $q$) is smaller than $p$. Now, we can write $a=qc+d$ and $b=qe+f$ with $0\le d<q$ and $0\le f<q$. If $d=0$, we can divide $a$ by $q$, and we have that $(a/q)b=p(k/q)$, contradicting the minimality of $k$. Similarly, we reach a contradiction if $f=0$. We can therefore assume that $d,f$ are positive integers, and both are strictly less than $q$. 
Finally, we have that $ab=pk$ is a multiple of $q$, but $ab=(qc+d)(qe+f)=qN+df$ for some $N$, and it follows that $q$ divides $df$. We have now contradicted the minimality of $p$ and, with this, concluded the proof.
