# Let A be the set of all d such that d &lt; 1 and d = a b </mfr

Let $A$ be the set of all $d$ such that $d<1$ and $d=\frac{a}{b}$ where ${a}^{2}+{b}^{2}={c}^{2}$ and $a,b,c\in \mathbb{I}$. Is A an infinite or finite set and how can that be proven?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Well, Pythagorean triples are given by
$\left(a,b,c\right)=\left(2pq,{p}^{2}-{q}^{2},{p}^{2}+{q}^{2}\right)$
for integers $p$ and $q$ with $p>q$ (to prove that these are all is irrelevant; we only need infinitely many triples, not necessarily all of them).
So let $q=1$ and $p$ an odd ', then we have the triple $\left(2p,{p}^{2}-1,{p}^{2}+1\right)$. The corresponding $d$ would then be $d=\frac{2p}{{p}^{2}-1}$ (since $2p<{p}^{2}-1$ for $p>2$), this is an irreducible fraction if we divide both sides by $2$ (that is $\frac{p}{\left({p}^{2}-1\right)/2}$, and this is irreducible because ${p}^{2}-1$ and $p$ are co') and so it's different for every $p$. Since there are infinitely many 's $p$, the set $A$ is infinitely large (it actually contains a lot more numbers than we've shown, but that is irrelevant for it infinite-ness).