I just rewatched some old math books for the fun of it and was able to observe the following which I want to formalize (could you help me with it):

Choose 2 numbers with distance 1 for each pair. Proceed with...

First Example $(i)$

(3,4) and (6,7)

$3\cdot 6+4\cdot 7=46$ and
$3\cdot 7 + 4\cdot 6 = 45$

EDIT: Second example $(ii)$

(distance between pairs don't need to be the same distance)

(5,6) and (9,10)

$5 \cdot 9 + 6 \cdot 10 = 105$ and
$5\cdot 10 + 6\cdot 9 = 104$

Therefore, the outcome of both equations in $(i)$ and $(ii)$ have a distance of $1^2$.


If you try this for different examples and for distances $n\in \mathbb{N}_0$, you'll most likely notice, that the observation is satisfied for all pairs $(a,b),(c,d)\mid a,b,c,d\in\mathbb{N}\setminus \{0\}$ with a specific distance $n\geq 0$. I want to formalize this fact, but I'm not really sure, if my following thoughts are correct.

Informal, we have: For all pairs $(a,b),(c,d) \in \mathbb{N}\setminus 0:$ exists one $n$ in $\mathbb{N}_0$ with $b$ is the successor of $a$ with distance $n$ and $d$ is the successor of $c$ with distance $n$ for which we obtain equivalent that the outcome of $(ac+bd)$ has distance $n^2$ to the outcome $(ad+bc)$.

I would formalize this as the following: $$\forall a,b,c,d \in \mathbb{N}\setminus 0 : \exists n \in \mathbb{N}_0:(b=a+n) \land (d=c+n) \iff \underbrace{(ac+bd)}_{=x+n^2}- \underbrace{(ad+bc)}_{=x}=n^2$$

Do you think this is correct? If not, could you share your thoughts about it? Could we use group-theory to discribe these equations easier?

  • $\begingroup$ "Formalization" does not mean "writing things with symbols". It means "fully explaining everything, without leaving gaps". This is a common misconception - using a bunch of symbols doesn't necessarily make your statement any more formal, useful, or understandable. $\endgroup$ – Deusovi Dec 14 '18 at 14:25
  • $\begingroup$ I would be glad if you share your thoughts with me. $\endgroup$ – Doesbaddel Dec 14 '18 at 14:54

If $b = a + n$ and $d = c + n$ then $$ (ac+bd)-(ad+bc) = (b-a)(d-c) = n^2. $$ For example, if the numbers are $(0,n)$ and $(0,n)$ then $0\cdot 0 + n \cdot n = n^2$ whereas $0 \cdot n + n \cdot 0 = 0$.

  • $\begingroup$ That would mean my formalization is wrong, right? $\endgroup$ – Doesbaddel Dec 14 '18 at 9:13
  • 2
    $\begingroup$ It's not your formalization that is wrong – it's your observation that is wrong. $\endgroup$ – Yuval Filmus Dec 14 '18 at 9:13
  • $\begingroup$ Oh, I see. I'm not really sure what the observation would be like otherwise. Could you give me a hint, please? $\endgroup$ – Doesbaddel Dec 14 '18 at 9:17
  • 1
    $\begingroup$ Note that $1^2=1$. $\endgroup$ – Yuval Filmus Dec 14 '18 at 9:24
  • 1
    $\begingroup$ My answer formalizes and proves your observation. $\endgroup$ – Yuval Filmus Dec 15 '18 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.