X.y=0 but x and y are non-zero numbers

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Homework Help Overview

The discussion revolves around the equation x.y=0 with the stipulation that both x and y are non-zero numbers. Participants explore the implications of this scenario within various mathematical contexts, including cyclic numbers, modulo arithmetic, and specific number systems like Tersonian and p-adic numbers.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants question the possibility of finding non-zero pairs (x, y) that satisfy the equation. Some suggest using cyclic numbers or modulo arithmetic, while others propose exploring specific mathematical structures, such as rings with zero divisors.

Discussion Status

The conversation is ongoing, with various interpretations and approaches being discussed. Some participants have provided insights into potential mathematical frameworks that could allow for the equation to hold true, while others express uncertainty about the problem's context and formulation.

Contextual Notes

There is a noted lack of clarity regarding the complete problem statement, which has led to speculation about the nature of x and y. Participants are attempting to navigate the implications of different mathematical systems and assumptions.

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Homework Statement



x and y are non-zero numbers.

Homework Equations


x.y=0


The Attempt at a Solution



I think we should use cyclic numbers but How?
 
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can't you even find one pair (x,y) such that xy = 0 and x<>0 and y<>0 ?
 


I think we should use cyclic numbers
 


Think modulo arithmetic, where the modulus is composite
 


x and y belong to some ring with zero divisors (since you obviously aren't interested in x=0 or y=0). Any idea what that ring is?
 


matal said:

Homework Statement


x and y are non-zero numbers.


Homework Equations



x.y= 0

The Attempt at a Solution

You're not provided the complete problem statement. If x and y are just plain old nonzero numbers (real or complex), then their product can't be zero. VeeEight and willem2 are having to guess at the context of this problem, which is information you should have provided.
 
What is the exact formulation of the question in the book?
 


As you know our decimal expansions have the form \pm d_n d_{n-1} \dots d_2 d_1 d_0.d_{-1}d_{-2}\dots where each d_i is in \{0,1,\dots,9\} We can have infinitely many nonzero digits after the decimal point, but we must have only finitely many nonzero digits before the decimal point.
In Tersonia they do just the opposite. Their decimal expansions have the form \dots t_3 d_2 t_1 t_0.t_{-1}t_{-2}\dots t_{-n} where each t_i is in \{0,1,\dots,9\} (Note that there is no minus sign!) They can have infinitely many nonzero digits before the decimal point, but they can only have finitely many digits after the decimal point.
Take a few minutes to convince yourself that Tersonians can add and multiply their decimal expansions just like we do without encountering any difficulty. we should find two Tersonian numbers .
 
If you use numbers mod 4 then x=2, y=2 would obviously do. If it's not something like that you'll probably find it's a trick question.
 
  • #10
Or if you take x and y to be 1 and i and interpret x.y as the dot product of the corresponding vectors.
 
  • #11
I think it cannot be solved by using complex numbers
 
  • #12
Ok so its Klingon arithmetic. If they write their numbers from right to left, the may use "9" to represent 0, "8" to represent 1 etc. so then you'd have x.y=0 with x=y=6.
 
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  • #13
I think this whole 'Tersonian' thing is actually p-adic numbers. In particular, 10-adic. Since 10 is composite, there are zero divisors. The product of two numbers can be zero because there can be an infinite carry to the left. I don't actually know much about the subject. But I know that much. There are examples if you troll the nets but I haven't checked out the details. It doesn't seem to be particularly elementary so I don't know why this whole thing is being cloaked in 'Tersonian' baby talk.
 
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  • #14
Yes, it is related to p- adic numbers. I found the solution in wikipedia. Thanks Dick=)
 

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