Proof that for 2 integers of which their product is 1, then each integer is 1. I am working through my Real Analysis' exercises and have run into one that conceptually is very straight forward but has proven to be somewhat difficult for me. I suppose it is less "Real Analysis" and more just learning to create proofs.
The claim is:
$$\forall p,q\in \mathbb{Z} \:(p>0 \wedge q>0 \wedge pq=1) \implies p=q=1$$
Now my issue is that I'm not really sure what it is appropriate to assume in this circumstance, as I suppose the goal is being able to prove it in the most fundamental way possible.
For example, we know for most elements in the integers, they do not possess a multiplicative inverse, except for 1 and -1. If we applied this, the proof would be pretty easy, but I would say we would be able to extend the proof of our claim to prove this fact itself.
Here's my attempt at a proof:
$$
Assume \:p,q\in\mathbb{Z} \: such \: that \: pq=1 \; and \; p>0 \; and \; q>0
\\Assume  \: for \; a\;contradiction \;that\;p\neq1 \; or \; q\neq1 \; or \; p\neq1 \; and\; q\neq1 
\\Then \; 1=|pq|=|p||q|\geq |p| \; and\; 1=|pq|=|p||q|\geq |q|
\\ Then \; 1\geq p\; and \; 1\geq q
\\ p=1 \; or \; p=0 \; and \; q=1\; or\; q=0
\\ But \; this \; is \; a \; contradiction \; as \; p>0\; and \; q>0\; and\; both\; p,q\; cannot \; simultaneously \; be\; 0.
\\ Hence \; \forall p,q\in \mathbb{Z} \:(p>0 \wedge q>0 \wedge pq=1) \implies p=q=1 \; $$
(Sorry can't figure out the QED symbol in LaTeX, clearly not very good at it).
Is this proof solid or would there be a more fundamental method of reaching the conclusion.
Also, would anyone have any good resources that assisted them in really grasping Real Analysis and Proofs in general?
Thanks!
 A: Comments:
You never actually use the hypothesis that $p,q\neq 1$ and you're argument doesn't require the use of absolute value signs, since by hypothesis, we have $p,q>0$. Besides, if you start off with the assumption that $p,q\neq 1$ and $p,q>0$, then you should conclude that $p,q\geq 2$. The contradiction you should have reached is $1\geq 2$, which actually utilizes the additional assumption that $p,q\neq 1$ and would make for a sound proof by contradiction.
Alternative Solution (1):
To be more general however, lets just say $p,q\in\mathbb{Z}$ and suppose we have $pq=1$. Aiming for a more direct approach, we notice that if one of $p$ or $q$ were $0$, then we would have $pq=0\neq 1$, so we can assume $p,q\neq 0$. Now, $|p|,|q|\geq 1$, since $p,q\in\mathbb{Z}$. From your argument, we have $$1=|pq|=|p||q|\geq |p|\geq 1\text{ and }1=|pq|=|p||q|\geq |q|\geq 1 $$
so we must have $|p|=|q|=1$.
Now we can immediately check off the four cases that satisfy the previous relation, those being $$p=1,q=1\tag{1}$$ $$p=-1,q=-1\tag{2}$$ $$p=1, q=-1\tag{3}$$ $$p=-1,q=1\tag{4}$$
and conclude that we must have either case $(1)$ or case $(2)$. The cases come directly from the definition of $|\cdot|:\mathbb{R}\to\mathbb{R}$, that being $|x|=\max\{x,-x\}$.
In fact, once you have $|p|=1$, then you can conclude that $p=1$ or $p=-1$ so that $1=q$ or $1=-q$.
Alternative Solution (2):
Let that $p,q\in\mathbb{Z}$ with $p,q>0$ and $pq=1$. We immediately see that $p,q\geq 1$. Assume for contradiction that  $p$ or $q$ is larger than $1$. Without loss of generality, suppose $p > 1$. Write $p$ in terms of its prime factors:
$$p=p_1^{k_1}\cdot p_2^{k_2}\cdots p_n^{k_n}$$
We must have $p_1\geq 2$ since $p_1$ is a positive prime, hence $$1=pq\geq p_1q\geq 2\cdot 1=2$$ a contradiction. Our assumption that $p$ or $q$ was greater than one must be false, hence $p=q=1$.
Alternative Solution (3):
Let $p,q\in\mathbb{Z}$ with $p,q>0$ and $pq=1$. Then $p,q\geq 1$ $\ln(p),\ln(q)\geq\ln(1)=0$. Now we see that $$0=\ln(1)=\ln(pq)=\ln(p)+\ln(q)\geq\ln(p)\geq 0$$
hence $\ln(p)=\ln(1)=0$. Since $\ln$ is injective, then we must have $p=1$. Now $1=pq=q$ so $q=1$.
Alternative Solution (4):
Assuming the same hypothesis as the first solution, we have that $p\mid q$ and $q\mid p$, so there exists non-zero integers $t_1,t_2$ such that $$p=t_1q\text{ and }q=t_2p$$Now, $$t_1q^2=1=t_2p^2$$ and we conclude that because $q^2,p^2>0$, then we must have $t_1,t_2>0$, or equivalently $t_1,t_2\geq 1$. But we also see that if $t_1,t_2>1$, then so are $t_1q^2$ and $t_2p^2$, hence $t_1=t_2=1$. Now we see that $pq=p^2=q^2=1$. Since the only solutions to the polynomial $f(x)=x^2-1$ are $x=\pm 1$, then the result follows.
Alternative Solution (5):
Let $p,q\in\mathbb{Z}$ with $pq=1$. Consider matricies
$$R
=
\begin{bmatrix}
1 && -p\\
q && 1
\end{bmatrix} 
,\,\,\,
R^{-1}
=
\frac{1}{1-(-pq)}
\begin{bmatrix}
1 && p\\
-q && 1
\end{bmatrix}
$$
Then $RR^{-1}=I$ so we have
$$RR^{-1}=
\frac{1}{2}
\begin{bmatrix}
2 && p-q\\
0 && 2
\end{bmatrix}
=
\begin{bmatrix}
1 && 0\\
0 && 1
\end{bmatrix}$$
Now we must have $p=q$, implying $p^2-1=0$ and the result follows.
Final Comment:
It would be cool to see a proof done using group or ring theory to show this result as well. I cannot fully see through a proof by those means as of right now, but I believe its likely that it can be done.
