Isomorphism and Binary Structures

  • #1
Let J be a set of all linear functions. Consider the set R^2 in the Euclidean plane. Define a binary operation * on R^2 in such a way that the two binary structures <J, +> and <R^2, *> will be isomorphic. Any thoughts?
If something is not clear please ask. Thank you.
 
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  • #2
Fredrik
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This forum's policy on homework requires you to show what you've done so far. (This applies to all questions that look like homework, even the ones that aren't). By the way, are you talking about linear functions from [itex]\mathbb R^2[/itex] into [itex]\mathbb R^2[/itex]?
 
  • #3
Thanks for the heads-up Fredrik; I am talking R^2 into R^2.
So my thoughts are I need to look at the linear function y = mx + b as determined by m and b then I could map any y = mx + b to the point (m, b). I believe the point (m_1, b_1)*(m_2, b_2) must represent the linear function (m_1x + b_1) + (m_2x + b_2). I think I need to manipulate the two slopes and two y-intercepts to get m and b of the linear function (m_1x + b_1) + (m_2x + b_2). This is where I get stuck and it could be my reasoning is all washed up. I feel this is just off on the edge of my understanding.
 
  • #4
Fredrik
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Did you mean that m and b are real numbers, and x and y are vectors? The function f defined by f(x)=mx+b for all x in [itex]\mathbb R^2[/itex] is not linear for arbitrary m and b. If U and V are vector spaces (like [itex]\mathbb R^2[/itex] with the standard definitions of addition and multiplication by a real number), then a function [itex]f:U\rightarrow V[/itex] is said to be linear if [itex]f(ax+by)=af(x)+bf(y)[/itex] for all real numbers a,b, and all x,y in U.

Are you sure you have stated the problem correctly? The natural way to represent linear operators on [itex]\mathbb R^2[/itex] is using four numbers, not two, so a bijective function into [itex]\mathbb R^2[/itex] would have to be pretty weird.
 
  • #5
Maybe I am stating it wrong:

What I am trying to do is define a binary operation * on R^2 such that the two binary structures <J, +> and <R^2, *> are isomorphic. J is a set of all linear functions and the set R^2 = R X R = {(a, b): a is in the set of real#s, b is in the set of real #s} in the Euclidean plane.

So, thus far I have looked at specific linear functions ie y1 = 2x + 3 and y2 = 5x + 10 and said y1 + y2 = 7x + 13 with that (2,3) and (5,10) = (2,3)*(5,13). This would make the general forms (m1x + b1) + (m2x + b2) for J and (m1, b1)*(m2, b2) for R^2.

I do not know if this is even close and I do not know the function (phe) that connects the two binary structures.
 
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  • #6
Office_Shredder
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To clarify: By linear functions you mean functions whose graph is a line correct?

If so you're close. How did the m's and b's interact when you added two linear functions? They should interact in the same way when you're *-ing points in R2
 
  • #7
To clarify: By linear functions you mean functions whose graph is a line correct?

If so you're close. How did the m's and b's interact when you added two linear functions? They should interact in the same way when you're *-ing points in R2

Yes, as in functions whose graphs are a line....

ms and bs..... they shifted upward by there sum?

Argh!
 
  • #8
Office_Shredder
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(mx+b)+(nx+c)=(m+n)x+(b+c).

So if we were representing these by (m,b) and (n,c), what should (m,b)*(n,c) be?
 
  • #9
So, is the property that "connects" the two (m+n, b+c)? Is it that simple?
 

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