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anonymous
 one year ago
Let A be the set of all functions with domain R and codomain [1,1].
1. Give 2 functions that are elements of A.
2. Let R = {(f,g) l f(0) = g(0)}. Determine if the given reflection is reflexive, symmetric, and/or transitive. Give brief explanations to go with your answers.
3. Describe in words the set given by [cosx]. (We did a problem before that defined the set [x] for a given relation R by the rule [x] = {y l (x,y) ∈ R}.)
anonymous
 one year ago
Let A be the set of all functions with domain R and codomain [1,1]. 1. Give 2 functions that are elements of A. 2. Let R = {(f,g) l f(0) = g(0)}. Determine if the given reflection is reflexive, symmetric, and/or transitive. Give brief explanations to go with your answers. 3. Describe in words the set given by [cosx]. (We did a problem before that defined the set [x] for a given relation R by the rule [x] = {y l (x,y) ∈ R}.)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since and cosine come to mind right away

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they leave so very quickly...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for your help; I really appreciate it. I realized I switched the domain and codomain for part a when I initially tried this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for b, does that mean f and g are each a function where their y intercepts are equal?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, ya it seems that way :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then it would definitely be reflexive since the same function would have the same points

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, I'm not sure if I'm missing something... because showing that it's an equivalence relation almost seems trivial. Transitive: If fRg then we need gRf. So if f(0)=g(0), then g(0)=f(0) is clearly true for any functions in A.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ya reflexive seems to work out clearly as well.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For part c, I assumed the set [cosx] just meant the range [1,1] of the function.. Is that the way to describe that in words?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm I dunno :d the notation is confusing lol.\[\large\rm [\color{orangered}{x}]=\{y~~(\color{orangered}{x},y)\in R\}\]Based on the previous exercise they referred to, I guess we have something like this,\[\large\rm [\color{orangered}{\cos x}]=\{y~~(\color{orangered}{\cos x},y)\in R\}\]So we need ... words? Like human words for this? Hmmm..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1This is the `set of y values such that cos x is in relation to y`. That's how I would read it. So ya, it's uhhh.. the entire range right? as you said? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1there's probably some pretty way to word that >.<

freckles
 one year ago
Best ResponseYou've already chosen the best response.0[cos(x)] is the set of functions f with domain all real numbers and codomain [1,1] and cos(0)=f(0)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0it has been a while since I looked at words codomain and range I think the range of the function can actually be a subset of the codomain and they used the word codomain earlier not range

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so I think one element of [cos(x)] would be f(x)=1 since cos(0)=f(0)=1 that is just one example what one of the elements I think we would have

freckles
 one year ago
Best ResponseYou've already chosen the best response.0also I'm under the assumption we are using the way R was defined in 2...

dan815
 one year ago
Best ResponseYou've already chosen the best response.0also circles and ellipses
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