# Heron's Formula and proof for a relation on the natural number triples

$$\sigma \left( x \right) =\cases{-1&x<0\cr \,\,\,\,\,0&x=0\cr \,\,\,\,\,1&x \gt 0\cr}\tag{Signum Definition}$$

We impose that:

$$\forall n \in {\{i,j,k}\} \subset \mathbb N$$ we have: $$\sigma_{{n}} \left( m \right) \in \left\{ \sigma \left( m \right) , \sigma \left( -m \right) \right\}$$

$$\tag{Condition A}$$

And then we define the following four arithmetic functions:

$$u_{0}(i,j,k)=\sigma_{{i}} \left( m \right) {m}^{i}+\sigma_{{j}} \left( m \right) {m}^{j}+\sigma_{{k}} \left( m \right) {m}^{k} \tag{i}$$

$$u_{1}(i,j,k)=\sigma_{{i}} \left( m \right) {m}^{i}-\sigma_{{j}} \left( m \right) {m}^{j}+\sigma_{{k}} \left( m \right) {m}^{k} \tag{ii}$$

$$u_{2}(i,j,k)=\sigma_{{i}} \left( m \right) {m}^{i}+\sigma_{{j}} \left( m \right) {m}^{j}-\sigma_{{k}} \left( m \right) {m}^{k} \tag{iii}$$

$$u_{3}(i,j,k)=-\sigma_{{i}} \left( m \right) {m}^{i}+\sigma_{{j}} \left( m \right) {m}^{j}+\sigma_{{k}} \left( m \right) {m}^{k} \tag{iv}$$

And finally the composite of the four, (which I had originally been referring to as Heron's formula for its likeness to the area formula as is commonly known):

$$\psi(m)=\sqrt{u_{0}(i,j,k)\cdot u_{1}(i,j,k)\cdot u_{2}(i,j,k)\cdot u_{3}(i,j,k)}\tag{Psi Definition}$$

Question:

I have been trying to prove that for some natural number $$q,r \in \mathbb N$$ and some square free number $$s$$, the following statements are true:

$$(i,j,k) \in S_a \land m \gt r\Rightarrow \Re(\psi(m))=q \sqrt{s} \,\land \, \Im(\psi(m))=0 \tag{0}$$

$$(i,j,k) \in S_b \land m \gt r\Rightarrow \Re(\psi(m))=0 \,\land \, \Im(\psi(m))=q \sqrt{s} \tag{1}$$

Attempt Thus Far

The following two would establish that we are dealing with an equivalence relation on $$\mathbb N^3$$:

$$S_a \cap S_b={\{\,}\} \tag{2}$$

$$S_a \cup S_b = \mathbb N^3 \tag{3}$$

And I also observed that:

$${\{(i,j,k) \in S_b \,:\,i,j,k \leq N}\} \cap {\{(i,j,k) \in \mathbb N^3:\,i,j,k \leq N \land (i \neq j \land i \neq k \land j \neq k)}\}={\{\,}\}\tag{4}$$

And I found that the ratio of the cardinalities for the subsets: $$\vartheta S_a(N)={\{(i,j,k) \in S_a \,:\,i,j,k \leq N}\}$$ $$\vartheta S_b(N)={\{(i,j,k) \in S_b \,:\,i,j,k \leq N}\}$$

Is a rational function of $$N$$:

$$\frac{|\vartheta S_a(N)|}{|\vartheta S_b(N)|}={\frac { \left( 2\,N-1 \right) \left( N-1 \right) }{3\,N-1}}\tag{5}$$

And in assuming (2) & (3) to assert we know that:

$$|\vartheta S_a(N)|+|\vartheta S_b(N)|=N^3\tag{6}$$

Substituting in (5) gives us their cardinalities as individual rational polynomial functions of $$N$$:

$$|\vartheta S_a(N)|={N}^{3}-\frac{3\,N^2}{2}+\frac{N}{2}\tag{7i}$$

$$|\vartheta S_b(N)| =\frac{3\,N^2}{2}-\frac{N}{2}\tag{7ii}$$

Given that the denominator of this rational is the number of triples with maximal element N that are not unique, and the numerator is the number of triples in $$\vartheta S_a(N)$$ that are not unique,the limit as $$N$$ tends to infinity establishes that the infinite set $$S_a$$ contains exactly half of all triples that have exactly two elements that are equal, therefore the other half are elements of $$S_b$$:

$$\lim _{N\rightarrow \infty }{\Biggl(\frac {{N}^{3}-\frac{3}{2}{N}^{2}+\frac{N}{2}-N \left( N-1 \right) \left( N-2 \right) }{{N}^{3}-N \left( N-1 \right) \left( N-2 \right) }\Biggr)}=\frac{1}{2}\tag{8}$$

To summarise the unanswered component of this question, given the above assertions, (particularly 8 & 4) are we able to say that the cardinality of the infinite set $$S_a$$ is greater than that of the infinite set $$S_b$$, the difference being the cardinality of the set : $${\{(i,j,k) \in \mathbb N^3:\ i \neq j \land i \neq k \land j \neq k}\}$$

I do still believe this would be a fallacy with respect to the continuum hypothesis, and it's much the same as with asserting there are more rationals than integers, or any other less specific examples of the fallacy.

