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?

lurtzslikgtgjd
2022-05-08
Answered

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?

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Madelyn Lynch

Answered 2022-05-09
Author has **15** answers

Well, Pythagorean triples are given by

$(a,b,c)=(2pq,{p}^{2}-{q}^{2},{p}^{2}+{q}^{2})$

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 $(2p,{p}^{2}-1,{p}^{2}+1)$. 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 $\frac{p}{({p}^{2}-1)/2}$, 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).

$(a,b,c)=(2pq,{p}^{2}-{q}^{2},{p}^{2}+{q}^{2})$

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 $(2p,{p}^{2}-1,{p}^{2}+1)$. 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 $\frac{p}{({p}^{2}-1)/2}$, 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).

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