• What are $S_a$ and $S_b$? Come to think of it, what are $a$ and $b$? – Blue Jan 18 at 5:14
• Haha the $a$ and $b$ are arbitrary But I will definitely accept a more insightful definition of $S_a$ and $S_b$ other than as I've stated, (0) & (1) as the predicates – Adam L Jan 18 at 5:22
• If you look on the other side of the Implicative arrow, you will understand what I am getting at, as in the triples on $\mathbb N$ are split into two subsets, and of course this is something that got my attention – Adam L Jan 18 at 5:24
• As for $(0)$ and $(1)$ ... The $q$ and $s$ appear to be a distraction. You seem to be saying that there is some $r$ such that, for $m>r$, the function $\psi(m)$ partitions $\mathbb{N}^3$ into two sets —I'll call them $R_m$ and $I_m$— such that $\psi(m)$ is a non-zero real for $(i,j,k)\in R_m$, and $\psi(m)$ is a non-zero imaginary for $(i,j,k)\in I_m$. So, effectively, this is an assertion that for "sufficiently large" $m$, the function $\psi(m)$ is non-zero for all $(i,j,k)$. This, in turn, becomes a question of whether the sum and/or difference of three powers of $m$ can be zero. Am I close? – Blue Jan 18 at 6:13
• ("Extended discussion" warning. This may be my last comment.) As for whether the $u$ functions can be made non-zero (for $m>0$): each has the form $\pm m^i \pm m^j\pm m^k$. Factoring-out the smallest power allows us to concentrate on $\pm1\pm m^p\pm m^q$. If $p=q=0$, we have $\pm1\pm1\pm1$, which is never zero. Otherwise, if $p=0$ and $q\neq 0$, then we have $0\pm m^q$ (non-zero for positive $m$) or $\pm 2\pm m^q$ (non-zero for $m>2$). For other $p$ and $q$, we have that $\pm m^p\pm m^q$ is necessarily even, which cannot cancel the $\pm 1$. So, we need only take $m>2$ to force $\psi(m)\neq 0$. – Blue Jan 18 at 6:50

This is not an answer, but an attempt to reformulate (the first section of) the question.

Lemma. For positive integers $$i$$, $$j$$, $$k$$, and for positive integer $$m \neq 2$$, we have $$\pm m^i \pm m^j \pm m^k \neq 0 \tag{1}$$ for any independent choices of $$\pm$$.

Proof of lemma: Factoring-out the smallest power allows us to concentrate on the nature of $$\pm 1 \pm m^p \pm m^q \tag{1'}$$ If $$p$$ and $$q$$ are both zero or both non-zero, then $$\pm m^p\pm m^q$$ is even, which cannot cancel the $$\pm 1$$. If only one of them, say $$p$$, is zero, then we have $$0\pm m^q$$, which is non-zero for positive $$m$$, or $$\pm 2\pm m^q$$, which is non-zero for $$m\neq2$$. $$\square$$

Now, for positive integers $$i$$, $$j$$, $$k$$, and positive integer $$m\neq2$$, and any choice of $$\sigma_1$$, $$\sigma_2$$, $$\sigma_3 = \pm 1$$, define \begin{align} u_0(i,j,k;m) &:=\phantom{-}\sigma_1 m^i + \sigma_2 m^j + \sigma_3 m^k \\ u_1(i,j,k;m) &:= -\sigma_1 m^i + \sigma_2 m^j + \sigma_3 m^k \\ u_2(i,j,k;m) &:=\phantom{-}\sigma_1 m^i - \sigma_2 m^j + \sigma_3 m^k \\ u_3(i,j,k;m) &:=\phantom{-}\sigma_1 m^i + \sigma_2 m^j - \sigma_3 m^k \end{align}

(where I'll suppress the $$(i,j,k;m)$$ parameters to avoid visual clutter).

We're interested in the (Heron-like) product $$u_0 u_1 u_2 u_3$$. Expanding, we find that the $$\sigma$$ factors all appear to even powers, making their signs irrelevant. Unless we need to consider the $$u$$-functions individually (and it doesn't appear that we do), we might as well take each $$\sigma$$ factor to be $$1$$ (that is, we can ignore them altogether). Be that as it may ...

Definition. For positive integer $$m\neq2$$, the function $$\phi_m(i,j,k) := u_0u_1u_2u_3$$ is non-zero on $$\mathbb{N}^3$$, partitioning that domain into two disjoint sets, $$S_m^+$$ and $$S_m^-$$, over which $$\phi_m$$ is, respectively, (strictly) positive or (strictly) negative.

(This is equivalent to OP's $$\psi$$, with its square root and use of its real- or imaginary-ness to determine the partition. The function itself doesn't seem to figure into later discussion, so replacing it with $$\phi_m$$ shouldn't be a problem.)

From here, OP seems to want to quantify the relative sizes of $$S_m^+$$ and $$S_m^-$$. I'm not clear on that part —I don't even know if what I've written above aligns with OP's intention— so I'll stop typing.

• Exactly what I think I am looking for, is equivalence relations splitting the elements of the partition of this equivalence relation, kind of like those Russian dolls that reveal another inside of except in the first reveal it's twins haha – Adam L Jan 18 at 8:35